Question

$\int \sqrt{\frac{cosecx-cotx}{cosecx+cotx}}.\frac{secx}{\sqrt{1+2secx}}dx$

• A. $si{n}^{-1}\left(\frac{1}{2}se{c}^2\frac{x}{2}\right)+c$
• B. $co{s}^{-1}\left(\frac{1}{2}se{c}^2\frac{x}{2}\right)+c$
• C. $co{s}^{-1}\left(\frac{1}{2}cose{c}^2\frac{x}{2}\right)+c$
• D. $si{n}^{-1}\left(\frac{1}{2}cose{c}^2\frac{x}{2}\right)+c$

Answer 1

Answer: (A)

$si{n}^{-1}\left(\frac{1}{2}se{c}^2\frac{x}{2}\right)+c$

Let $I=\int \sqrt{\frac{\csc x-\cot x}{\csc x+\cot x}}.\frac{\sec x}{\sqrt{1+2\sec x}}dx$
$=\int \tan \frac{x}{2}.\frac{{\sec }^2\frac{x}{2}}{\sqrt{(2-{\sec }^4\frac{x}{2})(2+{\sec }^2\frac{x}{2})}}dx=\int \frac{\tan \frac{x}{2}{\sec }^2\frac{x}{2}}{\sqrt{4-{\sec }^4\frac{x}{2}}}dx$
Substitute ${\sec }^2\frac{x}{2}=t\Rightarrow {\sec }^2\frac{x}{2}\tan \frac{x}{2}dx=dt$
$I=\int \frac{dt}{\sqrt{4-t^2}}=si{n}^{-1}\left(\frac{t}{2}\right)+c=si{n}^{-1}\left(\frac{1}{2}{\sec }^2\frac{x}{2}\right)+c.$