#### Please Solve RD Sharma Class 12 Chapter 18 Indefinite Integrals Exercise18.31 Question 11 Maths Textbook Solution.

$I=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{\tan ^{2} x-1}{\sqrt{3} \tan x}\right)+c$

Hint:The given integral function can be converted into,

$\int \frac{d x}{x^{2}+a^{2}}=\frac{1}{a} \tan ^{-1} \frac{x}{a}+c$

Given:

$I=\int \frac{1}{\sin ^{4} x+\sin ^{2} x \cos ^{2} x+\cos ^{4} x}$

Solution

\begin{aligned} I &=\int \frac{1}{\sin ^{4} x+\sin ^{2} x \cos ^{2} x+\cos ^{4} x} \\ &=\int \frac{\left(\tan ^{2 } x+1\right) \sec ^{2} x}{\tan ^{4} x+\tan ^{2} x+1} d x \end{aligned} (dividing N and D by $\cos ^{4}x$)

$= \int \frac{t^{2}+1}{t^{4}+t^{2}+1}dt$   where $\tan x= t$

$=\int \frac{1+\frac{1}{t^{2}}}{t^{2}+\frac{1}{t^{2}}+1} d t$         (Dividing N and D by $t^{2}$)

putting\; t-\frac{1}{t}=z\; so\; that \left(1+\frac{1}{t^{2}}\right) d t=d \; z\; and \; t^{2}+\frac{1}{t^{2}}=z^{2}+2\\\\ \begin{aligned} I &=\int \frac{d z}{z^{2}+(\sqrt{3})^{2}} \\ &=\frac{1}{\sqrt{3}} \tan ^{-} \frac{z}{\sqrt{3}}+c \\ &=\frac{1}{\sqrt{3}} \tan ^{-}\left(\frac{t^{2}-1}{\sqrt{3} t}\right)+c \end{aligned}

The required value of the given integration is,

$I=\frac{1}{\sqrt{3}} \tan ^{-1}\left(\frac{\tan ^{2} x-1}{\sqrt{3} \tan x}\right)+c$

where c is an integrating constant.