RD Sharma Class 12 Exercise 18.16 Indefinite Integrals Solutions Maths

RD Sharma Class 12 Exercise 18.16 Indefinite Integrals Solutions Maths - Download PDF Free Online

Edited By Kuldeep Maurya | Updated on Jan 24, 2022 12:16 PM IST

Countless students in India have chosen the RD Sharma class 12th exercise 18.16 solution to be their holy grail when it comes to exam preparations. It cannot be argued that RD Sharma Solutions have made a name for themselves among other NCERT solutions. The answers provided in the RD Sharma class 12 chapter 18 exercise 18.16 deserves special mention because the chapter is extremely important for students in class 12.

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RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise
Indefinite Integrals Excercise:18.16
RD Sharma Chapter wise Solutions

The RD Sharma class 12 solutions Indefinite Integrals ex 18.16 is arguably the best NCERT solution in the RD Sharma solutions series which deals with chapter 18 of the NCERT textbook. Students will find in this chapter concepts of Integrals, Graphs of indefinite integrals, Indefinite integrals of common functions,etc. Exercise 18.16 has 16 questions which tests students on basic concepts of the chapter. The RD Sharma class 12th exercise 18.16 solutions will help them solve these problems and get accurate answers.

RD Sharma Class 12 Solutions Chapter18 Indefinite Integrals - Other Exercise

Indefinite Integrals Excercise:18.16

Indefinite Integrals exercise 18.16 question 1

Answer:
$\frac{1}{2}log\left | \frac{1+tan\, x}{1-tan\, x} \right |+C$
Hint:
Use substitution method as well as special integration formula to solve this type of problem
Given:
$\int \frac{\sec ^{2} x}{1-\tan ^{2} x} d x$
Solution:
$Let\: \: \int \frac{\sec ^{2} x}{1-\tan ^{2} x} d x$
$Put\: \: tan\: x=t\Rightarrow sec^{2}x\: dx=dt$
$Then \: \: \: I=\int \frac{1}{1-t^{2}} d t=\int \frac{1}{1^{2}-t^{2}} d t$
$\begin{array}{ll} =\frac{1}{2 \times 1} \log \left|\frac{1+t}{1-t}\right|+C \quad & {\left[\because \int \frac{1}{a^{2}-x^{2}} d x=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+C\right]} \\ \\ =\frac{1}{2} \log \left|\frac{1+\tan x}{1-\tan x}\right|+C & {[\because t=\tan x]} \end{array}$

Indefinite Integrals exercise 18.16 question 2

Answer:
$tan^{-1}(e^{x})+C$
Hint:
Use substitution method as well as special integration formula to solve this type of problem
Given:
$\int \frac{e^{x}}{1+e^{2x}}dx$
Solution:
$Let\: \: I=\int \frac{e^{x}}{1+e^{2x}}dx$
$Put\: \: e^{x}=t\Rightarrow e^{x}dx=dt$
$Then\: \: I= \int \frac{1}{1+t^{2}}dx$
$\begin{array}{ll} =\tan ^{-1}\left(\frac{t}{1}\right)+C & {\left[\because \int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C\right]} \\ \\ =\tan ^{-1}\left(e^{x}\right)+C \quad & {\left[\because t=e^{x}\right]} \end{array}$

Indefinite Integrals exercise 18.16 question 3

Answer:
$tan^{-1}(sin\: x+2)+C$
Hint:
Use substitution method as well as special integration formula to solve this type of problem
Given:
$\int \frac{\cos x}{\sin ^{2} x+4 \sin x+5} d x$
Solution:
$Let\: \: I=\int \frac{\cos x}{\sin ^{2} x+4 \sin x+5} d x$
$Put\: \: sin\: x=t\Rightarrow cos\: x\: dx=dt$
$Then\: \: I=\int \frac{1}{t ^{2}+4 t+5} d x$
$\begin{aligned} &=\int \frac{1}{t^{2}+2 \cdot 2 \cdot t+2^{2}-2^{2}+5} d t \\ &=\int \frac{1}{(t+2)^{2}-4+5} d t \\ &=\int \frac{1}{(t+2)^{2}+1} d t \end{aligned}$
$Put\: \: t+2=u\Rightarrow dt=du$
$Then\: \: I=\int \frac{1}{u ^{2}+1} d u$
$\begin{array}{ll} =\tan ^{-1}\left(\frac{u}{1}\right)+C \quad & {\left[\because \int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C\right]} \\ \\=\tan ^{-1}\left(\frac{t+2}{1}\right)+C \quad & {[\because u=t+2]} \\ \\=\tan ^{-1}(\sin x+2)+C & {[\because t=\sin x]} \end{array}$

Indefinite Integrals exercise 18.16 question 4

Answer:
$log\left | \frac{e^{x}+2}{e^{x}+3} \right |+C$
Hint:
Use substitution method as well as special integration formula to solve this type of problem
Given:
$\int \frac{e^{x}}{e^{2 x}+5 e^{x}+6} d x$
Solution:
$Let\: \: I=\int \frac{e^{x}}{e^{2 x}+5 e^{x}+6} d x$
$Put\: \: e^{x}=t\Rightarrow e^{x}dx=dt$
$Then\: \: I=\int \frac{1}{t^{2}+5 t+6} d t$
$\begin{aligned} &=\int \frac{1}{t^{2}+2 \cdot t \cdot \frac{5}{2}+\left(\frac{5}{2}\right)^{2}-\left(\frac{5}{2}\right)^{2}+6} d t \\ \\ &=\int \frac{1}{\left(t+\frac{5}{2}\right)^{2}-\frac{25}{4}+6} d t=\int \frac{1}{\left(t+\frac{5}{2}\right)^{2}-\frac{25+24}{4}} d t \end{aligned}$
$\begin{aligned} &=\int \frac{1}{\left(t+\frac{5}{2}\right)^{2}-\frac{1}{4}} d t \\ &=\int \frac{1}{\left(t+\frac{5}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}} d t \end{aligned}$
$Put\: \: t+\frac{5}{2}=u\Rightarrow dt=du$
$\begin{aligned} =\int \frac{1}{u^{2}-\left(\frac{1}{2}\right)^{2}} d u \end{aligned}$
$\begin{aligned} &=\frac{1}{2 \times \frac{1}{2}} \log \left|\frac{u-\frac{1}{2}}{u+\frac{1}{2}}\right|+C \quad\left[\because \int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C\right] \\ \\ &=\log \mid \frac{t+\frac{5}{2}-\frac{1}{2}}{t+\frac{5}{2}-\frac{1}{2}} \mid+C \end{aligned}$
$=\log \left|\frac{\frac{2 t+5-1}{2}}{\frac{2 t+5+1}{2}}\right|+C \quad\left[\because u=t+\frac{5}{2}\right]$
. $\begin{aligned} &I=\log \left|\frac{2 t+4}{2 t+6}\right|+C=\log \left|\frac{t+2}{t+3}\right|+C\\ &I=\log \left|\frac{e^{x}+2}{e^{x}+3}\right|+C \end{aligned}$ $[\therefore t=e^{x}]$

Indefinite Integrals exercise 18.16 question 5

Answer:
$\frac{1}{36} \log \left|\frac{2 e^{3 x}-3}{2 e^{3 x}+3}\right|+C$
Hint:
Use substitution method as well as special integration formula to solve this type of problem
Given:
$\int \frac{e^{3 x}}{4 e^{6 x}-9} d x$
Solution:
$Let\: \: \int \frac{e^{3 x}}{4 e^{6 x}-9} d x$
$Put \: \: \: e^{3x}=t\Rightarrow 3e^{3x}dx=dt\Rightarrow e^{3x}dx=\frac{dt}{3}$
$Then\: \: \: I=\int \frac{1}{4 t^{2}-9} \frac{d t}{3}=\int \frac{1}{4\left(t^{2}-\frac{9}{4}\right)} \frac{d t}{3}$
$=\frac{1}{4 \times 3} \int \frac{1}{t^{2}-\frac{9}{4}} d t=\frac{1}{4 \times 3} \int \frac{1}{t^{2}-\left(\frac{3}{2}\right)^{2}} d t$
$=\frac{1}{4 \times 3}\left[\frac{1}{2 \times \frac{3}{2}} \log \left|\frac{t-\frac{3}{2}}{t+\frac{3}{2}}\right|\right]+C \quad\left[\because \int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C\right]$
$\begin{aligned} &=\frac{1}{4 \times 3 \times 3} \log \left|\frac{\frac{2 t-3}{2}}{\frac{2 t+3}{2}}\right|+C \\ \\ &=\frac{1}{36} \log \left|\frac{2 t-3}{2 t+3}\right|+C \\\\ &=\frac{1}{36} \log \left|\frac{2 e^{3 x}-3}{2 e^{3 x}+3}\right|+C \quad\left[\because t=e^{3 x}\right] \end{aligned}$

Indefinite Integrals exercise 18.16 question 6

Answer:
$tan^{-1}(e^{x})+C$
Hint:
Use substitution method as well as special integration formula to solve this type of problem
Given:
$\int \frac{1}{e^{x}+e^{-x}}dx$
Solution:
$Let\: \:I= \int \frac{1}{e^{x}+e^{-x}}dx=\int \frac{1}{e^{x}+\frac{1}{e^{x}}}dx$
$\begin{aligned} &=\int \frac{1}{\frac{e^{2 x}+1}{e^{x}}} d x \\ &=\int \frac{e^{x}}{e^{2 x}+1} d x \end{aligned}$
$Put\: \: e^{x}=t\Rightarrow e^{x}dx=dt$
$Then\: \: \begin{aligned} I=\int \frac{1}{t^{2}+1} d t \end{aligned}$
$\begin{array}{ll} =\tan ^{-1}(t)+C & {\left[\because \int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C\right]} \\\\ =\tan ^{-1}\left(e^{x}\right)+C & {\left[\because e^{x}=t\right]} \end{array}$

Indefinite Integrals exercise 18.16 question 7

Answer:
$\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{x^{2}+1}{\sqrt{2}}\right)+C$
Hint:
Use substitution method as well as special integration formula to solve this type of problem
Given:
$\int \frac{x}{x^{4}+2x^{2}+3}dx$
Solution:
$Let\: \: \: I=\int \frac{x}{x^{4}+2x^{2}+3}dx$
$Put\: \: x^{2}=t\Rightarrow 2x\, dx=dt\Rightarrow x\: dx=\frac{dt}{2}$
$Then\: \: \: I=\int \frac{1}{t^{2}+2t+3}\frac{dt}{2}$
$\begin{aligned} &=\frac{1}{2} \int \frac{1}{t^{2}+2 t+3} d t \\ &=\frac{1}{2} \int \frac{1}{t^{2}+2 \cdot t \cdot 1+1^{2}-1^{2}+3} d t \\ &=\frac{1}{2} \int \frac{1}{(t+1)^{2}-1+3} d t \\ &=\frac{1}{2} \int \frac{1}{(t+1)^{2}+2} d t \\ &=\frac{1}{2} \int \frac{1}{(t+1)^{2}+(\sqrt{2})^{2}} d t \end{aligned}$
$Put\: \: u=t+1\Rightarrow du=dt$
$Then\: \: \begin{aligned} I=\frac{1}{2} \int \frac{1}{u^{2}+(\sqrt{2})^{2}} d u \end{aligned}$
$\begin{array}{ll} =\frac{1}{2} \times \frac{1}{\sqrt{2}} \tan ^{-1}\left(\frac{u}{\sqrt{2}}\right)+C & {\left[\because \int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C\right]} \\ \\=\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{t+1}{\sqrt{2}}\right)+C \quad & {[\because u=t+1]} \\ \\=\frac{1}{2 \sqrt{2}} \tan ^{-1}\left(\frac{x^{2}+1}{\sqrt{2}}\right)+C & {\left[\because t=x^{2}\right]} \end{array}$

Indefinite Integrals exercise 18.16 question 8

Answer:
$\frac{1}{2}tan^{-1}(x^{6})+C$
Hint:
Use substitution method as well as special integration formula to solve this type of problem
Given:
$\int \frac{3x^{5}}{1+x^{12}}dx$
Solution:
$Let\: \: I=\int \frac{3x^{5}}{1+x^{12}}dx$
$=\int \frac{3x^{5}}{1+(x^{6})^{2}}dx$
$Put\: \: x^{6}=t\Rightarrow 6x^{5}dx=dt$
$Then\: \: I=\int \frac{3x^{5}}{1+t^{2}}\: \frac{dt}{6x^{5}}$
$\begin{aligned} &=\frac{1}{2} \int \frac{1}{1+t^{2}} d t \\ &=\frac{1}{2} \int \frac{1}{1^{2}+t^{2}} d t \\ &=\frac{1}{2} \cdot \frac{1}{1} \tan ^{-1}\left(\frac{t}{1}\right)+C \quad\left[\because \int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C\right] \\ &=\frac{1}{2} \tan ^{-1}(t)+C \\ &=\frac{1}{2} \tan ^{-1}\left(x^{6}\right)+C \quad\left[\because t=x^{6}\right] \end{aligned}$

Indefinite Integrals exercise 18.16 question 9

Answer:
$\frac{1}{6 a^{3}} \log \left|\frac{x^{3}-a^{3}}{x^{3}+a^{3}}\right|+C$
Hint:
Use substitution method as well as special integration formula to solve this type of problem
Given:
$\int \frac{x^{2}}{x^{6}-a^{6}}dx$
Solution:
$Let\: \: I=\int \frac{x^{2}}{x^{6}-a^{6}}dx$
$=\int \frac{x^{2}}{(x^{3})^{2}-(a^{3})^{2}}dx$
$Put\: \: x^{3}=t\Rightarrow 3x^{3}dx=dt\Rightarrow x^{2}dx=\frac{dt}{3}$
$Then\: \: I=\int \frac{1}{t^{2}-(a^{3})^{2}}\: \: \frac{dt}{2}$
$\begin{aligned} &=\frac{1}{3} \int \frac{1}{t^{2}-\left(a^{3}\right)^{2}} d t \\ &=\frac{1}{3} \times \frac{1}{2\left(a^{3}\right)} \log \left|\frac{t-a^{3}}{t+a^{3}}\right|+C \quad\left[\because \int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C\right] \\ &=\frac{1}{6 a^{3}} \log \left|\frac{x^{3}-a^{3}}{x^{3}+a^{3}}\right|+C \quad\left[\because t=x^{3}\right] \end{aligned}$

Indefinite Integrals exercise 18.16 question 10

Answer:
$\frac{1}{3 a^{3}} \tan ^{-1}\left(\frac{x^{3}}{a^{3}}\right)+C$
Hint:
Use substitution method as well as special integration formula to solve this type of problem
Given:
$\int \frac{x^{2}}{x^{6}+a^{6}}dx$
Solution:
$Let\: \: \: I=\int \frac{x^{2}}{x^{6}+a^{6}}dx$
$=\int \frac{x^{2}}{(x^{3})^{2}+(a^{3})^{2}}dx$
$Put \: \: x^{3}=t\Rightarrow 3x^{2}dx=dt\Rightarrow x^{2}dx=\frac{dt}{3}$
$Then\: \: I=\int \frac{1}{t^{2}+(a^{3})^{2}}dt$
$\begin{aligned} &=\frac{1}{3} \int \frac{1}{t^{2}+\left(a^{3}\right)^{2}} d t \\ &=\frac{1}{3} \times \frac{1}{a^{3}} \tan ^{-1}\left(\frac{t}{a^{3}}\right)+C \quad\left[\because \int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C\right] \\ &=\frac{1}{3 a^{3}} \tan ^{-1}\left(\frac{x^{3}}{a^{3}}\right)+C \quad\left[\because t=x^{3}\right] \end{aligned}$

Indefinite Integrals exercise 18.16 question 11

Answer:
$\frac{1}{6} \log \left|\frac{x^{6}}{x^{6}+1}\right|+C$
Hint:
Use substitution method as well as special integration formula to solve this type of problem
Given:
$\int \frac{1}{x\left(x^{6}+1\right)} d x$
Solution:
$Let\: \: I=\int \frac{1}{x\left(x^{6}+1\right)} d x$
$\begin{aligned} &=\int \frac{x^{5}}{x^{5}\left\{x\left(x^{6}+1\right)\right\}} d x \\ &=\int \frac{x^{5}}{x^{6}\left(x^{6}+1\right)} d x \end{aligned}$
$Put\: \: x^{6}=t\Rightarrow 6x^{5}dx=dt\Rightarrow x^{5}dx=\frac{dt}{6}$
$Then\: \: I=\int \frac{1}{t\left(t+1\right)} \frac{dt}{6}$
$\begin{aligned} &=\frac{1}{6} \int \frac{1}{t^{2}+t} d t \\ &=\frac{1}{6} \int \frac{1}{t^{2}+2 \cdot t \cdot \frac{1}{2}+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}} d t \\ &=\frac{1}{6} \int \frac{1}{\left(t+\frac{1}{2}\right)^{2}-\frac{1}{4}} d t \end{aligned}$
$Put\: \: t+\frac{1}{2}=u\Rightarrow dt=du$
$\begin{aligned} =\frac{1}{6} \int \frac{1}{u^{2}-(\frac{1}{2})^{2}} d t \end{aligned}$ $\begin{aligned} =\frac{1}{6} \int \frac{1}{u^{2}-(\frac{1}{2})^{2}} d u \end{aligned}$
$\begin{aligned} &=\frac{1}{6} \cdot \frac{1}{2 \times\left(\frac{1}{2}\right)} \log \left|\frac{u-\frac{1}{2}}{u+\frac{1}{2}}\right|+C \quad\left[\because \int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C\right] \\ &=\frac{1}{6} \log \left|\frac{\frac{2u-1}{2}}{\frac{2u+1}{2}}\right|+C \\ &=\frac{1}{6} \log \left|\frac{2 u-1}{2 u+1}\right|+C \end{aligned}$
$\begin{aligned} &=\frac{1}{6} \log \left|\frac{2\left(t+\frac{1}{2}\right)-1}{2\left(t+\frac{1}{2}\right)+1}\right|+C \quad\left[\because t+\frac{1}{2}=u\right] \\ &=\frac{1}{6} \log \left|\frac{2 t+1-1}{2 t+1+1}\right|+C \\ &=\frac{1}{6} \log \left|\frac{2 t}{2 t+2}\right|+C \\ &=\frac{1}{6} \log \left|\frac{t}{t+1}\right|+C \\ &=\frac{1}{6} \log \left|\frac{x^{6}}{x^{6}+1}\right|+C \quad\left[\because t=x^{6}\right] \end{aligned}$

Indefinite Integrals exercise 18.16 question 12

Answer:
$\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x^{2}-1}{\sqrt{3}}\right)+C$
Hint:
Use substitution method as well as special integration formula to solve this type of problem
Given:
$\int \frac{x}{x^{4}-x^{2}+1}dx$
Solution:
$Let\: \: I=\int \frac{x}{x^{4}-x^{2}+1}dx$
$=\int \frac{x}{(x^{2})^{2}-x^{2}+1}dx$
$Put\: \: x^{2}=t\Rightarrow 2x\, \, dx=dt\Rightarrow x\, dx=\frac{dt}{2}$
$=\int \frac{1}{t^{2}-t+1}\, \frac{dt}{2}$
$\begin{aligned} &=\frac{1}{2} \int \frac{1}{t^{2}-2 \cdot t \cdot \frac{1}{2}+\left(\frac{1}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}+1} d t \\ &=\frac{1}{2} \int \frac{1}{\left(t-\frac{1}{2}\right)^{2}-\frac{1}{4}+1} d t \\ &=\frac{1}{2} \int \frac{1}{\left(t-\frac{1}{2}\right)^{2}-\left(\frac{1-4}{4}\right)} d t \\ &=\frac{1}{2} \int \frac{1}{\left(t-\frac{1}{2}\right)^{2}-\left(\frac{-3}{4}\right)} d t \\ &=\frac{1}{2} \int \frac{1}{\left(t-\frac{1}{2}\right)^{2}+\frac{3}{4}} d t \end{aligned}$
$Let\: \: t-\frac{1}{2}=u\Rightarrow dt=du$
$Then \: \: I=\frac{1}{2} \int \frac{1}{u^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} d u$
$\begin{aligned} &=\frac{1}{2} \times \frac{1}{\left(\frac{\sqrt{3}}{2}\right)} \tan ^{-1}\left(\frac{u}{\frac{\sqrt{3}}{2}}\right)+C \quad\left[\because \int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C\right] \\ &=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{t-\frac{1}{2}}{\frac{\sqrt{3}}{2}}\right)+C \quad\left[\because u=t-\frac{1}{2}\right] \\ &=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{\frac{2 t-1}{2}}{\frac{\sqrt{3}}{2}}\right)+C \\ &=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 t-1}{\sqrt{3}}\right)+C \\ &=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{2 x^{2}-1}{\sqrt{3}}\right)+C \quad\left[\because t=x^{2}\right] \end{aligned}$

Indefinite Integrals exercise 18.16 question 13

Answer:
$\frac{\sqrt{3}}{48} \log \left|\frac{x^{2}-3-\frac{4}{\sqrt{3}}}{x^{2}-3+\frac{4}{\sqrt{3}}}\right|+C$
Hint:
Use substitution method as well as special integration formula to solve this type of problem
Given:
$\int \frac{x}{3 x^{4}-18 x^{2}+11} d x$
Solution:
$Let\: \: I=\int \frac{x}{3 x^{4}-18 x^{2}+11} d x$
$=\int \frac{x}{3 (x^{2})^{2}-18 x^{2}+11} d x$
$Put\: \: x^{2}=t\Rightarrow 2x\, dx=dt\Rightarrow x\, dx=\frac{dt}{2}$
$Then\: \: I=\int \frac{1}{3t^{2}-18 t+11} \, \frac{dt}{2}$
$\begin{aligned} &=\frac{1}{2} \int \frac{1}{3\left(t^{2}-6 t+\frac{11}{3}\right)} d t \\ &=\frac{1}{6} \int \frac{1}{t^{2}-2.3 . t+(3)^{2}-(3)^{2}+\frac{11}{3}} d t \\ &=\frac{1}{6} \int \frac{1}{(t-3)^{2}-9+\frac{11}{3}} d t \\ &=\frac{1}{6} \int \frac{1}{(t-3)^{2}-\left(\frac{27-11}{3}\right)} d t \\ &=\frac{1}{6} \int \frac{1}{(t-3)^{2}-\frac{16}{3}} d t \\ &=\frac{1}{6} \int \frac{1}{(t-3)^{2}-\left(\frac{4}{\sqrt{3}}\right)^{2}} d t \end{aligned}$
$Put\: \: t-3=u\Rightarrow dt=du$
$Then\: \: I =\frac{1}{6} \int \frac{1}{u^{2}-\left(\frac{4}{\sqrt{3}}\right)^{2}} du$
$\begin{aligned} &=\frac{1}{6} \cdot \frac{1}{2 \cdot \frac{4}{\sqrt{3}}} \log \left|\frac{u-\frac{4}{\sqrt{3}}}{u+\frac{4}{\sqrt{3}}}\right|+C \quad\left[\because \int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C\right] \\ &=\frac{\sqrt{3}}{6 \times 8} \log \left|\frac{u-\frac{4}{\sqrt{3}}}{u+\frac{4}{\sqrt{3}}}\right|+C \\ &=\frac{\sqrt{3}}{48} \log \left|\frac{t-3-\frac{4}{\sqrt{3}}}{t-3+\frac{4}{\sqrt{3}}}\right|+C \quad[\because u=t-3] \end{aligned}$
$=\frac{\sqrt{3}}{48} \log \left|\frac{x^{2}-3-\frac{4}{\sqrt{3}}}{x^{2}-3+\frac{4}{\sqrt{3}}}\right|+C\: \: \: \: \: \: \: [\because t=x^{2}]$

Indefinite Integrals exercise 18.16 question 14

Answer:
$\log \left|\frac{e^{x}+1}{e^{x}+2}\right|+C$
Hint:
Use substitution method as well as special integration formula to solve this type of problem
GIven:
$\int \frac{e^{x}}{\left(1+e^{x}\right)\left(2+e^{x}\right)} d x$
Solution:
$Let\: \: I=\int \frac{e^{x}}{\left(1+e^{x}\right)\left(2+e^{x}\right)} d x$
$Let\: \: e^{x}=t\Rightarrow e^{x}dx=dt$
$Then\: \: I=\int \frac{1}{\left(1+t\right)\left(2+t\right)} d t$
$\begin{aligned} &=\int \frac{1}{t^{2}+2 t+t+2} d t \\ &=\int \frac{1}{t^{2}+3 t+2} d t \\ &=\int \frac{1}{t^{2}+2 \cdot t \cdot \frac{3}{2}+\left(\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}+2} d t \\ &=\int \frac{1}{\left(t+\frac{3}{2}\right)^{2}-\frac{9}{4}+2} d t \\ &=\int \frac{1}{\left(t+\frac{3}{2}\right)^{2}-\left(\frac{9-8}{4}\right)} d t \\ &=\int \frac{1}{\left(t+\frac{3}{2}\right)^{2}-\frac{1}{4}} d t \\ &=\int \frac{1}{\left(t+\frac{3}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}} d t \end{aligned}$
$Put\: \: t+\frac{3}{2}=u\Rightarrow dt=du$
$Then\: \: I=\int \frac{1}{u^{2}-\left(\frac{1}{2}\right)^{2}} d u$
$\begin{array}{ll} =\frac{1}{2 \times \frac{1}{2}} \log \left|\frac{u-\frac{1}{2}}{u+\frac{1}{2}}\right|+C \quad\left[\because \int \frac{1}{x^{2}-a^{2}} d x=\frac{1}{2 a} \log \left|\frac{x-a}{x+a}\right|+C\right] \\ \\=\log \left|\frac{t+\frac{3}{2}-\frac{1}{2}}{t+\frac{3}{2}+\frac{1}{2}}\right|+C \quad\left[\because t+\frac{3}{2}=u\right] \\ \\=\log \left|\frac{2 t+2}{2 t+4}\right|+C \\ \\=\log \left|\frac{t+1}{t+2}\right|+C \\ \end{array}$
$=\log \left|\frac{e^{x}+1}{e^{x}+2}\right|+C\: \: \: \: \: \: \: [\because t=e^{x}]$

Indefinite Integrals exercise 18.16 question 15

Answer:
$\frac{1}{6} \log \left|\frac{1+\tan ^{3} x}{1-\tan ^{3} x}\right|+C$
Hint:
Use substitution method as well as special integration formula to solve this type of problem
Given:
$\int \frac{\tan ^{2} x \sec ^{2} x}{1-\tan ^{6} x} d x$
Solution:
$Let\: \: I=\int \frac{\tan ^{2} x \sec ^{2} x}{1-\tan ^{6} x} d x$
$=\int \frac{\tan ^{2} x \sec ^{2} x}{1-(\tan ^{3} x)^{2}} d x$
$\begin{aligned} &\text { Put } \tan ^{3} x=t \Rightarrow 3 \tan ^{2} x \cdot \sec ^{2} x d x=d t \\ &\text { Then } I=\int \frac{1}{1-t^{2}} \cdot \frac{d t}{3} \\ &=\frac{1}{3} \int \frac{1}{1^{2}-t^{2}} d t \\ &=\frac{1}{3} \cdot \frac{1}{2 \times 1} \log \left|\frac{1+t}{1-t}\right|+C \quad\left[\because \int \frac{1}{a^{2}-x^{2}} d x=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+C\right] \\ &=\frac{1}{6} \log \left|\frac{1+\tan ^{3} x}{1-\tan ^{3} x}\right|+C \quad\left[\because t=\tan ^{3} x\right] \end{aligned}$

Indefinite Integrals exercise 18.16 question 16

Answer:
$\frac{1}{2 \sqrt{3}} \log \left|\frac{\sqrt{3}+\sin x-\cos x}{\sqrt{3}-\sin x+\cos x}\right|-\tan ^{-1}(\sin x+\cos x)+C$
Hint:
Use substitution method as well as special integration formula to solve this type of problem
Given:
$\int \frac{1}{\cos x+\operatorname{cosec} x} d x$
Solution:
$Let\: \: I=\int \frac{1}{\cos x+\operatorname{cosec} x} d x$
$\begin{aligned} &=\frac{1}{\cos x+\frac{1}{\sin x}} d x \quad\left[\because \operatorname{cosec} x=\frac{1}{\sin x}\right] \\ &=\int \frac{1}{\frac{\cos x \cdot \sin x+1}{\sin x}} d x=\int \frac{\sin x}{\cos x \cdot \sin x+1} d x \\ &=\int \frac{2 \sin x}{2(\cos x \cdot \sin x+1)} d x=\int \frac{\sin x+\sin x}{2+2 \sin x \cdot \cos x} d x \\ &=\int \frac{\sin x+\cos x-\cos x+\sin x}{2+2 \sin x \cdot \cos x} d x \\ &=\int\left\{\frac{\sin x+\cos x}{2+2 \sin x \cdot \cos x}+\frac{\sin x-\cos x}{2+2 \sin x \cdot \cos x}\right\} d x \end{aligned}$
$\begin{aligned} &=\int \frac{\sin x+\cos x}{2+2 \sin x \cdot \cos x} d x+\int \frac{\sin x-\cos x}{2+2 \sin x \cdot \cos x} d x \\ &=\int \frac{\sin x+\cos x}{3-1+2 \sin x \cdot \cos x} d x+\int \frac{\sin x-\cos x}{1+1+2 \sin x \cdot \cos x} d x \\ &=\int \frac{\sin x+\cos x}{3-\left(\sin ^{2} x+\cos ^{2} x\right)+2 \sin x \cdot \cos x} d x+\int \frac{\sin x-\cos x}{1+\left(\sin x+\cos ^{2} x\right)+2 \sin x \cdot \cos x} d x \\ &{\left[\because 1=\sin ^{2} x+\cos ^{2} x\right]} \\ &=\int \frac{\sin x+\cos x}{3-\sin ^{2} x-\cos ^{2} x+2 \sin x \cdot \cos x} d x+\int \frac{\sin x-\cos x}{1+\left(\sin ^{2} x+\cos ^{2} x+2 \sin x \cdot \cos x\right)} d x \end{aligned}$
$\begin{aligned} &=\int \frac{\sin x+\cos x}{3-\left(\sin ^{2} x+\cos ^{2} x-2 \sin x \cdot \cos x\right)} d x+\int \frac{\sin x-\cos x}{1+(\sin x+\cos x)^{2}} d x \\ &=\int \frac{\sin x+\cos x}{3-(\sin x-\cos x)^{2}} d x+\int \frac{\sin x-\cos x}{1+(\sin x+\cos x)^{2}} d x \\ &\text { Let } I=I_{1}+I_{2}\: \: \: \: \: \: ...(i) \\ &\Rightarrow \int \frac{1}{\cos x+\operatorname{cosec} x} d x=\int \frac{\sin x+\cos x}{3-(\sin x-\cos x)^{2}} d x+\int \frac{\sin x-\cos x}{1+(\sin x+\cos x)^{2}} d x \\ &\text { Where } I_{1}=\int \frac{\sin x+\cos x}{3-(\sin x-\cos x)^{2}} d x \end{aligned}$
$\text { And } I_{2}=\int \frac{\sin x-\cos x}{1+(\sin x+\cos x)^{2}} d x$
Here first we find the integral of I₁ and I₂ then we put the value of I₁ and I₂ in I and then we get the required solution
$\begin{aligned} &\text { So } I_{1}=\int \frac{\sin x+\cos x}{3-(\sin x-\cos x)^{2}} d x \\ &\text { Put } t=\sin x-\cos x \Rightarrow d t=[\cos x-(-\sin x)] d x \Rightarrow d t=[\cos x+\sin x] d x \\ &\text { Then } I_{1}=\int \frac{1}{3-t^{2}} d t=\int \frac{1}{(\sqrt{3})^{2}-t^{2}} d t \\ &=\frac{1}{2 \sqrt{3}} \log \left|\frac{\sqrt{3}+t}{\sqrt{3}-t}\right|+C_{1} \quad\left[\because \int \frac{1}{a^{2}-x^{2}} d x=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+C\right] \\ &=\frac{1}{2 \sqrt{3}} \log \left|\frac{\sqrt{3}+\sin x-\cos x}{\sqrt{3}-(\sin x-\cos x)}\right|+C_{1} \quad[\because t=\sin x-\cos x] \\ &=\frac{1}{2 \sqrt{3}} \log \left|\frac{\sqrt{3}+\sin x-\cos x}{\sqrt{3}-\sin x+\cos x}\right|+C_{1}\: \: \: \: ......(ii) \\ &\text { And } I_{2}=\int \frac{\sin x-\cos x}{1+(\sin x+\cos x)^{2}} d x \end{aligned}$ $\begin{aligned} &\text { Put } u=\sin x+\cos x \Rightarrow d u=(\cos x-\sin x) d x \Rightarrow d u=-(\sin x-\cos x) d x\\ &\Rightarrow(\sin x-\cos x) d x=-d u\\ &\therefore I_{2}=\int \frac{1}{1+u^{2}}(-d u)=-\int \frac{1}{1^{2}+u^{2}} d u\\ &=-\tan ^{-1}\left(\frac{u}{1}\right)+C_{2} \quad\left[\because \int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C\right]\\ &=-\tan ^{-1}(\sin x+\cos x)+C_{2}\: \: \: \: .....\text { (iii) }[\because u=\sin x+\cos x]\\ &\text { Putting the value of (ii) and (iii) in (i) we get }\\ &I=I_{1}+I_{2}\\ &=\frac{1}{2 \sqrt{3}} \log \left|\frac{\sqrt{3}+\sin x-\cos x}{\sqrt{3}-\sin x+\cos x}\right|+C_{1}+\left\{-\tan ^{-1}(\sin x+\cos x)+C_{2}\right\}\\ &=\frac{1}{2 \sqrt{3}} \log \left|\frac{\sqrt{3}+\sin x-\cos x}{\sqrt{3}-\sin x+\cos x}\right|-\tan ^{-1}(\sin x+\cos x)+C \quad\left[\because C=C_{1}+C_{2}\right] \end{aligned}$

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