Question

Evaluate the following integrals:

$\int \frac{1}{{\sin }^{4}x+{\sin }^{2}x{\cos }^{2}x+{\cos }^{4}x}\mathrm{d}x$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{1}{{\sin }^{4}x+{\sin }^{2}x{\cos }^{2}x+{\cos }^{4}x}\mathrm{d}x \\ =\int \frac{1}{{\left({\sin }^{2}x+{\cos }^{2}x\right)}^2-{\sin }^{2}x{\cos }^{2}x}\mathrm{d}x \\ =\int \frac{1}{1-{\sin }^{2}x{\cos }^{2}x}\mathrm{d}x \\ =\int \frac{{\displaystyle \frac{1}{{\cos }^{4}x}}}{{\displaystyle \frac{1}{{\cos }^{4}x}}-{\displaystyle \frac{{\sin }^{2}x}{{\cos }^{2}x}}}\mathrm{d}x \\ =\int \frac{{\displaystyle {\sec }^{2}x\left(1+{\tan }^{2}x\right)}}{{\displaystyle {\sec }^{4}x-{\tan }^{2}x}}\mathrm{d}x \\ \mathrm{Let}\tan x=t \\ \mathrm{On}\mathrm{differentiating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ {\sec }^{2}x\mathrm{d}x=\mathrm{d}t \\ \therefore I=\int \frac{1+t^2}{{\left(1+t^2\right)}^2-t^2}\mathrm{d}t \\ =\int \frac{1+t^2}{\left(t^4+t^2+1\right)}\mathrm{d}t \\ =\int \frac{{\displaystyle \frac{1}{t^2}}+1}{\left(t^2+1+{\displaystyle \frac{1}{t^2}}\right)}\mathrm{d}t \\ =\int \frac{{\displaystyle \frac{1}{t^2}+1}}{{\left({\displaystyle t-\frac{1}{t}}\right)}^2+3}\mathrm{d}t \\ \mathrm{Let}\left(t-\frac{1}{t}\right)=u \\ \mathrm{On}\mathrm{differentiating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \left(1+\frac{1}{t^2}\right)\mathrm{d}t=\mathrm{d}u \\ \therefore I=\int \frac{{\displaystyle 1}}{{\left({\displaystyle u}\right)}^2+3}\mathrm{d}u \\ =\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{u}{\sqrt{3}}\right)+c \\ =\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{t-{\displaystyle \frac{1}{t}}}{\sqrt{3}}\right)+c \\ =\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{\tan x-{\displaystyle \cot x}}{\sqrt{3}}\right)+c \\ \mathrm{Hence},\int \frac{1}{{\sin }^{4}x+{\sin }^{2}x{\cos }^{2}x+{\cos }^{4}x}\mathrm{d}x=\frac{1}{\sqrt{3}}{\tan }^{-1}\left(\frac{\tan x-{\displaystyle \cot x}}{\sqrt{3}}\right)+c \end{gathered}$$