RD Sharma Class 12 Exercise 18.7 Indefinite Integrals Solutions Maths

RD Sharma Class 12 Exercise 18.7 Indefinite Integrals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 11:57 AM IST

Maths is a subject that requires practice and efficiency. RD Sharma books are the gold standard for CBSE students as they are highly informative, exam-oriented, and easy to understand. They contain simple explanations for questions that can be understood by the weakest of students. Many teachers use RD Sharma books for homework and exam paper questions. RD Sharma Class 12th Exercise 18.7 deals with the chapter ‘Indefinite Integrals.’ It has six Level 1 questions that are concept-based and easy to understand. They include topics like integral of Sin and Cos functions and their evaluation. As this exercise contains only six sums, students can finish this off with ease with the help of RD Sharma solutions provided by Brainly.

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RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise
Indefinite Integrals Excercise:18.7
RD Sharma Chapter wise Solutions

RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.7

Indefinite Integrals Excercise 18.7 Question 1

Answer:
$-\frac{1}{22}\cos 11x+\frac{1}{6}\cos3x+c$
Hint:
$2\sin A\cos B=\sin \left ( A+B\right )+\sin \left ( A-B \right )$
Given:
$\int \sin4x\cos7x dx$
Explanation:
$I=\frac{1}{2}\int2 \sin4x\cos7xdx=\frac{1}{2}\int\[sin\left ( 4x+7x \right )+\sin\left ( 4x-7x \right )]dx\\ =\frac{1}{2}\int \sin11x dx-\frac{1}{2}\int \sin 3x dx \\ =\frac{1}{2}\left [ \frac{\cos11x}{-11} \right ]-\frac{1}{2}\left [ \frac{\cos3x}{-3} \right ]+c\\ =\frac{-1}{22}\cos11x+\frac{1}{6}\cos3x+c$

Indefinite Integrals Excercise 18.7 Question 2

Answer:
$\frac{1}{14}\sin7x+\frac{1}{2}\sin x+c$
Hint:
$2\cos A\cos B=\cos\left ( A+B \right )+ cos \left ( A-B\right )$
Given:
$\int \cos3x\cos4xdx$
Explanation:
$\begin{aligned} & \int \cos 3 x \cos 4 x d x=\int \frac{1}{2}[\cos (3 x+4 x)+\cos (3 x-4 x)] d x \\ =& \frac{1}{2} \int \cos 7 x d x+\frac{1}{2} \int \cos x d x \\ =& \frac{1}{2} \times \frac{\sin 7 x}{7}+\frac{1}{2} \sin x+c \\ =& \frac{1}{14} \sin 7 x+\frac{1}{2} \sin x+c \end{aligned}$

Indefinite Integrals Excercise 18.7 Question 3

Answer:
$=\frac{1}{2}\left[\frac{\sin (m+n) x}{m+n}+\frac{\sin (m-n) x}{m-n}\right]+c$
Hint:
$2\cos A\cos B=\cos\left ( A+B \right )+\cos\left ( A-B \right )$
Given:
$\cos mx\cos nx dx,m\neq n$
Explanation:
$\begin{aligned} &\int(\cos m x \cos n x) d x=\frac{1}{2} \int[\cos (m+n) x+\cos (m-n) x] d x \\ &=\frac{1}{2}\left[\frac{\sin (m+n) x}{m+n}+\frac{\sin (m-n) x}{m-n}\right]+c \end{aligned}$

Indefinite Integrals Excercise 18.7 Question 4

Answer:
$\frac{1}{2}\left[\frac{-\cos (m+n) x}{m+n}-\frac{\cos (m-n) x}{m-n}\right]+c$
Hint:
$2\sin A\cos B=\sin \left ( A+B \right )+\sin\left ( A-B\right )$
Given:
$\sin mx\cos nxdx,m\neq n$
Explanation:
$\begin{aligned} &\int \sin m x \cos n x d x=\frac{1}{2} \int[\sin (m+n) x d x+\sin (m-n) x d x] \\ &=\frac{1}{2} \int \sin (m+n) x d x+\frac{1}{2} \int \sin (m-n) x d x \end{aligned}$
$\begin{aligned} &=\frac{1}{2}\left[\frac{-\cos (m+n) x}{m+n}\right]-\frac{1}{2}\left[\frac{\cos (m-n) x}{m-n}\right]+c \\ &=\frac{1}{2}\left[\frac{-\cos (m+n) x}{m+n}-\frac{\cos (m-n) x}{m-n}\right]+c \end{aligned}$

Indefinite Integrals Excercise 18.7 Question 5

Answer:
$\frac{-1}{32} \cos 8 x+\frac{1}{48} \cos 12 x-\frac{1}{6} \cos 4 x+c$
Hint:
$2 \sin A \sin B=\cos (A+B)-\cos (A-B)$
Given:
$\int \sin 2 x \sin 4 x \sin 6 x d x$
Explanation:
$\begin{gathered} \int \sin 2 x \sin 4 x \sin 6 x d x=\frac{1}{2} \int \sin 2 x(2 \sin 4 x \sin 6 x d x) \\ =\frac{1}{2} \int \sin 2 x[\cos 10 x-\cos 2 x] d x \\ =\frac{1}{2} \int \sin 2 x \cos 10 x d x-\frac{1}{2} \int \sin 2 x \cos 2 x d x \end{gathered}$
$\begin{aligned} &=\frac{1}{4} \int[-\sin 12 x+\sin 8 x] d x+\frac{1}{4} \int[\sin 4 x-\sin 0 x] d x \\ &=\frac{1}{4}\left[\frac{\cos 12 x}{12}-\frac{\cos 8 x}{8}\right]-\frac{1}{4} \frac{\cos 4 x}{4}+c \\ &=\frac{-1}{32} \cos 8 x+\frac{1}{48} \cos 12 x-\frac{1}{6} \cos 4 x+c \end{aligned}$

Indefinite Integrals Excercise 18.7 Question 6

Answer:
$\frac{x}{4}+\frac{1}{16}\sin4x-\frac{1}{24}\sin6x-\frac{1}{8}\sin2x+c$
Hint:
$2\sin A\cos B=\sin\left ( A+B\right )+\sin\left ( A-B \right )$
Given:
$\sin x\cos2x\sin3xdx$
Explanation:
$\sin x\cos 2x\sin3xdx=\frac{1}{2}[2 \sin x \cos 2x]\sin3x$
$=\frac{1}{2}(\sin^{2} 3x-\sin x \sin3x) \quad \quad \quad \quad [2 \sin A \cos B = \sin(A + B) + \sin(A - B)]$
$\begin{aligned} &=\frac{1}{2} \int \sin ^{2} 3 x d x-\frac{1}{2} \int \sin x \sin 3 x d x \\ &=\frac{1}{2} \int \frac{1-\cos 6 x}{2} d x-\frac{1}{4} \int[-\cos 4 x+\cos 2 x] d x \\ &=\frac{1}{4} \int d x-\frac{1}{4} \int \cos 6 x d x+\frac{1}{4} \int \cos 4 x d x-\frac{1}{4} \int \cos 2 x d x \\ &=\frac{x}{4}+\frac{1}{16} \sin 4 x-\frac{1}{24} \sin 6 x-\frac{1}{8} \sin 2 x+c \end{aligned}$

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RD Sharma Chapter wise Solutions

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