$$ \int \frac{1}{(\sin x+\cos x+\sqrt{2} \sqrt{\sin 2 x})^2} d x= $$

$$ \int \frac{1}{(\sin x+\cos x+\sqrt{2} \sqrt{\sin 2 x})^2} d x= $$
  1. $\frac{-(1+3 \sqrt{\tan x})}{\left(3+\tan ^2 x\right)^3}+C$
  2. $\frac{-(1+3 \sqrt{\tan x})}{3(1+\sqrt{\tan x})^3}+C$
  3. $\frac{-(1+\sqrt{\tan x})}{3(1+3 \sqrt{\tan x})^2}+C$
  4. $\frac{1}{(1+3 \sqrt{\tan x})^3}+C$

Solution

$ \begin{aligned} & \text { } \int \frac{1}{(\sin x+\cos x+\sqrt{2} \sqrt{\sin 2 x})^2 d x} \\ & \int \frac{1}{\left(\sqrt{\sin x}+\sqrt{\cos x)^4 d x}\right.}=\int \frac{\sec ^2 x}{(\sqrt{\tan x}+1)} d x \end{aligned} $ Let $\tan x=t^2 \Rightarrow \sec ^2 x d x=2 t d t$ $\begin{aligned} & \int \frac{2 t}{(t+1)} d t=2 \int\left(\frac{1}{(t+1)^3}-\frac{1}{(t+1)^4}\right) d t \\ & =2\left[\frac{1}{3(t+1)^3}-\frac{1}{2(t+1)^2}\right]+c \\ & =\frac{-[3 t+1]}{3(t+1)^3}+c=\frac{-(1+3 \sqrt{\tan x})}{3(1+\sqrt{\tan x})^3}+c\end{aligned}$

Asked in: AP EAMCET 2022 (06 Jul Shift 1)

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