RD Sharma Class 12 Exercise 18.11 Indefinite Integrals Solutions Maths - Download PDF Free Online

# RD Sharma Class 12 Exercise 18.11 Indefinite Integrals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 12:06 PM IST

RD Sharma books are well known and widely used for their Maths materials. They offer the best solutions that are comprehensive and easy to learn. Many teachers prefer RD Sharma books to assign homework and to set up question papers. This is why it is the perfect choice for students for Maths.

RD Sharma Class 12th Exercise 18.11 explains the chapter ‘Indefinite Integrals.’ The exercise contains 12 sums that are of Level 1 difficulty and concept-oriented. However, they are pretty easy to understand how to integrate a function w.r.t to x and also cover trigonometric integrals of values like Tan, Sec, Cosec, Cot. Students should refer to the theorems of those trigonometric values to solve these questions.

## Indefinite Integrals Excercise:18.11

Indefinite Integrals Exercise 18.11 Question 2

Answer: $\frac{1}{2} \tan ^{2} x+\frac{1}{4} \tan ^{4} x+C$
Hint :- Use substitution method to solve this integral.
Given :- $\int \tan x \cdot \sec ^{4} x d x$
Sol : - Let $I=\int \tan x \sec ^{4} x d x$
Re-Write $\mathrm{I}=\int \tan x \sec ^{2} x \cdot \sec ^{2} x d x$
$\mathrm{I}=\int \tan x\left(1+\tan ^{2} x\right) \sec ^{2} x d x$ $\text { (if, } \left.\sec ^{2} x=1+\tan ^{2} x\right)$
$\mathrm{I}=\int\left(\tan x+\tan ^{3} x\right) \sec ^{2} x d x$
Substitute $\tan x=t \rightarrow \sec ^{2} x d x=d t$ then
$\mathrm{I}=\int\left(\mathrm{t}+t^{3}\right) \sec ^{2} x \frac{d t}{\sec ^{2} x} \quad \text { (if, } \tan \mathrm{x}=\mathrm{t} \text { ) }$
$=\int\left(\mathrm{t}+t^{3}\right) d t=\int \mathrm{t} d t+\int t^{3} d t$
$=\frac{t^{1+1}}{1+1}+\frac{t^{3+1}}{3+1}+C$ $\text { (if, } \left.\int x^{n} d x=\frac{x^{n+1}}{n+1}+\mathrm{C}\right)$ (i)
$I=\frac{t^{2}}{2}+\frac{t^{4}}{4}+C$
$I=\frac{\tan ^{2} x}{2}+\frac{\tan ^{4} x}{4}+\mathrm{C}$ $\text { (if, } t=\tan x)$

Indefinite Integrals Exercise 18.11 Question 3

Answer: $\frac{1}{6} \tan ^{6} x+\frac{1}{8} \tan ^{8} x+C$
Hint: Use substitution method to solve this integral.
Given: $\int \tan ^{5} x \cdot \sec ^{4} x d x$
Solution: Let,$\mathrm{I}=\int \tan ^{5} x \sec ^{4} x d x$
Re-Write $I=\int \tan ^{5} x \sec ^{2} \cdot \sec ^{2} x d x$
$\left.I=\int \tan ^{5} x\left(1+\tan ^{2} x\right) \cdot \sec ^{2} x d x \quad \text { (if, } \sec ^{2} x=1+\tan ^{2} x\right)$
$\mathrm{I}=\int\left(\tan ^{5} x+\tan ^{7} x\right) \sec ^{2} x d x$
$\text { Substitute } \tan \mathrm{x}=\mathrm{t} \rightarrow \sec ^{2} \mathrm{x} \mathrm{d} \mathrm{x}=\mathrm{dt} \text { , then }$
$\left.I=\int\left(\mathrm{t}^{5}+t^{7}\right) \sec ^{2} x \frac{d t}{\sec ^{2} x} \quad \text { (if, } \tan \mathrm{x}=\mathrm{t}\right)$
$=\int\left(\mathrm{t}^{5}+t^{7}\right) d t=\int \mathrm{t}^{5} d t+\int t^{7} d t$
$=\frac{t^{5+1}}{5+1}+\frac{t^{7+1}}{7+1}+C$ $\text { (if, } \left.\int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right)$......(i)
$\mathrm{I}=\frac{t^{6}}{6}+\frac{t^{8}}{8}+\mathrm{C}$
$\left.\mathrm{I}=\frac{\tan ^{6} x}{6}+\frac{\tan ^{8} x}{8}+\mathrm{C}\: \: \: \: \: \: \: \: \: \quad \text { (if, } \mathrm{t}=\tan \mathrm{x}\right)$

Indefinite Integrals Exercise 18.11 Question 4

Answer:$\frac{1}{6} \sec ^{6} x+C$
Hint:- Use substitution method to solve this integral.
Given:$\int \sec ^{6} x \cdot \tan x d x$
Solution:Let,$\mathrm{I}=\int \sec ^{6} x \cdot \tan x d x$
Re-write,$I=\int \sec ^{5} x \sec x \cdot \tan x d x$$\mathrm{I}=\int \sec ^{5} x(\sec x \cdot \tan x) d x$

Substitute, $\sec x=t \Rightarrow \sec x \cdot \tan x d x=d t$, then,

$I=\int t^{5} \cdot \mathrm{dt} \quad(\text { if }, \sec x=t)$

$=\frac{t^{5+1}}{5+1}+C$ $\text { (if, } \left.\int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right)$

$\left.=\frac{t^{6}}{6}+c=\frac{\sec ^{6} x}{6}+C \quad \text { (if, } \sec x=t\right)$

Indefinite Integrals Exercise 18.11 Question 5

Answer:$\frac{\tan ^{4} x}{4}-\frac{\tan ^{2} x}{2}+\log |\sec \times| C$
Hint:- Use substitution method to solve this integral.
Given:$\int \tan ^{5} x d x$
Solution: Let,$\mathrm{I}=\int \tan ^{5} x d x$
Re-Write,$\mathrm{I}=\int \tan ^{3} x \cdot \tan ^{2} x \cdot d x$
$I=\int \tan ^{3} x\left(\sec ^{2} x-1\right) d x$ $\text { (if, } \left.\tan ^{2} x=\sec ^{2} x-1\right)$
$\mathrm{I}=\int\left(\tan ^{3} x \cdot \sec ^{2} x-\tan ^{3}\right) d x$
$I=\int \tan ^{3} x \cdot \sec ^{2} x \cdot d x-\int \tan ^{3} x d x$
$=\int \tan ^{3} x \cdot \sec ^{2} x \cdot d x-\int \tan ^{2} x \cdot \tan x d x$
$\inline \left.=\int \tan ^{3} x \cdot \sec ^{2} x \cdot d x-\int\left(\sec ^{2} x-1\right) \cdot \tan x d x \quad \text { (if, } \sec ^{2} x-\tan ^{2} x=1\right)$
$\inline =\int \tan ^{3} x \cdot \sec ^{2} x \cdot d x-\int\left(\sec ^{2} x \tan x-\tan x\right) d x$
$=\int \tan ^{3} x \cdot \sec ^{2} x \cdot d x-\int \sec ^{2} x \tan x d x+\int \tan x d x$
Put $tan x = t \Rightarrow \sec^{2}x d x = d t , then$
$\mathrm{I}=\int \mathrm{t}^{3} d t-\int t d t+\int \tan x d x$
$=\frac{t^{8+1}}{3+1}+\frac{t^{1+1}}{1+1}+\log |\sec \times| C$ $\text { (if, } \left.\int x^{n} d x=\frac{x^{n+1}}{n+1}+C \& \int \tan x d x=\log |\sec x|+c\right)$
$=\frac{t^{4}}{4}-\frac{t^{2}}{2}+\log |\sec x| C$
$\left.=\frac{\tan ^{4} x}{4}-\frac{\tan ^{2} x}{2}+\log |\sec x| C\: \: \: \: \: \: \quad \text { (if, } t=\tan x\right)$

Indefinite Integrals Exercise 18.11 Question 6

Answer: $\frac{2}{3} \tan ^{\frac{3}{2}} x+\frac{2}{7} \tan ^{\frac{7}{2}} x+C$
Hint: Use substitution method to solve this integral.
Given:$\int \sqrt{\tan x} \cdot \sec ^{4} x d x$
Solution:Let, $I=\int \sqrt{\tan x} \cdot \sec ^{4} x d x$
Re-Write $\mathrm{I}=\int \sqrt{\tan x} \cdot \sec ^{2} x \cdot \sec ^{2} x d x$
$I=\int \sqrt{\tan x}\left(1+\tan ^{2} x\right) \cdot \sec ^{2} x d x \quad\left(\because \sec ^{2} x-\tan ^{2}=1\right)$
$I=\int\left(\sqrt{\tan x}+\sqrt{\tan x} \cdot \tan ^{2} x\right) \sec ^{2} x d x$
$I=\int\left(\tan ^{\frac{1}{2}} x+\tan ^{\frac{1}{2}+2} x\right) \cdot \sec ^{2} x d x$
$I=\int\left(\tan ^{\frac{1}{2}} x+\tan ^{\frac{5}{2}} x\right) \cdot \sec ^{2} x d x$
$\text { Substitute } \tan \mathrm{x}=\mathrm{t} \rightarrow \sec ^{2} \mathrm{x} \mathrm{d} \mathrm{x}=\mathrm{d} \mathrm{t} \text { , then }$
$\begin{array}{ll} \mathrm{I}=\int\left(t^{\frac{1}{2}}+t^{5 / 2}\right) d t \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: & (\because \tan \mathrm{x}=\mathrm{t}) \end{array}$
$=\frac{t^{\frac{1}{2}+1}}{\frac{1}{2}+1}+\frac{t^{\frac{5}{2}}+1}{\frac{5}{2}+1}+C \: \: \: \: \: \: \: \: \: \: \: \: \: \: \quad\left(\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right)$
$=\frac{\tan ^{3 / 2} x}{3 / 2}+\frac{\tan ^{7 / 2} x}{7 / 2}+\mathrm{C} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \quad(\because \mathrm{t}=\tan \mathrm{x})$
$=\frac{2}{3} \tan ^{3 / 2} x+\frac{2}{7} \tan ^{7 / 2} x+C$

Indefinite Integrals Exercise 18.11 Question 7

\begin{aligned} &\text { Answer }:-\frac{1}{2} \tan 2 x+\frac{1}{6} \tan ^{3} 2 x+C\\ &\text { Hint :- Use substitution method to solve this integral. }\\ &\text { Given :- } \int \sec ^{4} 2 x d x\\ &\text { Sol: - Let } I=\int \sec ^{4} 2 x d x\\ &\text { Re-writing, } I=\int \sec ^{2} 2 x \cdot \sec ^{2} 2 x d x \end{aligned}
\begin{aligned} &\left.I=\int\left(1+\tan ^{2} 2 x\right) \sec ^{2} 2 x d x\: \: \: \: \: \: \: \: \: \: \: \quad \text { (if, } \sec ^{2} x=1+\tan ^{2} x\right) \\ &I=\int\left(\sec ^{2} 2 x+\tan ^{2} 2 x \cdot \sec ^{2} 2 x\right) d x \end{aligned}
\begin{aligned} &\text { Substitute } \tan 2 \mathrm{x}=\mathrm{t}\\ &\sec ^{2} 2 x .2 d x=d t, \text { then } \end{aligned}
\begin{aligned} &\mathrm{I}=\int \sec ^{2} 2 x \mathrm{~d} x+\int \tan ^{2} 2 x \cdot \sec ^{2} 2 x \mathrm{dx} \\ &=\int \sec ^{2} 2 x \mathrm{~d} x+\int t^{2} \cdot \frac{d t}{2} \: \: \: \: \: \: \: \: \: \: \quad(\text { since }, \tan 2 x=t) \end{aligned}
$\inline =\frac{\tan 2 x}{2}+\frac{1}{2} \frac{t^{2+1}}{2+1}+C$ $\inline \text { (if, } \int x^{n} d x=\frac{x^{n+1}}{n+1}+c \quad \& \int \sec ^{2} d x=\tan x+\mathrm{c}$
\inline \begin{aligned} &=\frac{\tan 2 x}{2}+\frac{1}{2} \cdot \frac{t^{3}}{3}+C \\ &\left.=\frac{\tan 2 x}{2}+\frac{1}{6} \cdot \tan ^{3} 2 x+C \: \: \: \: \: \: \: \: \: \: \quad \text { (if, } t=\tan 2 x\right) \end{aligned}

Indefinite Integrals Exercise 18.11 Question 8

Answer: $-\frac{1}{9} \cot ^{3} 3 x-\frac{1}{3} \cot 3 x+C$
Hint :- Use substitution method to solve this integral.
Given: $\int \operatorname{cosec}^{4} 3 x d x$
Solution:Let $\mathrm{I}=\int \operatorname{cosec}^{4} 3 x d x$
Re-Write $\mathrm{I}=\int \operatorname{cosec}^{2} 3 x \cdot \operatorname{cosec}^{2} 3 x d x$
$\left.I=\int\left(1+\cot ^{2} 3 x\right) \cdot \operatorname{cosec}^{2} 3 x d x \: \: \: \: \: \: \: \: \quad \text { (if, } \operatorname{cosec}^{2} x-\cot ^{2} x=1\right)$
\begin{aligned} &I=\int\left(\operatorname{cosec}^{2} 3 x+\cot ^{2} 3 x \cdot \operatorname{cosec}^{2} 3 x\right) d x \\ &I=\int \operatorname{cosec}^{2} 3 x \cdot \cot ^{2} 3 x d x+\int \operatorname{cosec}^{2} 3 x d x \end{aligned}
Substitute, cot3x = t
\begin{aligned} &\operatorname{cosec}^{2} 3 x \cdot 3 d x=d t, \text { then } \\ &I=\int t^{2} \frac{d t}{3}+\int \operatorname{cosec}^{2} 3 x d x \\ &=-\frac{1}{3} \int t^{2} d t+\int \operatorname{cosec}^{2} 3 x d x \end{aligned}
$=-\frac{1}{3} \frac{t^{2+1}}{2+1}+\frac{-\cot 3 x}{3}+C$ $\text { (if, } \int x^{n} d x=\frac{x^{n+1}}{n+1}+c \quad \& \int \operatorname{cosec}^{2} x d x=-\cot x+c$
$=-\frac{1}{3} \cdot \frac{t^{3}}{3}-\frac{\cot 3 x}{3}+C$
$=-\frac{1}{9} \cot ^{3} 3 x-\frac{1}{3} \cdot \cot 3 x+\mathrm{C} \: \: \: \: \: \: \: \: \quad(\text { if, } \cot 3 \mathrm{x}=\mathrm{t})$

Indefinite Integrals Exercise 18.11 Question 9

Answer: $-\frac{\cot ^{n+1} x}{n+1}+\mathrm{C}$
Hint:Use substitution method to solve this integral.
Given:$\int \cot ^{n} x \cdot \operatorname{cocec}^{2} x d x, \mathrm{n} \neq-1$
Solution:let,$\mathrm{I}=\int \cot ^{n} x \cdot \operatorname{cocec}^{2} x d x$
Substitute, cot x = t
$\operatorname{cosec}^{2} x d x=\text { dt then }$
$\begin{array}{ll} \mathrm{I}=\int t^{n} \cdot \operatorname{cosec}^{2} \mathrm{x} \cdot \frac{d t}{-\operatorname{cosec}^{2} x} &\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \text { (if, } \cot \mathrm{x}=\mathrm{t}) \end{array}$
$=-\int t^{n}$
$=-\left[\frac{t^{n+1}}{n+1}\right]+C$ $\text { (if, } \left.\int x^{n} d x=\frac{x^{n+1}}{n+1}+c\right)$
$\left.=-\frac{\cot ^{n+1} x}{n+1}+C \: \: \: \: \: \: \: \quad \text { (if, } \mathrm{t}=\cot \mathrm{x}\right)$

Indefinite Integrals Exercise 18.11 Question 10

Answer: $-\frac{1}{6} \cot ^{6} x-\frac{1}{8} \cot ^{8} x+C$
Hint :- Use substitution method to solve this integral.
Given:$\int \cot ^{5} x \cdot \operatorname{cosec}^{4} x d x$
Solution:Let, $\mathrm{I}=\int \cot ^{5} x \cdot \operatorname{cosec}^{4} x d x$
Re-Write,$\mathrm{I}=\int \cot ^{5} x \cdot \operatorname{cosec}^{2} x \cdot \operatorname{cosec}^{2} x d x$
$\mathrm{I}=\int \cot ^{5} x\left(1+\cot ^{2} x\right) \cdot \operatorname{cosec}^{2} x d x$ $\text { (if, } \left.\operatorname{cosec}^{2} x-\cot ^{2} x=1\right)$
$\mathrm{I}=\int\left(\cot ^{5} x+\cot ^{5} x \cdot \cot ^{2}\right) \operatorname{cosec}^{2} x d x$
$\mathrm{I}=\int\left(\cot ^{5} x+\cot ^{7} x\right) \operatorname{cosec}^{2} x d x$

Substitute, cotx = t

$-\operatorname{cosec}^{2} x d x=d t, \text { then }$

$I=\int\left(t^{5}+t^{7}\right) \operatorname{cosec}^{2} x \cdot \frac{d t}{-\operatorname{cosec}^{2} x} \mathrm{dx} \: \: \: \: \: \: \: \: \quad(\text { as } \cot x=t)$

\begin{aligned} &=-\int\left(t^{5}+t^{7}\right) \mathrm{dt} \\ &=-\int t^{5} d t-\int t^{7} \mathrm{dt} \end{aligned}

$=-\frac{t^{5+1}}{5+1}-\frac{t^{7+1}}{7+1}+C$ $\text { (if, } \left.\int x^{n} d x=\frac{x^{n+1}}{n+1}+\mathrm{c}\right)$

$=-\frac{t^{6}}{6}-\frac{t^{8}}{8}+C$

$=-\frac{\cot ^{6} x}{6}-\frac{\cot ^{8} x}{8}+\mathrm{C}$ $(\text { if, } t=\cot x)$

Indefinite Integrals Exercise 18.11 Question 11

Answer: $-\frac{1}{4} \cot ^{4} x+\frac{1}{2} \cot ^{2} x+\log |\sin x|+C$
Hint :- Use substitution method to solve this integral.
Given: $\int \cot ^{5} x d x$
Solution: Let $I=\int \cot ^{5} x d x$
Re-Write,$\mathrm{I}=\int \cot ^{5} x d x$
$=\int \cot ^{3} x \cdot \cot ^{2} x d x$
$\mathrm{I}=\int \cot ^{3} x\left(\operatorname{cosec}^{2} x-1\right) d x$ $\text { (if, } \operatorname{cosec}^{2} x-\cot ^{2} x=1 \text { ) }$
\begin{aligned} &I=\int\left(\cot ^{3} x \cdot \operatorname{cosec}^{2} x-\cot ^{3} x\right) d x \\ &I=\int \cot ^{3} x \cdot \operatorname{cosec}^{2} x d x-\int \cot ^{3} x d x \\ &I=\int \cot ^{3} x \cdot \operatorname{cosec}^{2} x d x-\int \cot ^{2} x \cot x d x \end{aligned}
\begin{aligned} &\mathrm{I}=\int \cot ^{3} x \cdot \operatorname{cosec}^{2} x d x-\int\left(\operatorname{cosec}^{2} x-1\right) \cot x d x \\ &\mathrm{I}=\int \cot ^{3} x \cdot \operatorname{cosec}^{2} x d x-\int\left(\operatorname{cosec}^{2} x \cdot \cot x-\cot x\right) d x \\ &\mathrm{I}=\int \cot ^{3} x \cdot \operatorname{cosec}^{2} x d x-\int \operatorname{cosec}^{2} x \cdot \cot x d x+\int \cot x d x \end{aligned}

Substitute, cotx = t

\begin{aligned} &-\operatorname{cosec}^{2} x d x=d t, \text { then } \\ &I=\int t^{3}(-d t)-\int t(-d t)+\int \cot x d x \\ &=-\int t^{3} d t+\int t d t+\int \cot x d x \end{aligned}

$=-\frac{t^{8+1}}{3+1}+\frac{t^{1+1}}{1+1}+\log |\sin x|+C$ $\inline \text { (if, } \left.\int x^{n} d x=\frac{x^{n+1}}{n+1}+\mathrm{c} \& \int \cot x d x=\log |\sin x|+c\right)$

\inline \begin{aligned} &=-\frac{1}{4} t^{4}+\frac{1}{2} t^{2}+\log |\sin x|+C \\ &=-\frac{1}{4} \cot ^{4} x+\frac{1}{2} \cot ^{2} x+\log |\sin x|+C \end{aligned} $\inline \left ( as,t=\cot x \right )$

Indefinite Integrals Exercise 18.11 Question 12

Answer: $-\frac{1}{5} \cot ^{5} x+\frac{1}{3} \cot ^{3} x-\cot x-x+C$
Hint :- Use substitution method to solve this integral.
Given: $\int \cot ^{6} x d x$
Solution: Let,$I=\int \cot ^{6} x d x$
Re-Write,$\mathrm{I}=\int \cot ^{6} x d x$
$\mathrm{I}=\int \cot ^{2} x \cdot \cot ^{4} x d x$
$\mathrm{I}=\int\left(\operatorname{cosec}^{2} x-1\right) \cot ^{4} x d x$ $\text { (if, } \operatorname{cosec}^{2} x-\cot ^{2} x=1 \text { ) }$
\begin{aligned} &I=\int \operatorname{cosec}^{2} x \cdot \cot ^{4} x d x-\int \cot ^{4} x d x \\ &I=\int \operatorname{cosec}^{2} x \cdot \cot ^{4} x d x-\int \cot ^{2} x \cdot \cot ^{2} x d x \end{aligned}
$\inline \left.I=\int \operatorname{cosec}^{2} x \cdot \cot ^{4} x d x-\int \cot ^{2} x\left(\operatorname{cosec}^{2} x-1\right) d x \quad \text { (if, } \operatorname{cosec}^{2} x-\cot ^{2} x=1\right)$
\inline \begin{aligned} &1=\int \operatorname{cosec}^{2} x \cdot \cot ^{4} x d x-\int\left(\cot ^{2} x \cdot \operatorname{cosec}^{2} x-\cot ^{2} x\right) d x \\ &\mathrm{I}=\int \operatorname{cosec}^{2} x \cdot \cot ^{4} x d x-\int \cot ^{2} x \cdot \operatorname{cosec}^{2} x d x+\int \cot ^{2} x d x \end{aligned}
$\left.I=\int \operatorname{cosec}^{2} x \cdot \cot ^{4} x d x-\int \cot ^{2} x \cdot \operatorname{cosec}^{2} x d x+\int\left(\operatorname{cosec}^{2} x-1\right) d x \text { (if, } \cot ^{2} x=\operatorname{cosec}^{2} x-1\right)$
$\mathrm{I}=\int \operatorname{cosec}^{2} x \cdot \cot ^{4} x d x-\int \cot ^{2} x \cdot \operatorname{cosec}^{2} x d x+\int \operatorname{cosec}^{2} x d x-\int 1 d x$
Substitute, cotx = t
$-\operatorname{cosec}^{2} x d x=d t, \text { then }$
\begin{aligned} &\mathrm{I}=\int t^{4}(-d t)-\int t^{2}(-d t)+\int \operatorname{cosec}^{2} x d \mathrm{x}-\int 1 \mathrm{dx} \\ &=-\int t^{4} d t+\int t^{2} d t+\int \operatorname{cosec}^{2} x d \mathrm{x}-\int x^{0} \mathrm{dx} \end{aligned}
$=-\frac{t^{4+1}}{4+1}+\frac{t^{2+1}}{2+1}-\cot x-\frac{x^{0+1}}{0+1}+C$ $\text { (if, } \left.\int x^{n} d x=\frac{x^{n+1}}{n+1}+c \& \int \operatorname{cosec}^{2} x d x=-\cot x+c\right)$
\begin{aligned} &=-\frac{t^{5}}{5}+\frac{t^{3}}{3}-\cot x-x+C \\ &=-\frac{\cot ^{5} x}{5}+\frac{\cot ^{8} x}{3}-\cot x-x+C \end{aligned} $\left ( if,\cot x=t \right )$

RD Sharma Class 12th Exercise 18.11 solutions provided by Career360 is the best choice for students for exam preparation as it covers the entire syllabus. Students who find it confusing to study from the RD Sharma book can refer to this material to simplify their preparation. Furthermore, this material complies with the CBSE syllabus, which means that students can directly refer to it to understand their class lectures better.

As there are hundreds of sums in this chapter, teachers can't cover all of them in their lectures and usually give a huge chunk of questions as homework. Therefore, RD Sharma Class 12th Exercise 18.11 material is specifically designed for students to get a good understanding of the subject which will help them finish their homework and get good marks in exams.

All the solutions from this material go through quality tests to ensure that every answer is accurate. Additionally, as the answers are expert-created they are specifically prepared to help students find the best method of solving questions. RD Sharma Solution As there are different methods to solve a problem students can gain knowledge on all of them and use whatever suits them best.

Therefore, students can rest assured that RD Sharma Class 12th Exercise 18.11 solutions provides the best quality answers and is perfect for reference. Once the students are through with this material, there is no other source they would need to follow as it covers all the topics.

RD Sharma Class 12th Exercise 18.11 material contains step-by-step answers that are simple and direct. Students can take advantage of this material as it is convenient and free to study from Career360’s website. They can access this material through any device using a browser and internet connection and learn all the concepts in a paperless manner. Thus, they can prepare from the comfort of their homes and score well in exams.

## RD Sharma Chapter wise Solutions

1. Does this material have all topics?

Yes, RD Sharma Class 12 Chapter 18 Exercise 18.11 material covers all concepts and contains step-by-step solutions to help students understand better.

Career360 offers a variety of solutions on its website. For example, students can search the book name and exercise number to find the results of their desired material.

3. Can I score good marks in exams with this material?

Class 12 RD Sharma Chapter 18 Exercise 18.11 Solutions are easy to understand and cover all concepts. In addition, it helps students score better marks using its exam-oriented solutions.

4. What are the limits in Integration?

Limits are the values that are applied after the calculation of integral value. The upper limit subtracts the lower limit after Integration. To know more about integrals, check RD Sharma Class 12 Solutions Indefinite Integrals Ex 18.11.

5. What are the additional costs for this material?

The solutions provided by Career360 in RD Sharma Class 12 solutions Chapter 18 Ex 18.11 are absolutely free with no hidden costs.

## Upcoming School Exams

#### National Means Cum-Merit Scholarship

Application Date:01 August,2024 - 16 September,2024

Exam Date:19 September,2024 - 19 September,2024