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Question: The value of \(\int_{}^{}\frac{\sin x}{\sin 4x}\)dx is –...

The value of ∫sin⁡xsin⁡4x\int_{}^{}\frac{\sin x}{\sin 4x}dx is –

A

14\frac{1}{4}log ∣sin⁡x−1sin⁡x+1∣\left| \frac{\sin x - 1}{\sin x + 1} \right| – 12\frac{1}{\sqrt{2}}log ∣2sin⁡x−12sin⁡x+1∣\left| \frac{\sqrt{2}\sin x - 1}{\sqrt{2}\sin x + 1} \right|+ c

B

18\frac{1}{8}log∣cos⁡x−1cos⁡x+1∣\left| \frac{\cos x - 1}{\cos x + 1} \right| – 122\frac{1}{2\sqrt{2}}log ∣2cos⁡x−12cos⁡x+1∣\left| \frac{\sqrt{2}\cos x - 1}{\sqrt{2}\cos x + 1} \right|+ c

C

18\frac{1}{8}log∣sin⁡x−1sin⁡x+1∣\left| \frac{\sin x - 1}{\sin x + 1} \right| – 142\frac{1}{4\sqrt{2}}log ∣2sin⁡x−12sin⁡x+1∣\left| \frac{\sqrt{2}\sin x - 1}{\sqrt{2}\sin x + 1} \right|+ c

D

None of these

Answer

18\frac{1}{8}log∣sin⁡x−1sin⁡x+1∣\left| \frac{\sin x - 1}{\sin x + 1} \right| – 142\frac{1}{4\sqrt{2}}log ∣2sin⁡x−12sin⁡x+1∣\left| \frac{\sqrt{2}\sin x - 1}{\sqrt{2}\sin x + 1} \right|+ c

Explanation

Solution

The given integrand can be written as

14cos⁡xcos⁡2x\frac{1}{4\cos x\cos 2x}= cos⁡x4(1−sin⁡2x)(1−2sin⁡2x)\frac{\cos x}{4(1 - \sin^{2}x)(1 - 2\sin^{2}x)}, hence

∫sin⁡xsin⁡4x\int_{}^{}\frac{\sin x}{\sin 4x}dx= 14\frac { 1 } { 4 } ∫dt(1−t2)(1−2t2)\int_{}^{}\frac{dt}{(1 - t^{2})(1 - 2t^{2})} (t = sin x)

= 14\frac { 1 } { 4 } ∫[1t2−1−1t2−1/2]\int_{}^{}\left\lbrack \frac{1}{t^{2} - 1} - \frac{1}{t^{2} - 1/2} \right\rbrackt

= 14\frac { 1 } { 4 } [12log⁡∣t−1t+1∣−12log⁡∣t−1/2t+1/2∣]\left\lbrack \frac{1}{2}\log\left| \frac{t - 1}{t + 1} \right| - \frac{1}{\sqrt{2}}\log\left| \frac{t - 1/\sqrt{2}}{t + 1/\sqrt{2}} \right| \right\rbrack + c

=18\frac{1}{8}log∣sin⁡x−1sin⁡x−1∣\left| \frac{\sin x - 1}{\sin x - 1} \right|–142\frac{1}{4\sqrt{2}}log ∣2sin⁡x−12sin⁡x+1∣\left| \frac{\sqrt{2}\sin x - 1}{\sqrt{2}\sin x + 1} \right|+ c