Question

If $I=\int \frac{dx}{{\cos }^{4}x-{\cos }^{2}x{\sin }^{2}x+{\sin }^{4}x}$, then I equals

• A. $2{\tan }^{-1}\left(\frac{\tan 2x}{2}\right)+C$
• B. ${\tan }^{-1}\left(\frac{\tan 2x}{\sqrt{2}}\right)+C$
• C. ${\tan }^{-1}\left(\frac{\tan 2x}{2}\right)+C$
• D. $\sqrt{2}{\tan }^{-1}\left(\frac{\tan 2x}{\sqrt{2}}\right)+C$

Answer 1

The correct option is C ${\tan }^{-1}\left(\frac{\tan 2x}{2}\right)+C$

$I=\int \frac{dx}{{\cos }^{4}x-{\cos }^{2}x{\sin }^{2}x+{\sin }^{4}x}I=\int \frac{dx}{{\cos }^{4}x(1-{\tan }^{2}x+{\tan }^{4}x)}let\tan x=t\therefore {\sec }^{2}xdx=dtI=\int \frac{\left(\frac{dt}{{\sec }^{2}x}\right)}{{\cos }^{4}x(1-t^2+t^4)}I=\int \frac{{\sec }^{2}xdt}{(1-t^2+t^4)}=\int \frac{(1+t^2)dt}{(1-t^2+t^4)}$

We'll now divide the numerator and denominator by $t^2$

$I=\int \frac{(1+\frac{1}{t^2})}{(t^2+\frac{1}{t^2}-1)}dtI=\int \frac{(1+\frac{1}{t^2})}{(t^2+\frac{1}{t^2}-2+1)}dtI=\int \frac{(1+\frac{1}{t^2})}{{(t-\frac{1}{t})}^2+1}dtletz=(t-\frac{1}{t})\therefore dz=(1+\frac{1}{t^2})dt\therefore I=\int \frac{dz}{z^2+1}={\tan }^{-1}z+Cz=t-\frac{1}{t}=\tan x-\frac{1}{\tan x}z=\frac{{\tan }^{2}x-1}{\tan x}=-2{\left(\frac{2\tan x}{1-{\tan }^{2}x}\right)}^{-1}z=\frac{-2}{\tan 2x}{\tan }^{-1}z={\tan }^{-1}\left[\frac{-2}{\tan 2x}\right]=-{\tan }^{-1}\left[\frac{2}{\tan 2x}\right]={\tan }^{-1}\left[\frac{\tan 2x}{2}\right]\because -{\tan }^{-1}z={\tan }^{-1}\left(\frac{1}{z}\right)\therefore I={\tan }^{-1}\left[\frac{\tan 2x}{2}\right]+C$