Question

If $\int \frac{\sin x}{\sin 4x}dx=A\log \left|\frac{1+\sin x}{1-\sin x}\right|+B\log \left|\frac{1+\sqrt{2}\sin x}{1-\sqrt{2}\sin x}\right|+C$

• A. $A=\frac{1}{8},B=\frac{1}{4\sqrt{2}}$
• B. $A=-\frac{1}{8},B=-\frac{1}{4\sqrt{2}}$
• C. $A=-\frac{1}{8},B=\frac{1}{4\sqrt{2}}$
• D. $A=\frac{1}{8},B=-\frac{1}{4\sqrt{2}}$

Answer 1

The correct option is C $A=-\frac{1}{8},B=\frac{1}{4\sqrt{2}}$

Let $I=\int \frac{\sin x}{\sin 4x}dx=\frac{1}{4}\int \frac{\cos x}{{\cos }^{2}x\cos 2x}dx$
$=\frac{1}{4}\int \frac{\cos x}{(1-{\sin }^{2}x)(1-2{\sin }^{2}x)}dx$
Substitute $\sin x=t$
$I=\frac{1}{4}\int \frac{dt}{(1-t^2)(1-2t^2)}=\frac{1}{4}\int (\frac{2}{1-2t^2}-\frac{1}{1-t^2})dt$
$=\frac{1}{2}.\frac{1}{2\sqrt{2}}\log \left(\frac{1+\sqrt{2}t}{1-\sqrt{2}t}\right)-\frac{1}{8}\log \left(\frac{1+t}{1-t}\right)+c$
$=\frac{1}{4\sqrt{2}}\log \left(\frac{1+\sqrt{2}\sin x}{1-\sqrt{2}\sin x}\right)-\frac{1}{8}\log \left(\frac{1+\sin x}{1-\sin x}\right)+c$
$\Rightarrow A=-\frac{1}{8},B=\frac{1}{4\sqrt{2}}$