Question

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

• A. $\frac{1}{2}log\left|\frac{1+\sqrt{1-ta{n}^{2}x}}{1-\sqrt{1-ta{n}^{2}x}}\right|-\frac{1}{\sqrt{2}}\left|\frac{\sqrt{2}+\sqrt{1-ta{n}^{2}x}}{\sqrt{2}-\sqrt{1-ta{n}^2}}\right|+C$
• B. $\frac{1}{2}log\left|\frac{1+\sqrt{1-ta{n}^{2}x}}{1-\sqrt{1-ta{n}^{2}x}}\right|-\frac{1}{\sqrt{2}}\left|\frac{\sqrt{2}+\sqrt{1-ta{n}^{2}x}}{\sqrt{2}-\sqrt{1-tanx}}\right|+C$
• C. $\frac{1}{2}log\left|\frac{1+\sqrt{1-ta{n}^{2}x}}{1-\sqrt{1-ta{n}^{2}x}}\right|-\frac{1}{\sqrt{2}}\left|\frac{\sqrt{2}+\sqrt{1-ta{n}^{2}x}}{\sqrt{2}-\sqrt{1-ta{n}^{2}x}}\right|+C$
• D. $\frac{1}{2}log\left|\frac{1+\sqrt{1-ta{n}^{2}x}}{1+\sqrt{1-ta{n}^{2}x}}\right|-\frac{1}{\sqrt{2}}\left|\frac{\sqrt{2}+\sqrt{1-ta{n}^{2}x}}{\sqrt{2}-\sqrt{1-ta{n}^{2}x}}\right|+C$

Answer 1

The correct option is B $\frac{1}{2}log\left|\frac{1+\sqrt{1-ta{n}^{2}x}}{1-\sqrt{1-ta{n}^{2}x}}\right|-\frac{1}{\sqrt{2}}\left|\frac{\sqrt{2}+\sqrt{1-ta{n}^{2}x}}{\sqrt{2}-\sqrt{1-tanx}}\right|+C$

$\int \frac{\sqrt{\cos 2x}}{\sin x}=\int \frac{\sqrt{\frac{1+ta{n}^{2}x}{1+ta{n}^{2}x}}}{\sin x}dx=\int \frac{\sqrt{1+ta{n}^{2}x}}{secx\sin x}dx=\int \frac{1-ta{n}^{2}x}{tanx}dx=\int \frac{\sqrt{1-ta{n}^{2}x}}{tanx(1+ta{n}^{2}x)}de{c}^{2}xdx=\int \frac{\sqrt{1-ta{n}^{2}x}}{ta{n}^{2}x(1+ta{n}^{2}x)}tanxse{c}^{2}xdx$

Let $\sqrt{1-ta{n}^{2}x}=1-ta{n}^{2}x=t^2$

Differentiate both sides with respect to x

$-2tanxse{c}^{2}xdx=2tstI=-\int \frac{t}{(1-t^2)(2-t^2)}tdt=-\int \frac{t^2}{(1-t^2)(2-t^2)}dt$

Let

$\Rightarrow \frac{t^2}{(1+t^2)(2+2)}=\frac{At+B}{1-t^2}+\frac{Ct+D}{2-t^2}\Rightarrow \frac{t^2}{(1-t^2)(2-t^2)}=\frac{At(2-t^2)+B(2-t^2)+Ct(1-t^2)+D(1-t^2)}{(1-t^2)(2-t^2)}\Rightarrow t^2=2At-A{t}^2+2B-B{t}^2+Ct-C{t}^3+D-D{t}^2\Rightarrow t^2=-t^3(A+C)-t^2(B+D)+(2A+C)t+2B+D$

Equating co-efficient both sides we get

$A+C=0B+D=-12A+C=02B+D=0$

Solving above equation

$A=0C=0B=1D=-2$

$R=\int \frac{1}{1-t^2}dt-2\int \frac{1}{2-t^2}dt=\frac{1}{2}log|\frac{1+t}{1-t}|-\frac{2}{2\sqrt{2}}log|\frac{\sqrt{2}+t}{\sqrt{2}-t}|+C=\frac{1}{2}log|\frac{1+\sqrt{1-ta{n}^{2}x}}{1-\sqrt{1-ta{n}^{2}x}}|-\frac{1}{\sqrt{2}}log|\frac{\sqrt{2}+\sqrt{1-ta{n}^{2}x}}{\sqrt{2}-\sqrt{1-ta{n}^{2}x}}|+C$