A) \[\frac{1}{4}\log \left| \frac{\sin x-1}{\sin x+1} \right|-\frac{1}{\sqrt{2}}\log \left| \frac{\sqrt{2}\sin x-1}{\sqrt{2}\sin x+1} \right|+C\]
B) \[\frac{1}{8}\log \left| \frac{\cos x-1}{\cos x+1} \right|-\frac{1}{2\sqrt{2}}\log \left| \frac{\sqrt{2}\cos x-1}{\sqrt{2}\cos x+1} \right|+C\]
C) \[\frac{1}{8}\log \left| \frac{\sin x-1}{\sin x+1} \right|-\frac{1}{4\sqrt{2}}\log \left| \frac{\sqrt{2}\sin x-1}{\sqrt{2}\sin x+1} \right|+C\]
D) none of these
Correct Answer: C
Solution :
\[I=\int_{{}}^{{}}{\frac{\sin x}{\sin 4x}dx}\] \[I=\int_{{}}^{{}}{\frac{\sin x}{2\sin 2x\cos 2x}dx=\int_{{}}^{{}}{\frac{dx}{4\cos x\cos 2x}}}\] \[=\int_{{}}^{{}}{\frac{\cos x}{4(1-{{\sin }^{2}}x)(1-2{{\sin }^{2}}x)}dx}\] Put \[\sin x=t\Rightarrow \cos xdx=dt\] \[\therefore \]\[I=\int_{{}}^{{}}{\frac{dt}{4(1-{{t}^{2}})(1-2{{t}^{2}})}=-\frac{1}{4}\int_{{}}^{{}}{\left( \frac{1}{{{t}^{2}}-1}-\frac{1}{{{t}^{2}}-\frac{1}{2}} \right)dt}}\]\[=\frac{1}{4}\left[ \frac{1}{2}\log \left| \frac{t-1}{t+1} \right|-\frac{1}{\sqrt{2}\log }\left| \frac{1-\frac{1}{\sqrt{2}}}{t+\frac{1}{\sqrt{2}}} \right| \right]+C\]
You need to login to perform this action.
You will be redirected in
3 sec
