Question

Assertion :Assume: $I=\int \frac{\sqrt{\cos 2x}}{\sin x}dx$

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

Reason: $\tan x=\sin \theta \to I=\int \frac{{\cos }^2\theta d\theta }{\sin \theta (1+{\sin }^2\theta )}$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

$I=\int \frac{\sqrt{\cos 2x}}{\sin x}dx$

Put $\tan x=\sin \theta \Rightarrow {\sec }^{2}xdx=\cos \theta d\theta$

$\Rightarrow (1+{\tan }^{2}x)dx=\cos \theta d\theta \Rightarrow dx=\frac{\cos \theta }{(1+{\sin }^2\theta )}d\theta$

$\sqrt{\cos 2x}=\sqrt{\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}}=\sqrt{\frac{1-{\sin }^2\theta }{1+{\sin }^2\theta }}=\frac{\cos \theta }{\sqrt{1+{\sin }^2\theta }}$

$\sin x=\frac{\sin \theta }{\sqrt{1+{\sin }^2\theta }}$

$\therefore I=\int \frac{\frac{\cos \theta }{\sqrt{1+{\sin }^2\theta }}}{\frac{\sin \theta }{\sqrt{1+{\sin }^2\theta }}}.\frac{\cos \theta }{(1+{\sin }^2\theta )}d\theta =\int \frac{{\cos }^2\theta }{\sin \theta (1+{\sin }^2\theta )}d\theta$

$=\int (\frac{1}{\sin \theta }-\frac{2\sin \theta }{2-{\cos }^2\theta })d\theta =\ln (\csc \theta -\cot \theta )+\frac{1}{\sqrt{2}}\ln \left(\frac{\sqrt{2}+\cos \theta }{\sqrt{2}-\cos \theta }\right)+c$

$=\ln \left[\left(\frac{1-\sqrt{1-{\tan }^{2}x}}{\tan x}\right){\left(\frac{\sqrt{2}+\sqrt{1-{\tan }^{2}x}}{\sqrt{2}-\sqrt{1-{\tan }^{2}x}}\right)}^{\frac{1}{\sqrt{2}}}\right]+c$