Question

Evaluate: $\int \frac{1}{1-{\tan }^{2}x}dx$

• A. $\frac{x}{2}-\frac{1}{4}\log |\tan (\frac{\pi }{4}+x)|+c$
• B. $\frac{x}{2}+\frac{1}{4}\log |\tan (\frac{\pi }{4}+x)|+c$
• C. $\frac{x}{2}+\frac{1}{4}\log |\tan (\frac{\pi }{4}+\frac{x}{2})|+c$
• D. $\frac{x}{2}-\frac{1}{4}\log |\tan (\frac{\pi }{4}+\frac{x}{2})|+c$

Answer 1

The correct option is B $\frac{x}{2}+\frac{1}{4}\log |\tan (\frac{\pi }{4}+x)|+c$

Substitute $\tan \left(x\right)=$ $t$,

$\therefore {\sec }^{2}xdx=dt$

${\sec }^{2}x=1+{\tan }^{2}x=1+t^2$

$\therefore I=\int \frac{1}{1-{\tan }^{2}x}dx=\int \frac{1}{1-t^2}\frac{dt}{1+t^2}$

Now we need to split the denominator by Using Partial Fractions;

$\frac{1}{(1-t^2)(1+t^2)}=\frac{1}{2}\frac{(1-t^2)+(1+t^2)}{(1-t^2)(1+t^2)}=\frac{1}{2}[\frac{1}{1+t^2}+\frac{1}{1-t^2}]\frac{1}{1-t^2}=\frac{1}{2}[\frac{1}{1+t}+\frac{1}{1-t}]$

So we have,

$\frac{1}{(1-t^2)(1+t^2)}=\frac{1}{2}[\frac{1}{1+t^2}+\frac{1}{2}[\frac{1}{1+t}+\frac{1}{1-t}]]\therefore \int \frac{1}{(1-t^2)(1+t^2)}dt=\int \frac{1}{2}\frac{1}{1+t^2}+\frac{1}{4}\frac{1}{1+t}+\frac{1}{4}\frac{1}{1-t}dt\Rightarrow I=\frac{1}{2}\int \frac{1}{1+t^2}dt+\frac{1}{4}\int \frac{1}{1+t}dt+\frac{1}{4}\int \frac{1}{1-t}dt$

$I=\frac{1}{2}{\tan }^{-1}t+\frac{1}{4}\log (1+t)-\frac{1}{4}\log (1-t)+C$

put $t=\tan \left(x\right)$

$I=\frac{x}{2}+\frac{1}{4}\log \left[\frac{1+\tan \left(x\right)}{1-\tan \left(x\right)}\right]+C$

ALso,

$\tan (\frac{\pi }{4}+x)=\frac{1+\tan \left(x\right)}{1-\tan \left(x\right)}$

$\therefore I=\frac{x}{2}+\frac{1}{4}\log \left[\tan (\frac{\pi }{4}+x)\right]+C$