Question

Evaluate $\int \frac{(cos2x{)}^{1/2}}{sinx}dx$.

• A. $\frac{1}{\sqrt{3}}log\left[\frac{\sqrt{2}-\sqrt{1-ta{n}^{2}x}}{\sqrt{2}-\sqrt{1-ta{n}^{2}x}}\right]-log(cotx+\sqrt{co{t}^{2}x-1})+c$
• B. $\frac{1}{\sqrt{2}}log\left[\frac{\sqrt{2}+\sqrt{1-ta{n}^{2}x}}{\sqrt{2}-\sqrt{1-ta{n}^{2}x}}\right]-log(cotx+\sqrt{co{t}^{2}x-1})+c$
• C. $\frac{1}{2}log\left[\frac{\sqrt{2}+\sqrt{1-ta{n}^{2}x}}{\sqrt{2}-\sqrt{1-ta{n}^{2}x}}\right]-log(cotx+\sqrt{co{t}^{2}x-1})+c$
• D. none of these

Answer 1

The correct option is B $\frac{1}{\sqrt{2}}log\left[\frac{\sqrt{2}+\sqrt{1-ta{n}^{2}x}}{\sqrt{2}-\sqrt{1-ta{n}^{2}x}}\right]-log(cotx+\sqrt{co{t}^{2}x-1})+c$

$I=\int \frac{\sqrt{cos2x}}{sinx}dx$$=\int \frac{\sqrt{co{s}^{2}x-si{n}^{2}x}}{sinx}dx=\int \sqrt{co{t}^{2}x-1dx}$

Put $cotx=sec\theta$$\Rightarrow -cose{c}^{2}xdx=sec\theta tan\theta d\theta$

$cose{c}^{2}x=1+co{t}^{2}x$ and put $co{t}^{2}x=se{c}^2\theta$

to get dx, then

$\therefore I=\int \sqrt{se{c}^2\theta -1}\frac{si{n}^2\theta }{cos\theta +co{s}^3\theta }d\theta$

$=-\int \frac{sec\theta {\tan }^2\theta }{1+se{c}^2\theta }d\theta =-\int \frac{si{n}^2\theta }{cos\theta +co{s}^3\theta }d\theta$

$=\int \frac{(1+co{s}^2\theta )-2co{s}^2\theta }{cos\theta (1+co{s}^2\theta )}d\theta$

$=-\int sec\theta d\theta +2\int \frac{cos\theta }{1+co{s}^2\theta }d\theta$

$=-log|sec\theta +tan\theta |+2\int \frac{cos\theta }{2-si{n}^2\theta }d\theta$

$=-log|sec\theta +tan\theta |+2\int \frac{dt}{2-t^2}(putsin\theta =t)$

$=-log|sec\theta +tan\theta |+2\frac{1}{2\sqrt{2}}log\left|\frac{\sqrt{2}+sin\theta }{\sqrt{2}-sin\theta }\right|+C$ .

............. Using $\int \frac{1}{a^2-x^2}dx=\frac{1}{2a}\log |\frac{a+x}{a-x}|+c$

$=-log|cotx+\sqrt{co{t}^{2}x-1}|+\frac{1}{\sqrt{2}}log\left|\frac{\sqrt{2}+\sqrt{1-ta{n}^{2}x}}{\sqrt{2}-\sqrt{1-ta{n}^{2}x}}\right|+C$