Question

Solve:
$\int \frac{\sin x}{{\cos }^{2}x\sqrt{\cos 2x}}dx$=

• A. $C-\sqrt{1-{\tan }^{2}x}$
• B. $C-\sqrt{1-{\sec }^{2}x}$
• C. $C-\sqrt{1+{\cos }^{2}x}$
• D. $C-\sqrt{1+{\sin }^{2}x}$

Answer 1

The correct option is A $C-\sqrt{1-{\tan }^{2}x}$

$\begin{array}{cc}\int \frac{\sin x}{{\cos }^{2}x\sqrt{\cos 2x}}dx& \\ \int \frac{\tan x}{\cos x.\sqrt{\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}}}dx& \left[\cos 2x=\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}\right]\\ \Rightarrow \int \frac{\tan x}{\frac{\cos x}{\sec x\sqrt{1-{\tan }^{2}x}}}dx& \\ \Rightarrow \int \tan x.{\sec }^{2}x& \\ Let1-{\tan }^{2}x=t& \\ -2\tan x.{\sec }^{2}xdx=dt& \\ dx=\frac{dt}{-2\tan .{\sec }^{2}x}& \\ \Rightarrow \int \frac{\tan x.{\sec }^{2}x}{\sqrt{t}}\times \frac{dt}{-2\tan .{\sec }^{2}x}& \\ \Rightarrow \frac{1}{2}\int \frac{dt}{\sqrt{t}}& \\ \Rightarrow \frac{1}{2}\int t^{-\frac{1}{2}}dt& \\ \Rightarrow \frac{-\frac{1}{2}{t}^{\frac{1}{2}}}{\frac{1}{2}}+c& \\ \Rightarrow 2.\sqrt{t}+c& \\ \Rightarrow -\frac{1}{2}\times 2\sqrt{1-{\tan }^{2}x}+c& \\ -\sqrt{1-{\tan }^{2}x}+c.& \end{array}$