Question

Evaluate : $\int \frac{sinx}{sin4x}$dx

• A. $\frac{1}{2\sqrt{2}}ln\left|\frac{1+\sqrt{2}sinx}{1-\sqrt{2}sinx}\right|+\frac{1}{8}ln\left|\frac{1+sinx}{1-sinx}\right|+c$
• B. $\frac{1}{2\sqrt{2}}ln\left|\frac{1+\sqrt{2}sinx}{1-\sqrt{2}sinx}\right|-\frac{1}{8}ln\left|\frac{1+sinx}{1-sinx}\right|+c$
• C. $\frac{1}{4\sqrt{2}}ln\left|\frac{1+\sqrt{2}sinx}{1-\sqrt{2}sinx}\right|+\frac{1}{8}ln\left|\frac{1+sinx}{1-sinx}\right|+c$
• D. $\frac{1}{4\sqrt{2}}ln\left|\frac{1+\sqrt{2}sinx}{1-\sqrt{2}sinx}\right|-\frac{1}{8}ln\left|\frac{1+sinx}{1-sinx}\right|+c$

Answer 1

The correct option is D $\frac{1}{4\sqrt{2}}ln\left|\frac{1+\sqrt{2}sinx}{1-\sqrt{2}sinx}\right|-\frac{1}{8}ln\left|\frac{1+sinx}{1-sinx}\right|+c$

$I=\int \frac{sinx}{sin4x}dx=\int \frac{sinx}{2sinx\cos x\cos 2x}dx=\frac{1}{4}\int \frac{dx}{cosx\cdot cos2x}$

$=\frac{1}{4}\int \frac{cosx}{(1-si{n}^{2}x)(1-2si{n}^{2}x)}dx$

Substitute t = sinx, we get dt = cosx.dx

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

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

$=\frac{1}{4}\frac{1}{\sqrt{2}}ln\left|\frac{\frac{1}{\sqrt{2}}+t}{\frac{1}{\sqrt{2}}-t}\right|-\frac{1}{8}ln\left|\frac{1+t}{1-t}\right|-C$

$=\frac{1}{4}\frac{1}{\sqrt{2}}ln\left|\frac{\frac{1}{\sqrt{2}}+sinx}{\frac{1}{\sqrt{2}}-sinx}\right|-\frac{1}{8}ln\left|\frac{1+sinx}{1-sinx}\right|-C$