RD Sharma Class 12 Exercise 18.12 Indefinite Integrals Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Exercise 18.12 Indefinite Integrals Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 18.12 Indefinite Integrals Solutions Maths - Download PDF Free Online

Kuldeep MauryaUpdated on 24 Jan 2022, 12:09 PM IST

RD Sharma class 12th exercise 18.12 is a top-rated NCERT solution that students will find among study materials. The brand has made a name for itself as numerous students have placed their trust in RD Sharma class 12 chapter 18 exercise 18.12 and greatly benefitted from the answers in the book. High school students who are in class 12 especially should use the RD Sharma class 12th exercise 18.12 for their exam preparations and home practice. Maths is a tough subject and to master it students need to practice answering questions daily. Self-study at home will be pretty easy with the help of RD Sharma solutions to master it. If you want to sharpen your skills in maths, thr class 12 RD Sharma chapter 18 exercise 18.12 solution will assist you in this endeavor.

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RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise
Indefinite Integrals Excercise:18.12
RD Sharma Chapter wise Solutions

RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.12

Indefinite Integrals Exercise 18.12 Question 1

Answer: - $\frac{1}{5} \sin ^{5} x-\frac{1}{7} \sin ^{7} x+C$
Hint: - Use substitution method to solve this integral
Given:-$\int \sin ^{4} x \cos ^{3} x d x$
Solution: - Let, $I=\int \sin ^{4} x \cos ^{3} x d x$
The exponent of $\cos x$ is odd, so we substitute $\sin x=t\Rightarrow \cos dt$then,
$I=\int t^{4} \cos ^{3} x \frac{d t}{\cos x} \quad\quad\quad\quad\quad\quad\quad[\because \sin x=t]$
$=\int t^{4} \cos ^{2} x d t$
$\begin{array}{ll} =\int t^{4}\left(1-\sin ^{2} x\right) d t &\quad\quad\quad\quad\quad\quad {\left[\because \sin ^{2} x=\cos ^{2} x=1\right]} \\ \\=\int t^{4}\left(1-t^{2}\right) d t & \quad\quad\quad\quad\quad\quad{[\because \sin x=t]} \end{array}$
$\begin{aligned}\\ &=\int\left(t^{4}-t^{4} t^{2}\right) d t \\ &=\int\left(t^{4}-t^{6}\right) d t \\ &=\int t^{4} d t-\int t^{6} d t \end{aligned}$
$=\frac{t^{4+1}}{4+1}-\frac{t^{6+1}}{6+1}+C$ $\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right]$
$=\frac{t^{5}}{5}-\frac{t^{7}}{7}+C$
$=\frac{\sin ^{5} x}{5}-\frac{\sin ^{7}}{7}+C$ $\left [ \because t=\sin x \right ]$

Indefinite Integrals Exercise 18.12 Question 2

Answer:$-\cos x+\frac{2}{3} \cos ^{3} x-\frac{1}{5} \cos ^{5} x+C$
Hint: Use substitution method to solve this integral
Given:$\int \sin^{5 }xdx$
Solution: Let, $I=\int \sin^{5 }xdx$
Re-writing,
$\begin{aligned} &I=\int \sin ^{3} x \sin ^{2} x d x \\ &=\int \sin ^{3} x\left(1-\cos ^{2} x\right) \quad \quad \quad \quad \quad \quad\left[\because \sin ^{2} x+\cos ^{2} x=1\right] \end{aligned}$
$\begin{aligned} &=\int\left(\sin ^{3} x-\sin ^{3} x \cos ^{2} x\right) d x \\ &=\int\left(\sin ^{2} x \sin x-\sin ^{3} x \cos ^{2} x\right) d x \\ &\left.=\int\left(1-\cos ^{2} x\right) \sin x-\sin ^{3} x \cos ^{2} x\right) d x \quad\quad\quad\quad\quad\quad\left[\because \sin ^{2} x=1-\cos ^{2} x\right] \end{aligned}$
$\begin{aligned} &=\int\left(\sin x-\sin x \cos ^{2} x-\sin ^{3} x \cos ^{2} x\right) d x \\ &=\int \sin x d x-\int \sin x \cos ^{2} x-\int \sin ^{3} x \cos ^{2} x d x \end{aligned}$
Substitute $\cos x=t \Rightarrow-\sin x d x \Rightarrow d t$ in second and third integral, then
$\begin{aligned} I &=\int \sin x d x-\int t^{2}(-d t)-\int t^{2} \sin ^{3} x \frac{d t}{-\sin x} \\ &=\int \sin x d x+\int t^{2} d t+\int \sin ^{2} x t^{2} d t \\ &=\int \sin x d x+\int t^{2} d t+\int\left(1-\cos ^{2} x\right) t^{2} d t\quad\quad\quad\quad\quad\quad&\left[\because \sin ^{2} x=1-\cos ^{2} x\right] \\ \end{aligned}$
$\begin{aligned} &=\int \sin x d x+\int t^{2} d t+\int t^{2}\left(1-t^{2}\right) d t \\ &=\int \sin x d x+\int t^{2} d t+\int t^{2}-t^{2} \cdot t^{2} d t \\ &=\int \sin x d x+\int t^{2} d t+\int t^{2} d t-\int t^{4} d t \end{aligned}$
$\begin{aligned} &=-\cos x+\frac{t^{2+1}}{2+1}+\frac{t^{2+1}}{2+1}-\frac{t^{4+1}}{4+1}+C \quad\quad\quad\quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+c \& \int \sin x d x=-\cos x\right] \\ &=-\cos x+\frac{t^{2+1}}{2+1}+\frac{t^{2+1}}{2+1}-\frac{t^{5}}{5}+C \end{aligned}$
$\begin{aligned} &=-\cos x+\frac{t^{2+1}}{2+1}+\frac{t^{2+1}}{2+1}-\frac{t^{5}}{5}+C \\ &=-\cos x+\frac{t^{3}}{3}+\frac{t^{3}}{3}-\frac{t^{5}}{5}+C \\ &=-\cos x+\frac{2 t^{3}}{3}-\frac{t^{5}}{5}+C \\ &=-\cos x+\frac{2}{3} \cos ^{3} x-\frac{1}{5} \cos ^{5}+C \quad \quad \quad \quad \quad \quad \quad[\because t=\cos x] \end{aligned}$

Indefinite Integrals Exercise 18.12 Question 3

Hint: Use substitution method to solve this integral
Given:$\int \cos ^{5} x d x$
Solution: Let $I=\int \cos ^{5} x d x$
Re-writing, $I=\int \cos ^{5} x d x$
$\begin{aligned} I &=\int \cos ^{3} x \cos ^{2} x d x \\ &=\int \cos ^{3} x\left(1-\sin ^{2} x\right) d x\quad\quad\quad\quad\quad\quad\quad\left[\because \sin ^{2} x+\cos ^{2} x=1\right] \end{aligned}$
$\begin{aligned} &=\int\left(\cos ^{3} x-\cos ^{3} x \sin ^{2} x\right) d x \\ &=\int \cos ^{3} x d x-\int \cos ^{3} x \sin ^{2} x d x \\ &=\int \cos ^{2} x \cos x d x-\int \cos ^{2} x \cos x \sin ^{2} x d x \end{aligned}$
$\begin{aligned} &=\int \cos x\left(1-\sin ^{2} x\right) d x-\int\left(1-\sin ^{2} x\right) \sin ^{2} x \cos x d x \quad \quad \quad \quad \quad \quad \quad\left[\because \cos ^{2} x=1-\sin ^{2} x\right] \\ &=\int \cos x d x-\int \cos x \sin ^{2} x d x-\int \sin ^{2} x \cos x d x+\int \sin ^{4} x \cos x d x \\ &=\int \cos x d x-2 \int \sin ^{2} x \cos x d x+\int \sin ^{4} x \cos x d x \end{aligned}$
Substitute $\sin x=t\Rightarrow \cos xdx=dt$ in second and third integral, then
$I =\int \cos x d x-2 \int t^{2} d t+\int t^{4} d t \quad\quad\quad\quad\quad\quad\left [ \because \int \cos xdx=\sin x+c \right ]$
$\begin{aligned} &=\sin x-2 \frac{t^{2+1}}{2+1}+\frac{t^{4+1}}{4+1}+C \\ &=\sin x-2 \frac{t^{3}}{3}+\frac{t^{5}}{5}+C \\ &=\sin x-2 \frac{\sin ^{3} x}{3}+\frac{\sin ^{5} x}{5}+C \end{aligned}$

Indefinite Integrals Exercise 18.12 Question 4

Answer:- $\frac{1}{6} \sin ^{6} x+C$
Hint: - Use substitution method to solve this integral.
Given:- $\int \sin ^{5} x \cdot \cos x d x$
Solution: - Let $\int \sin ^{5} x \cdot \cos x d x$
Substitute,
$\begin{aligned} \operatorname{Sin} x &=t \Rightarrow \cos x d x=d t, \text { then } \\ \quad I &=\int t^{5} \cdot d t \\ \quad &=\frac{t^{5}+1}{5+1}+C \quad\quad\quad\quad\quad\quad\quad\quad\left[\therefore \int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right] \end{aligned}$
$\begin{aligned} &=\frac{t^{6}}{6}+C \\ &=\frac{\sin ^{6} x}{6}+C \quad\quad\quad\quad\quad\quad\quad\quad[\because t=\sin x] \end{aligned}$

Indefinite Integrals Exercise 18.12 Question 5

Answer:-$\frac{-1}{7} \cos ^{7} x+\frac{1}{9} \cos ^{9} x+c$
Hint: - We substitute method to solve this integral.
Given:- $\int \sin ^{3} x \cdot \cos ^{6} x d x$
Solution: - Let $I=\int \sin ^{3} x \cdot \cos ^{6} x d x$
The exponent of $\sin x$ is odd, so we substitute
$\operatorname{Cos} x=t \Rightarrow-\operatorname{Sin} x d x=d t$, then
$\begin{aligned} &I=\int \sin ^{3} x t^{6} \cdot \frac{d t}{-\operatorname{Sin} x}\quad\quad\quad\quad\quad\left [ t=Cos x \right ]\\ &I=-\int \operatorname{Sin}^{2} x t^{6} d t \end{aligned}$
$\begin{aligned} &=-\int\left(1-\operatorname{Cos}^{2} x\right) \cdot t^{6} d t \quad\quad\quad\quad\quad\quad\left[\because \operatorname{Sin}^{2} x+\operatorname{Cos}^{2} x=1\right] \\ &=-\int t^{6}\left(1-t^{2}\right) d t \end{aligned}$
$\begin{aligned} &=-\int t^{6}\left(1-t^{2}\right) d t \\ &=-\int\left(t^{6}-t^{6} \cdot t^{2}\right) d t \\ &=-\int\left(t^{6}-t^{8}\right) d t \\ &=-\int t^{6} d t+\int t^{8} d t \end{aligned}$
$\begin{aligned} &=-\frac{t^{6+1}}{6+1}+\frac{t^{8+1}}{8+1}+C \quad\quad\quad\quad\quad\quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right] \\ &=-\frac{t^{7}}{7}+\frac{t^{9}}{9}+C \\ &=-\frac{\operatorname{Cos}^{7} x}{7}+\frac{\operatorname{Cos}^{9} x}{9}+C \quad\quad\quad\quad\quad[\because t=\operatorname{Cos} x] \end{aligned}$

Indefinite Integrals Exercise 18.12 Question 6

Answer:- $\operatorname{Sin} x-\operatorname{Sin}^{3} x+\frac{3}{5} \operatorname{Sin}^{5} x-\frac{1}{7} \operatorname{Sin}^{7} x+c$
Hint: - Use substitute method to solve this integral.
Given:-$\int \operatorname{Cos}^{7} x d x$
Solution: - Let $I=\int \operatorname{Cos}^{7} x d x$
Re-writing,
$\begin{aligned} &I=\int \operatorname{Cos}^{6} x \cdot \operatorname{Cos} x d x \\ &\Rightarrow I=\int\left(\operatorname{Cos}^{2} x\right)^{3} \cdot \operatorname{Cos} d x \end{aligned}$
$\begin{array}{ll} \Rightarrow I=\int\left(1-\operatorname{Sin}^{2} x\right)^{3} \cdot \operatorname{Cos} d x &\quad\quad\quad\quad\quad {\left[\because \operatorname{Sin}^{2} x+\operatorname{Cos}^{2} x=I\right]} \\\\ \Rightarrow I=\int\left[1-\operatorname{Sin}^{6} x-3 \operatorname{Sin}^{2} x+3 \operatorname{Sin}^{4} x\right] \cdot \operatorname{Cos} d x &\quad\quad\quad\quad {\left[\because(a-b)^{3}=a^{3}-b^{3}-3 a^{2} b+3 a b^{b}\right]} \end{array}$
$\begin{aligned} &\Rightarrow I=\int\left(\operatorname{Cos} x-\operatorname{Sin}^{6} x \cdot \operatorname{Cos} x-3 \operatorname{Sin}^{2} x \cdot \operatorname{Cos} x+3 \operatorname{Sin}^{4} x \cdot \operatorname{Cos} x\right) d x \\ &\Rightarrow I=\int \operatorname{Cos} d x-\int \operatorname{Sin}^{6} x \cdot \operatorname{Cos} d x-\int 3 \operatorname{Sin}^{2} x \cdot \operatorname{Cos} d x+3 \int \sin ^{4} x \cdot \cos x d x \end{aligned}$
Substitute $\operatorname{Sin} x=t \Rightarrow \operatorname{Cos} x d x=d t$ in 2nd, 3rd and 4th integral, then
$\begin{aligned} &I=\int \operatorname{Cos} d x-\int t^{6} d t-3 \int t^{2} d t+3 \int t^{4} \cdot d t \\ &=\operatorname{Sin} x-\frac{t^{6}+1}{6+1}-3 \frac{t^{2+1}}{2+1}+3 \frac{t^{4+1}}{4+1}+C \quad \quad \quad \quad \quad \quad \quad\left[\because \int \operatorname{Cos} x d x=\operatorname{Sin} x+c \& \int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right] \end{aligned}$
$\begin{aligned} &=\operatorname{Sin} x-\frac{t^{7}}{7}-3 \frac{t^{3}}{3}+3 \cdot \frac{t^{5}}{5}+C \\ &=\operatorname{Sin} x-\frac{\operatorname{Sin}^{7} x}{7}-\operatorname{Sin}^{3} x+\frac{3}{5} \operatorname{Sin}^{5} x+C \\ &=\operatorname{Sin} x-\operatorname{Sin}^{3} x+\frac{3}{5} \operatorname{Sin}^{5} x-\frac{1}{7} \operatorname{Sin}^{7} x+C \quad\quad\quad\quad\quad\quad\quad[\because \operatorname{Sin} x=t] \end{aligned}$

Indefinite Integrals Exercise 18.12 Question 7

Answer: -$-\frac{1}{8} \operatorname{Cos}^{4} x^{2}+C$
Hint: - Use substitution method to solve this integral.
Given: -$\int x \operatorname{Cos}^{3} x^{2} \operatorname{Sin} x^{2} d x$
Solution: - Let $I=\int x \operatorname{Cos}^{3} x^{2} \operatorname{Sin} x^{2} d x$
Substitute, $\operatorname{Cos} x^{2}=t \Rightarrow-\operatorname{Sin} x^{2} 2 x d x=d t \text { , then }$
$\begin{aligned} I &=\int t^{3} x \operatorname{Sin} x^{2} \cdot \frac{d t}{-\left(\sin x^{2}\right)(x)^{2}}\quad\quad\quad\quad &\left[\because \operatorname{Cos} x^{2}=t\right] \\ &=-\frac{1}{2} \int t^{3} d t=\frac{-1}{2} \cdot \frac{t^{3+1}}{3+1}+C &\quad\quad\quad\quad\quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right] \end{aligned}$
$\begin{aligned} &=\frac{-1}{2} \cdot \frac{t^{4}}{4}+C=\frac{-1}{8} t^{4}+C \\ &=\frac{-1}{8} \cos ^{4} x^{2}+C \quad \quad \quad \quad \quad \quad\left[\because t=\cos x^{2}\right] \end{aligned}$

Indefinite Integrals Exercise 18.12 Question 8

Answer: -$-\cos x+\cos ^{3} x-\frac{3}{5} \cos ^{5} x+\frac{1}{7} \cos ^{7} x+C$
Hint: - Use substitution method to solve this integral.
Given: -$\int \sin ^{7} x d x$
Solution: - Let $I=\int \sin ^{7} x d x$
Re-writing,
$\begin{aligned} &I=\int \sin ^{6} x \cdot \sin x d x \\ &\Rightarrow I=\int\left(\sin ^{2} x\right)^{3} \sin x d x \\ &\Rightarrow I=\int\left(1-\cos ^{2} x\right) \sin x d x \quad\quad\quad\quad\quad\quad\quad\left[\because \sin ^{2} x+\cos ^{2} x=I\right] \end{aligned}$
$\begin{aligned} &\Rightarrow I=\int\left[1-\cos ^{6} x-3 \cos ^{2} x+3 \cos ^{4} x\right] \sin x d x \\ &\Rightarrow I=\int\left[\sin x-\cos ^{6} x \sin x-3 \cos ^{2} x \sin x+3 \cos ^{4} x \sin x\right] \\ &\Rightarrow I=\int \sin d x-\int \cos ^{6} \cdot \sin x d x-3 \int \cos ^{2} x \sin x d x+3 \int \cos ^{4} x \sin x d x \end{aligned}$
Substitute,$\cos x=t \Rightarrow-\sin x d x=d t$ in 2nd, 3rd and 4th integral, then
$\begin{aligned} I &=\int \sin x d x-\int t^{6}(-d t)-3 \int t^{2}(-d t)+3 \int t^{4}(-d t) \\ &=-\cos x+\int t^{6} d t+3 \int t^{2} d t-3 \int t^{4} d t &\quad\quad\quad\quad\quad\left [ \because \int \sin xdx=-\cos x +c \right ]\\ \end{aligned}$
$\begin{aligned} &=-\cos x+\frac{t^{6+1}}{6+1}+3 \frac{t^{2+1}}{2+1}-3 \frac{t^{4+1}}{4+1}+C \quad\quad\quad\quad\quad\quad\quad\quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right] \\ &=-\cos x+\frac{t^{7}}{7}+3 \cdot \frac{t^{3}}{3}-3 \frac{t^{5}}{5}+C \end{aligned}$
$\begin{aligned} &=-\cos x+\frac{\cos ^{7} x}{7}-\frac{3}{5} \cos ^{5} x+\cos ^{3} x+C \quad\quad\quad\quad\quad\quad\quad[\because t=\cos x] \\ &=-\cos x+\cos ^{3} x-\frac{3}{5} \cos ^{5} x+\frac{\cos ^{7} x}{7}+C \end{aligned}$

Indefinite Integrals Exercise 18.12 Question 10

Answer: -$\frac{-1}{3} \cot ^{3} x-2 \cot x+\tan x+C$
Hint: - Use substitution method to solve this integration.
Given: -$\int \frac{1}{\sin ^{4} x \cdot \cos ^{2} x} d x$
Solution: - Let $I=\int \frac{1}{\sin ^{4} x \cdot \cos ^{2} x} d x$
The power of denominator is $4+2=6$so we divide both numerator and denominator by $\cos ^{6}x$so that the integral $I$can come in integral able form.
$\begin{aligned} &\therefore I=\int \frac{\frac{1}{\cos ^{6} x}}{\frac{\sin ^{4} x \cdot \cos ^{2} x}{\cos ^{6} x}} d x=\int \frac{\sec ^{6} x}{\frac{\sin ^{4} x}{\cos ^{4} x}} d x \\ &\Rightarrow=\int \frac{\sec ^{6} x}{\tan ^{4} x} d x \end{aligned}$
Re-writing,
$\begin{aligned} I &=\int \frac{\sec ^{4} x \cdot \sec ^{2} x}{\tan ^{4} x} d x \\ &=\int \frac{\left(\sec ^{2} x\right)^{2} \sec ^{2} x}{\tan ^{4} x} d x \end{aligned}$
$\begin{aligned} &=\int \frac{\left(1+\tan ^{2} x\right)^{2} \sec ^{2} x}{\tan ^{4} x} d x \quad\quad\quad\quad\quad\quad\quad\quad\left[\because \sec ^{2} x-\tan ^{2} x=1\right] \\ &=\int \frac{\left(1+\tan ^{4} x+2 \tan ^{2} x\right) \sec ^{2} x}{\tan ^{4} x} \mathrm{dx} \quad\quad\quad\quad\quad\quad\left[\because(a+b)^{2}=a^{2}+b^{2}+2 a b\right] \end{aligned}$
Substitute $\tan x=t \Rightarrow \sec ^{2} x d x=d t$ then
$\begin{aligned} I &=\int \frac{\left(1+t^{4}+2 t^{2}\right)}{t^{4}} d t \\ &=\int\left(\frac{1}{t^{4}}+\frac{t^{4}}{t^{4}}+\frac{2 t^{2}}{t^{4}}\right) d t \\ &=\int\left(\frac{1}{t^{4}}+1+2 \cdot \frac{1}{t^{2}}\right) d t=\int\left(t^{-4}+t^{0}+2 t^{-2}\right) d t \end{aligned}$
$\begin{aligned} &=\int\left(\frac{1}{t^{4}}+1+2 \cdot \frac{1}{t^{2}}\right) d t=\int\left(t^{-4}+t^{0}+2 t^{-2}\right) d t \\ &=\frac{t^{-4+1}}{-4+1}+\frac{t^{0+1}}{0+1}+2 \frac{t^{-2+1}}{-2+1}+C \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\left[\because \int x^{n} d x=\frac{x^{n+1}}{n+1}+C\right] \\ &=\frac{t^{-3}}{-3}+t+2 \frac{t^{-1}}{(-1)}+C \end{aligned}$
$\begin{aligned} &=-\frac{t^{3}}{3}+t-2 t^{-1}+C=-\frac{1}{3 t^{3}}+t-\frac{2}{t}+C \\ &=-\frac{1}{3 \tan ^{3} x}+\tan x-\frac{2}{\tan x}+C \quad\quad\quad\quad\quad\quad\left [ \because t=\tan x \right ]\\ &=-\frac{1}{3} \cot ^{3} x+\tan x-2 \cot x+C\quad\quad\quad\quad\quad\quad\left [ \because \tan x=\frac{1}{\cot x} \right ] \\ &=-\frac{1}{3} \cot ^{3} x-2 \cot x+\tan x+C \end{aligned}$

Indefinite Integrals Exercise 18.12 Question 12

Answer: -$-\frac{1}{2 \tan ^{2} x}+\log |\tan x|+C$
Hint: - Use substitution method to solve this integral.
Given: -$\int \frac{1}{\sin ^{3} x \cdot \cos x} d x$
Solution: - Let $I=\int \frac{1}{\sin ^{3} x \cdot \cos x} d x$
The exponent of denominator is $3+1=4$ to solve this type of integral; uses have divided both numerator and denominator by $\cos ^{4}x$ then
$\begin{aligned} I &=\int \frac{\frac{1}{\cos ^{4} x}}{\frac{\sin ^{3} x \cdot \cos x}{\cos ^{4} x}} d x \\ &=\int \frac{\left(\sec ^{4} x\right)}{\frac{\sin ^{3} x}{\cos ^{3} x}} d x \end{aligned}$
$\begin{aligned} &=\int \frac{\sec ^{4} x}{\tan ^{3} x} d x=\int \frac{\sec ^{2} x \cdot \sec ^{2} x}{\tan ^{3} x} d x \\ &=\int \frac{\left(1+\tan ^{2} x\right) \sec ^{2} x}{\tan ^{3} x} d x\quad\quad\quad\quad\quad\quad\quad\quad\left[\because 1+\tan ^{2} x=\sec ^{2} x\right] \\ \end{aligned}$
Substitute $\tan x=t \Rightarrow \sec ^{2} x d x=d t$ then
$\begin{aligned} I &=\int \frac{1+t^{2}}{t^{3}} d t \\ &=\int\left(\frac{1}{t^{3}}+\frac{t^{2}}{t^{3}}\right) d t \\ &=\int\left(t^{-3}+\frac{1}{t}\right) d t \end{aligned}$
$=\frac{t^{-3+1}}{-3+1}+\log |t|+C \quad\quad\quad\quad\quad\quad\quad\left[\because \int x^{n} d x=x \frac{n+1}{n+1}+c \text { and } \int \frac{1}{x} d x=\log |x|+C\right]$
$\begin{aligned} &=\frac{t^{-2}}{-2}+\log |t|+C \\ &=-\frac{1}{2 t^{2}}+\log |t|+C \\ &=-\frac{1}{2 \tan ^{2} x}+\log |\tan x|+C \quad\quad\quad\quad\quad\quad\quad[\because t=\tan x] \end{aligned}$

Indefinite Integrals Exercise 18.12 Question 13

Answer:$\frac{1}{2} \tan ^{2} x+\log |\tan x|+C$
Hint: Use substitution method to solve this integral.
Given:$\int \frac{1}{\sin x \cdot \cos ^{3} x} d x$
Solution: Let $I=\int \frac{1}{\sin x \cdot \cos ^{3} x} d x$
Re-writing,
$\begin{aligned} I &=\int \frac{\sin ^{2} x+\cos ^{2} x}{\sin x \cdot \cos ^{3} x} d x \quad\left[\therefore 1=\sin ^{2} x+\cos ^{2} x\right] \\ &=\int\left(\frac{\sin ^{2} x}{\sin x \cos ^{3} x}+\frac{\cos ^{2} x}{\sin x \cdot \cos ^{3} x}\right) d x \end{aligned}$
$\begin{aligned} &=\int\left(\frac{\sin x}{\cos x} \cdot \frac{1}{\cos ^{2} x}+\frac{1}{\sin x \cdot \cos x}\right) d x \\ &=\int\left(\tan x \cdot \sec ^{2} x+\frac{\cos ^{2} x}{\cos ^{2} x} \cdot \frac{1}{\sin x \cdot \cos x}\right) d x \end{aligned}$
$=\int\left(\tan x \cdot \sec ^{2} x+\frac{1}{\cos ^{2} x} \cdot \frac{1}{\frac{\sin x \cdot \cos x}{\cos ^{2} x}}\right) d x$
$=\int\left(\tan x \cdot \sec ^{2} x+\sec ^{2} x \cdot \frac{1}{\frac{\sin x}{\cos x}}\right) d x$
$\begin{aligned} I &=\int\left(\tan x \cdot \sec ^{2} x+\frac{\sec ^{2} x}{\tan x}\right) d x \\ & \Rightarrow I=\int \tan x \cdot \sec ^{2} x d x+\int \frac{\sec ^{2} x}{\tan x} d x \end{aligned}$
Substitute $\tan x=t \Rightarrow \sec ^{2} x d x=d t$ then
$\begin{aligned} I &=\int t d t+\int_{t}^{1} d t \\ &=\frac{t^{1+1}}{1+1}+\log |t|+C \quad \quad \quad \quad \quad \quad \quad \quad\left[\because \int x^{2} d x=\frac{x^{n+1}}{n+1}+c, \int \frac{1}{x} d x=1\right] \end{aligned}$
$\begin{aligned} &=\frac{t^{2}}{2}+\log |t|+C \\ &=\frac{\tan ^{2} x}{2}+\log |\tan x|+C \quad\quad\quad\quad\quad\quad\quad[\because t=\tan x] \end{aligned}$

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