Question

If $I=\int \frac{\sin x+{\sin }^{3}x}{\cos 2x}dx=A\cos x+B\log |f\left(x\right)|+C$, then:

• A. $A=\frac{1}{4},B=-\frac{1}{2},f\left(x\right)=\frac{\sqrt{2}\cos x-1}{\sqrt{2}\cos x+1}$
• B. $A=\frac{1}{2},B=-\frac{3}{4}\sqrt{2}$
• C. $A=-\frac{1}{2},B=\frac{3}{\sqrt{5}},f\left(x\right)=\frac{\sqrt{2}cosx+1}{\sqrt{2}\cos x-1}$
• D. $A=\frac{1}{2},B=-\frac{3}{4\sqrt{2}},f\left(x\right)=\frac{\sqrt{2}\cos x-1}{\sqrt{2}\cos x+1}$

Answer 1

Answer: (D)

$A=\frac{1}{2},B=-\frac{3}{4\sqrt{2}},f\left(x\right)=\frac{\sqrt{2}\cos x-1}{\sqrt{2}\cos x+1}$

Given:
$I=\int \frac{\sin x+{\sin }^{3}x}{\cos 2x}dx=A\cos x+B\log |f\left(x\right)|+C$
To find:- $A,B$ and $f\left(x\right)$
$I=\int \frac{\sin x(1+{\sin }^{2}x)dx}{(2{\cos }^{2}x-1)}$
Put $t=\cos x\Rightarrow dt=-\sin xdx$

$\therefore I=\int \frac{(1-t^2)+1}{(2t^2-1)}(-dx)$ ...... $(\because {\sin }^{2}x=1-{\cos }^{2}x)$

$=\int \frac{(t^2-2)}{(2t^2-1)}dt$

$=\frac{1}{2}\int \frac{t^2-2}{t^2-\frac{1}{2}}dt$

$=\frac{1}{2}[\int (1-\frac{3}{2}\times \frac{1}{(t^2-\frac{1}{2})})dt]$

$I=\frac{1}{2}\int dt-\frac{3}{4}\int \frac{dt}{t^2-\frac{1}{2}}$

$I=\frac{1}{2}t-\frac{3}{4}\times \frac{1}{\sqrt{2}}\ln \left(\frac{t-\frac{1}{\sqrt{2}}}{t+\frac{1}{\sqrt{2}}}\right)+c$

$\therefore I=\frac{1}{2}\cos x-\frac{3}{4}\times \frac{1}{\sqrt{2}}\ln \left(\frac{\sqrt{2}\cos x-1}{\sqrt{2}\cos x+1}\right)+c$

Hence, $A=\frac{1}{2},B=\frac{-3}{4\sqrt{2}}$ and $f\left(x\right)=\frac{\sqrt{2}\cos x-1}{\sqrt{2}\cos x+1}$