Question

Integrate $\int \frac{1}{{\tan }^{2}x+{\sec }^{2}x}dx=$

• A. $\frac{1}{\sqrt{2}}{\tan }^{-1}(\sqrt{2}\tan x)-{\tan }^{-1}\left(\tan x\right)+c$
• B. None of these
• C. $\sqrt{2}{\tan }^{-1}(\sqrt{2}+\tan x)-{\tan }^{-1}\left(\tan x\right)+c$
• D. $\sqrt{2}{\tan }^{-1}(\tan x)-{\tan }^{-1}\left(\sqrt{2}\tan x\right)+c$

Answer 1

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

$\int \frac{1}{{\tan }^{2}x+{\sec }^{2}x}dx$

$=\int \frac{1}{2{\tan }^{2}x+1}dx$ Let $({\sec }^{2}x={\tan }^{2}x+1)$

$=\int \frac{1}{2{\tan }^{2}x+1}dx$ Let $\tan x=t$ On differentiating w.r.t $t$ we have

$=\int \frac{1}{(2t^2+1)(1+t^2)}dt=dx=\frac{dt}{1+{\tan }^{2}x}=dt1+t^2$

$=\int \left[\frac{1}{(2t^2+1)(1+t^2)}\right]dt$ [ by Rationalizing the term ]

$=\int \left[\frac{1}{(2t^2+1)(1+t^2)}\right]dt$

$=\int \frac{1}{t^2+{\left(\frac{1}{\sqrt{2}}\right)}^2}-\int \frac{1}{t^2+1}dt$

$=\frac{1}{\frac{1}{\sqrt{2}}}{\tan }^{-1}\frac{t}{\frac{1}{\sqrt{2}}}-{\tan }^{-1}t+c$

$=\sqrt{2}{\tan }^{-1}(\sqrt{2}+\tan x)-{\tan }^{-1}\tan x+c$

Hence, the answer is $\sqrt{2}{\tan }^{-1}(\sqrt{2}+\tan x)-{\tan }^{-1}\tan x+c.$