NCERT Solutions for Exercise 7.3 Class 12 Maths Chapter 7

NCERT Solutions for Exercise 7.3 Class 12 Maths Chapter 7 - Integrals

Edited By Ramraj Saini | Updated on Dec 03, 2023 09:49 PM IST | #CBSE Class 12th

NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.3

NCERT Solutions for Exercise 7.3 Class 12 Maths Chapter 7 Integrals are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. NCERT solutions for Class 12 Maths chapter 7 exercise 7.3 is another exercise of the NCERT syllabus chapter integrals. It exposes students to some higher level of problems of integrals which includes complex trigonometric functions. Exercise 7.3 Class 12 Maths can be directly solved by students here only to understand the concepts well. NCERT Solutions for Class 12 Maths chapter 7 exercise 7.3 provided below are holistic in nature and have been prepared by experienced faculties. NCERT solutions for exercise 7.3 Class 12 Maths chapter 7 cannot be neglected to perform better in exams like JEE Main.

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12th class Maths exercise 7.3 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Integrals Class 12 Chapter 7 Exercise: 7.3

Question:1 Find the integrals of the functions $\sin ^ 2 ( 2x+ 5 )$

Answer:

$\sin ^ 2 ( 2x+ 5 )$

using the trigonometric identity

$sin^2x=\frac{1-cos2x}{2}$

we can write the given question as

= $\frac{1-\cos 2(2x+5)}{2}$ $= \frac{1-\cos (4x+10)}{2}$
$\\=\int \frac{1-\cos (4x+10)}{2}dx\\ =\frac{1}{2}\int dx - \frac{1}{2}\int \cos(4x+10)dx\\ =\frac{x}{2}-\frac{1}{2}[\sin(4x+10)/4]\\ =\frac{x}{2}-\frac{\sin(4x+10)}{8}+C$

Question:2 Find the integrals of the functions $\sin 3x \cos 4x$

Answer:

Using identity $\sin A\cos B = 1/2[sin(A+B)+sin(A-B)]$

, therefore the given integral can be written as

$\int \sin 3x\cos 4x=\frac{1}{2}\int sin(3x+4x)+sin(3x-4x)\ dx$

$=\frac{1}{2}\int sin(7x)-sin(x)\ dx\\ =1/2[\int \sin (7x) dx-\int \sin x\ dx]\\ =\frac{1}{2}[(-1/7)\cos 7x+\cos x+ C]\\ = \frac{\cos x}{2}-\frac{\cos 7x}{14}+C$

Question:3 Find the integrals of the functions $\cos 2x \cos 4x \cos 6x$

Answer:

Using identity
$cosAcosB = \frac{1}{2}[cos(A+B)+cos(A-B)]$

$\int \cos 2x.\cos 4x.\cos 6x = \int \cos 2x. \frac{1}{2}[(\cos 10x)+\cos 2x]dx$

Again use the same identity mentioned in the first line

$\\= \frac{1}{2}\int (\cos 2x.\cos 10x+\cos 2x. \cos 2x)dx\\ =\frac{1}{2}\int\frac{1}{2}({\cos12x +\cos 8x})dx+\frac{1}{2}\int (\frac{1+\cos 4x}{2})dx\\ =\frac{\sin 12x}{48}+\frac{\sin 8x}{32}+\frac{\sin 4x}{16}+ x/4+C$

Question:4 Find the integrals of the functions $\sin ^ 3 ( 2x +1 )$

Answer:

$\int \sin^3(2x+1)dx = \int \sin^2(2x+1).\sin(2x+1)dx$

The integral can be written as

$= \int (1-\cos^2(2x+1)).\sin(2x+1)dx$
Let
$\\\cos (2x+1) =t\\ \sin (2x+1)dx = -dt/2$

$\\=\frac{-1}{2}\int (1-t^2)dt\\ =\frac{-1}{2}[t-t^3/3]\\ =\frac{t^3}{6}-\frac{t}{2}$

Now, replace the value of t, we get;

$=\frac{\cos^3(2x+1)}{6}-\frac{\cos(2x+1)}{2}+C$

Question:5 Find the integrals of the functions $\sin ^3 x \cos ^ 3 x$

Answer:

$I = \int \sin^3x.\cos^3x\ dx$

rewrite the integral as follows

$\\=\int cos^3x.sin^2x.\sin x\ dx\\ =\int cos^3x(1-\cos^2x)\sin x\ dx$
Let $\cos x = t \Rightarrow \sin x dx =-dt$

$\\=-\int t^3(1-t^2)dt\\ =-\int(t^3-t^5)dt\\ =-[\frac{t^4}{4}]+[\frac{t^6}{6}] +C\\ =\frac{\cos^6x}{6}-\frac{cos^4x}{4}+C$ ......(replace the value of t as $cos\ x$ )

Question:6 Find the integrals of the functions $\sin x \sin 2x \sin 3x$

Answer:

Using the formula
$sinAsinB=\frac{1}{2}(cos(A-B)-cos(A+B))$

we can write the integral as follows

$\int \sin x.\sin 2x\sin 3x\ dx = \int \sin x\frac{1}{2}[\cos x-\cos 5x]dx$
$\\=\frac{1}{2} \int [\sin x.\cos x-\sin x.\cos 5x]dx\\ =\frac{1}{2}\int \frac{\sin 2x}{2}dx-\frac{1}{2}\int \sin x. \cos 5x\ dx\\ =-\frac{\cos 2x}{8}-\frac{1}{4}\int[\sin 6x -\sin 4x]\\ =-\frac{\cos 2x}{8}-\frac{1}{4}[\frac{-\cos 6x}{6}+\frac{\cos 4x}{4}]\\ =-\frac{\cos 2x}{8}+\frac{\cos 6x}{24}-\frac{\cos 4x}{16}+C$

Question:7 Find the integrals of the functions $\sin 4x \sin 8x$

Answer:

Using identity

$sinAsinB=\frac{1}{2}(cos(A-B)-cos(A+B))$

we can write the following integral as

$\sin 4x \sin 8x$ =
$\\=\frac{1}{2}\int(\cos 4x - \cos 12x) dx\\ =\frac{1}{2} [\int\cos 4x\ dx - \int \cos 12x\ dx]\\ =\frac{\sin 4x}{8}-\frac{\sin 12x}{24}+C$

Question:8 Find the integrals of the functions $\frac{1- \cos x }{1+ \cos x }$

Answer:

We know the identities

$\\1+\cos 2A = 2\cos^2A\\ 1-\cos 2A = 2\sin^2 A$

Using the above relations we can write

$\frac{1-\cos x}{1+\cos x}=\frac{\sin^2x/2}{\cos^2x/2} = \tan^2x/2$

$=\int \tan^2x/2 =\int (\sec^2x/2-1)dx$
$\\=\int (\sec^2x/2)dx-\int dx\\ = 2[\tan x/2]-{x}+C$

Question:9 Find the integrals of the functions $\frac {\cos x }{1 + \cos x }$

Answer:

The integral is rewritten using trigonometric identities

$\frac{\cos x}{1+ \cos x}= \frac{\cos^2x/2-\sin^2x/2}{2\cos^2x/2} =\frac{1}{2}[1-\tan^2x/2]$
$\\=\int \frac{1}{2}[1-\tan^2x/2] dx\\ =\frac{1}{2}\int 1-[sec^2\frac{x}{2}-1]=\frac{1}{2}\int 2-sec^2\frac{x}{2}\\=x-tan\frac{x}{2}+c$

Question:10 Find the integrals of the functions $\sin ^ 4 x$

Answer:

$\sin ^ 4 x$ can be written as follows using trigonometric identities
$\\=\sin^2x.\sin^2x\\ =\frac{1}{4}(1-\cos 2x)^{2}\\ =\frac{1}{4}(1+\cos^22x-2\cos 2x)\\ =\frac{1}{4}(1+\frac{1}{2}(1+\cos 4x)-2\cos 2x)\\ =3/8+\frac{\cos 4x}{8}-\frac{\cos 2x}{2}$

Therefore,
$\Rightarrow \int \sin^4x\ dx = \int \frac{3}{8}dx+\frac{1}{8}\int\cos 4x\ dx -\frac{1}{2}\int\cos 2x\ dx$
$= \frac{3x}{8}+\frac{\sin 4x}{32} -\frac{\sin 2x}{4}+C$

Question:11 Find the integrals of the functions $\cos ^ 4 2x$

Answer:

$cos^42x=cos^32xcos2x$

now using the identity

$cos^3x=\frac{cos3x+3cosx}{4}$

$cos^32xcos2x=\frac{cos6x +3cos2x}{4}cos2x=\frac{cos6xcos2x+3cos^22x}{4}$

now using the below two identities

$cosacosb=\frac{cos(a+b)+cos(a-b)}{2}\\and\ cos^22x=\frac{1+cos4x}{2}\\$

the value

$cos^42x=cos^32xcos2x\\=\frac{cos6xcos2x+3cos^22x}{4}=\frac{cos 4x+cos8x}{8}+\frac{3}{4}\frac{1+cos4x}{2}$ .

the integral of the given function can be written as

$\int cos^42x=\int \frac{cos 4x+cos8x}{8}+\int \frac{3}{4}\frac{1+cos4x}{2}\\ \\=\frac{3}{8}x+\frac{sin4x}{8}+\frac{sin8x}{64}+C$

Question:12 Find the integrals of the functions $\frac{\sin ^ 2x }{1+ \cos x }$

Answer:

Using trigonometric identities we can write the given integral as follows.

$\frac{\sin ^ 2x }{1+ \cos x }$

$\\=\frac{4\sin^2\frac{x}{2}\cos^2\frac{x}{2}}{2\cos^2\frac{x}{2}}\\ =2\sin^2\frac{x}{2}\\ =1-\cos x$

$\therefore \int \frac{sin^22x}{1+\cos x} = \int (1-\cos x)dx$
$\\= \int 1dx-\int\cos x\ dx\\ =x-\sin x+C$

Question:13 Find the integrals of the functions $\frac{\cos 2x - \cos 2 \alpha }{\cos x - \cos \alpha }$

Answer:

We know that,

$\cos A-\cos B = -2\sin(\frac{A+B}{2})\sin(\frac{A-B}{2})$

Using this identity we can rewrite the given integral as

$\frac{\cos 2x-\cos 2\alpha }{\cos x-\cos\alpha}=\frac{-2\sin\frac{2x+2\alpha}{2}\sin\frac{2x-2\alpha}{2}}{-2\sin\frac{x+\alpha}{2}\sin\frac{x-\alpha}{2}}$

$\\=\frac{\sin(x+\alpha)\sin(x-\alpha)}{\sin\frac{x+\alpha}{2}\sin\frac{x-\alpha}{2}}\\ =\frac{[2\sin\frac{x+\alpha}{2}\cos \frac{x+\alpha}{2}][2\sin\frac{x-\alpha}{2}\cos\frac{x-\alpha}{2}]}{\sin\frac{x+\alpha}{2}\sin\frac{x-\alpha}{2}}\\ =4\cos\frac{x+\alpha}{2}\cos\frac{x-\alpha}{2}\\ =2[\cos x+\cos \alpha]$

$\therefore \int\frac{\cos 2x-\cos 2\alpha }{\cos x-\cos\alpha}=\int 2\cos x\ dx +\int 2\cos \alpha\ dx$
$=2[\sin x + x\cos \alpha]+C$

Question:14 Find the integrals of the functions $\frac{\cos x - \sin x }{1+ \sin 2x }$

Answer:

$\frac{\cos x-\sin x}{1+2\sin x}=\frac{\cos x-\sin x}{(sin^2x+cos^2x)+2 sin x.\cos x}$
$=\frac{\cos x-\sin x}{(\sin x+\cos x)^2}$

$\\sin x+\cos x =t\\ \therefore (\cos x-\sin x)dx = dt$

Now,
$=\int \frac{dt}{t^2}\\ =\int t^-2\ dt\\ =-t^{-1}+C\\ =-\frac{1}{(\sin x+\cos x)}+C$

Question:15 Find the integrals of the functions $\tan ^ 3 2x \sec 2x$

Answer:

$\tan^32x.\sec 2x = \tan^22x.\tan 2x.\sec 2x$
$\\= (\sec^22x-1).\tan 2x.\sec 2x\\ =\sec^22x.\tan 2x-\tan 2x.\sec 2x$

Therefore integration of $\tan ^ 3 2x \sec 2x$ =
$\\=\int\sec^22x.\tan 2x\ dx-\int\tan 2x.\sec 2x\ dx\\ =\int\sec^22x.\tan 2x\ dx-\sec 2x/2+C\\$ .....................(i)
Let assume

$\sec 2x = t$
So, that $2\sec 2x.\tan 2x\ dx =dt$
Now, the equation (i) becomes,

$\\\Rightarrow \frac{1}{2}\int t^2\ dt-\frac{\sec 2x}{2}+C\\ \Rightarrow \frac{t^3}{6}-\frac{\sec 2x}{2}+C\\ =\frac{(\sec 2x)^3}{6}-\frac{\sec 2x}{2}+C$

Question:16 Find the integrals of the functions $\tan ^ 4x$

Answer:

the given question can be rearranged using trigonometric identities

$tan^4x=(\sec^2x-1).\tan^2x\\ =\sec^2x.\tan^2x-\tan^2x\\ =\sec^2x.\tan^2x-\sec^2x+1$

Therefore, the integration of $\tan^4x$ = $\\=\int \sec^2x.\tan^2x\ dx-\int\sec^2x\ dx+\int dx\\ =(\int \sec^2x.\tan^2x\ dx)-\tan x+x+C\\$ ...................(i)
Considering only $\int \sec^2x.\tan^2x\ dx$
let $\tan x =t\Rightarrow \sec^2x\ dx =dt$

$\int \sec^2x\tan^2x\ dx = \int t^2\ dt = t^3/3=\frac{\tan^3x}{3}$

now the final solution is,

$\int \tan^4x =\frac{\tan^3x}{3}-\tan x+x+C$

Question:17 Find the integrals of the functions $\frac{\sin ^ 3x + \cos ^ 3x }{\sin ^ 2 x \cos ^2 x }$

Answer:

$\frac{\sin ^ 3x + \cos ^ 3x }{\sin ^ 2 x \cos ^2 x }$

now splitting the terms we can write

$\\=\frac{\sin^3x}{\sin^2x.\cos^2x}+\frac{\cos^3x}{\sin^2x.\cos^2x}\\ =\frac{\sin x}{cos^2x}+\frac{\cos x}{\sin^2x}\\ =\tan x.\sec x+\cot xcosec x$

Therefore, the integration of
$\frac{\sin ^ 3x + \cos ^ 3x }{\sin ^ 2 x \cos ^2 x }$

$\\=\int (\tan x\sec x+\cot xcosec x)dx\\ =\sec x-cosec\ x+C$

Question:18 Find the integrals of the functions $\frac{\cos 2 x + 2 \sin ^ 2x }{\cos ^ 2 x }$

Answer:

The integral of the above equation is

$\\=\int (\frac{\cos 2x+2\sin^2x}{\cos^2x})dx\\ =\int (\frac{\cos 2x+(1-\cos 2x)}{\cos^2x}\\ =\int\frac{1}{\cos^2x}\\ =\int \sec^2x\ dx =\tan x+C$

Thus after evaluation, the value of integral is tanx+ c

Question:19 Find the integrals of the functions $\frac{1}{\sin x \cos ^3 x }$

Answer:

Let
We can write 1 = $\sin^2x +\cos^2x$
Then, the equation can be written as
$I =\frac{\sin^2x +\cos^2x}{\sin x\cos^3x}$

$I =\int (\tan x+\frac{1}{\tan x})\sec^2 x dx$
put the value of tan $x$ = t
So, that $\sec^2xdx =dt$

$\\\Rightarrow I=\int (t+\frac{1}{t})dt\\ =\frac{t^2}{2}+\log\left | t \right |+C\\ =\log\left | \tan x \right |+\frac{1}{2}\tan^2x+C$

Question:20 Find the integrals of the functions $\frac{\cos 2x }{( \cos x + \sin x )^2}$

Answer:

we know that $cos2x= cos^2x-sin^2x$
therefore,

$\frac{\cos 2x }{( \cos x + \sin x )^2}$
$\frac{\cos 2x}{1+\sin 2x}\\ \Rightarrow \int \frac{\cos 2x}{1+\sin 2x}\\$ let $1+sin2x =t \Rightarrow 2cos2x\ dx = dt$
Now the given integral can be written as

$\therefore \int \frac{\cos 2x}{(\cos x+\sin x)^2}=\frac{1}{2}\int \frac{1}{t}dt$
$\\\Rightarrow \frac{1}{2}\log\left | t \right |+C\\ \Rightarrow \frac{1}{2}\log\left | 1+\sin 2x \right |+C\\=log|sin^2x+cos^2x+2sinxcosx|+C\\=\frac{1}{2}log|(sinx+cosx)^2|+C=log|sinx+cosx|+C$

Question:21 Find the integrals of the functions $\sin ^ { -1} ( \cos x )$

Answer:

using the trigonometric identities we can evaluate the following integral as follows

$\dpi{100} \int \sin^{-1}(\cos x)dx = \int \sin^{-1}(sin(\frac{\pi}{2}-x))dx\\=\int(\frac{\pi}{2}-x)dx=\frac{\pi x}{2}-\frac{x^2}{2}+C$

Question:22 Find the integrals of the functions $\frac{1}{\cos ( x-a ) \cos ( x-b )}$

Answer:

Using the trigonometric identities following integrals can be simplified as follows

$\frac{1}{\cos(x-a)\cos(x-b)}=\frac{1}{\sin(a-b)}[\frac{\sin(a-b)}{\cos(x-a)\cos(x-b)}]$

$=\frac{1}{\sin(a-b)}[\frac{\sin[(x-b)-(x-a)]}{\cos(x-a)\cos(x-b)}]$

$=\frac{1}{\sin(a-b)}[\frac{\sin(x-b)\cos(x-a)-\cos(x-b)\sin(x-a)}{\cos(x-a)\cos(x-b)}]$

$=\frac{tan(x-b)-\tan (x-a)}{\sin(a-b)}$

$=\frac{1}{\sin(a-b)}\int tan(x-b)-\tan (x-a)dx$
$\\=\frac{1}{\sin(a-b)}[-\log\left | \cos(x-b) \right |+\log\left | \cos(x-a) \right |]\\ =\frac{1}{\sin(a-b)}(\log\left | \frac{\cos(x-a)}{\cos(x-b)} \right |)$

Question:23 Choose the correct answer

$\int \frac{\sin ^ 2 x - \cos ^ 2 x dx }{\sin ^ 2 x \cos ^ 2x } dx \: \:is \: \:equal \: \: to$ stgdrffd

Answer:

The correct option is (A)

On reducing the above integral becomes $\sec^2x-csc^2x$
$\int\sec^2x-csc^2x\ dx = \tan x+ \cot x+C$

Question:24 Choose the correct answer $\int \frac{e ^x ( 1+x)}{\cos ^ 2 ( e ^xx )} dx \: \: equals$

$\\(A) -\cot (ex^x) + C \\\\ (B) \tan (xe^x) + C\\\\ (C) \tan (e^x) + C \\\\ (D) \cot (e^x) + C$

Answer:

The correct option is (B)

Let $e^xx = t$ .
So, $(e^x.x+ 1.e^x)dx = dt$
(1+ $x$ ) $e^x\ dx =dt$

therefore,

$\int \frac{e^x(1+x)}{\cos^2(e^x.x)}dx =\int\frac{dt}{\cos^2t}$
$\\=\int \sec^2t dt\\ =\tan t +C\\ =\tan(e^x.x)+C$

Benefits of NCERT Solutions for Class 12 Maths Chapter 7 Exercise 7.3

The Class 12th Maths chapter 7 exercise provided here is of best quality.
Students can skip questions from exercise 7.3 Class 12 Maths which are repeated in concept to save their time.
These Class 12 Maths chapter 7 exercise 7.3 solutions are helpful for NEET, JEE as well as Board examinations.
NCERT Solutions for class 12 maths chapter 7 exercise 7.3 are highly recommended to students. Theory also can be understood with the help of exercise solutions only.

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Key Features Of NCERT Solutions for Exercise 7.3 Class 12 Maths Chapter 7

Comprehensive Coverage: The solutions encompass all the topics covered in ex 7.3 class 12, ensuring a thorough understanding of the concepts.
Step-by-Step Solutions: In this class 12 maths ex 7.3, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
Accuracy and Clarity: Solutions for class 12 ex 7.3 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
Conceptual Clarity: In this 12th class maths exercise 7.3 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
Inclusive Approach: Solutions for ex 7.3 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
Relevance to Curriculum: The solutions for class 12 maths ex 7.3 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

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Happy learning!!!

Frequently Asked Questions (FAQs)

1. What are the topics asked in NCERT solutions for Class 12 Maths chapter 7 exercise 7.3 ?

Topics like finding out integrals of trigonometric functions, square root functions etc. are asked in the examination.

2. Where is the use of Integration ?

Integration can be used to find the area, volume etc. in physics and other disciplines.

3. Is exercise 7.3 important for Board examinations ?

Yes, it is significant for the board examinations. 5 markers are generally asked from this exercise only.

4. What is the level of questions in exercise 7.3 class 12 Maths ?

Moderate level of questions are there in this exercise,

5. Is this helpful for Physics Topics also?

Yes, some functions like sinx, cosx etc. are used in physics calculations. Hence can be referred for Physics topics also.

6. How many questions are there in Class 12th Maths chapter 7 exercise 7.3 ?

In total there are 24 questions in this exercise

7. Is it necessary to solve every question of the exercise?

No, just try to solve maximum and cover maximum concepts. Some questions which are based on the same concepts can be skipped.

8. Should I refer to any other book for this chapter for Boards?

No, as NCERT questions are more than sufficient for Boards. You can revise these questions only.

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quartely question paper for mathematics cbse

Hello there! Thanks for reaching out to us at Careers360.

Ah, you're looking for CBSE quarterly question papers for mathematics, right? Those can be super helpful for exam prep.

Unfortunately, CBSE doesn't officially release quarterly papers - they mainly put out sample papers and previous years' board exam papers. But don't worry, there are still some good options to help you practice!

Have you checked out the CBSE sample papers on their official website? Those are usually pretty close to the actual exam format. You could also look into previous years' board exam papers - they're great for getting a feel for the types of questions that might come up.

If you're after more practice material, some textbook publishers release their own mock papers which can be useful too.

Let me know if you need any other tips for your math prep. Good luck with your studies!

Read Complete Answer

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Possible steps:

Re-evaluate Your Study Strategies:
- Identify Weak Areas: Pinpoint the specific topics or concepts that caused difficulties.
- Seek Clarification: Reach out to teachers, tutors, or online resources for additional explanations.
- Practice Regularly: Consistent practice is key to mastering chemistry.
Consider Professional Help:
- Tutoring: A tutor can provide personalized guidance and support.
- Counseling: If you're feeling overwhelmed or unsure about your path, counseling can help.
Explore Alternative Options:
- Retake the Exam: If you're confident in your ability to improve, consider retaking the chemistry compartment exam.
- Change Course: If you're not interested in pursuing chemistry further, explore other academic options that align with your interests.
Focus on NEET 2025 Preparation:
- Stay Dedicated: Continue your NEET preparation with renewed determination.
- Utilize Resources: Make use of study materials, online courses, and mock tests.
Seek Support:
- Talk to Friends and Family: Sharing your feelings can provide comfort and encouragement.
- Join Study Groups: Collaborating with peers can create a supportive learning environment.

Remember: This is a temporary setback. With the right approach and perseverance, you can overcome this challenge and achieve your goals.

I hope this information helps you.

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Kanishka kaushikii 17 September,2024

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

Qualifications:
Age: As of the last registration date, you must be between the ages of 16 and 40.
Qualification: You must have graduated from an accredited board or at least passed the tenth grade. Higher qualifications are also accepted, such as a diploma, postgraduate degree, graduation, or 11th or 12th grade.
How to Apply:
Get the Medhavi app by visiting the Google Play Store.
Register: In the app, create an account.
Examine Notification: Examine the comprehensive notification on the scholarship examination.
Sign up to Take the Test: Finish the app's registration process.
Examine: The Medhavi app allows you to take the exam from the comfort of your home.
Get Results: In just two days, the results are made public.
Verification of Documents: Provide the required paperwork and bank account information for validation.
Get Scholarship: Following a successful verification process, the scholarship will be given. You need to have at least passed the 10th grade/matriculation scholarship amount will be transferred directly to your bank account.

Scholarship Details:

Type A: For candidates scoring 60% or above in the exam.

Type B: For candidates scoring between 50% and 60%.

Type C: For candidates scoring between 40% and 50%.

Cash Scholarship:

Scholarships can range from Rs. 2,000 to Rs. 18,000 per month, depending on the marks obtained and the type of scholarship exam (SAKSHAM, SWABHIMAN, SAMADHAN, etc.).

Since you already have a 12th grade qualification with 84%, you meet the qualification criteria and are eligible to apply for the Medhavi Scholarship exam. Make sure to prepare well for the exam to maximize your chances of receiving a higher scholarship.

Hope you find this useful!

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Yuvan 30 August,2024

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

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Ashvadha 12 July,2024

Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Read Complete Answer

kumardurgesh1802 10 July,2024

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Central Board of Secondary Education Class 12th Examination

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Solutions for Exercise 7.3 Class 12 Maths Chapter 7 - Integrals

NCERT Solutions For Class 12 Maths Chapter 7 Exercise 7.3

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Answer:

Question:2 Find the integrals of the functions

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Question:3 Find the integrals of the functions

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Question:4 Find the integrals of the functions

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Question:5 Find the integrals of the functions

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Question:6 Find the integrals of the functions

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Question:7 Find the integrals of the functions

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Question:8 Find the integrals of the functions

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Question:9 Find the integrals of the functions

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Question:10 Find the integrals of the functions

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Question:11 Find the integrals of the functions

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Question:12 Find the integrals of the functions

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Question:13 Find the integrals of the functions

Answer:

Question:14 Find the integrals of the functions

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Question:15 Find the integrals of the functions

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Question:16 Find the integrals of the functions

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Question:17 Find the integrals of the functions

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Question:18 Find the integrals of the functions

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Question:19 Find the integrals of the functions

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Question:20 Find the integrals of the functions

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Question:21 Find the integrals of the functions

Answer:

Question:22 Find the integrals of the functions

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Question:23 Choose the correct answer

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Question:24 Choose the correct answer

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CBSE

Question:2 Find the integrals of the functions $\sin 3x \cos 4x$

Question:3 Find the integrals of the functions $\cos 2x \cos 4x \cos 6x$

Question:4 Find the integrals of the functions $\sin ^ 3 ( 2x +1 )$

Question:5 Find the integrals of the functions $\sin ^3 x \cos ^ 3 x$

Question:6 Find the integrals of the functions $\sin x \sin 2x \sin 3x$

Question:7 Find the integrals of the functions $\sin 4x \sin 8x$

Question:8 Find the integrals of the functions $\frac{1- \cos x }{1+ \cos x }$

Question:9 Find the integrals of the functions $\frac {\cos x }{1 + \cos x }$

Question:10 Find the integrals of the functions $\sin ^ 4 x$

Question:11 Find the integrals of the functions $\cos ^ 4 2x$

Question:12 Find the integrals of the functions $\frac{\sin ^ 2x }{1+ \cos x }$

Question:13 Find the integrals of the functions $\frac{\cos 2x - \cos 2 \alpha }{\cos x - \cos \alpha }$

Question:14 Find the integrals of the functions $\frac{\cos x - \sin x }{1+ \sin 2x }$

Question:15 Find the integrals of the functions $\tan ^ 3 2x \sec 2x$

Question:16 Find the integrals of the functions $\tan ^ 4x$

Question:17 Find the integrals of the functions $\frac{\sin ^ 3x + \cos ^ 3x }{\sin ^ 2 x \cos ^2 x }$

Question:18 Find the integrals of the functions $\frac{\cos 2 x + 2 \sin ^ 2x }{\cos ^ 2 x }$

Question:19 Find the integrals of the functions $\frac{1}{\sin x \cos ^3 x }$

Question:20 Find the integrals of the functions $\frac{\cos 2x }{( \cos x + \sin x )^2}$

Question:21 Find the integrals of the functions $\sin ^ { -1} ( \cos x )$

Question:22 Find the integrals of the functions $\frac{1}{\cos ( x-a ) \cos ( x-b )}$

Question:24 Choose the correct answer $\int \frac{e ^x ( 1+x)}{\cos ^ 2 ( e ^xx )} dx \: \: equals$