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Question

Mathematics Question on integral

Evaluate : ∫sinxsin4xdx\int\frac{sin x}{sin 4x} dx

A

−18log∣1+sinx1−sinx∣+142log∣1+2sinx1−2sinx∣+C-\frac{1}{8}log\left|\frac{1+sin x}{1-sin x}\right|+\frac{1}{4\sqrt{2}}log \left|\frac{1+\sqrt{2}sin x}{1-\sqrt{2}sin x}\right| +C

B

18log∣1+sinx1−sinx∣−142log∣1+2sinx1−2sinx∣+C\frac{1}{8}log\left|\frac{1+sin x}{1-sin x}\right|-\frac{1}{4\sqrt{2}}log \left|\frac{1+\sqrt{2}sin x}{1-\sqrt{2}sin x}\right| +C

C

−18log∣1+sinx1−sinx∣+142log∣1−2sinx1+2sinx∣+C-\frac{1}{8}log\left|\frac{1+sin x}{1-sin x}\right|+\frac{1}{4\sqrt{2}}log \left|\frac{1-\sqrt{2}sin x}{1+\sqrt{2}sin x}\right| +C

D

None of these

Answer

−18log∣1+sinx1−sinx∣+142log∣1+2sinx1−2sinx∣+C-\frac{1}{8}log\left|\frac{1+sin x}{1-sin x}\right|+\frac{1}{4\sqrt{2}}log \left|\frac{1+\sqrt{2}sin x}{1-\sqrt{2}sin x}\right| +C

Explanation

Solution

We have, I=∫sinxsin4xdx=∫sinx2sin2xcos2xdx I= \int\frac{sinx}{sin 4x} dx = \int\frac{sinx}{2sin 2x cos2x} dx =∫sinx4sinxcosxcos2xdx =\int\frac{sinx}{4sinx cosx cos2x }dx ⇒I=14∫1cosxcos2xdx=14∫cosxcos2xcos2xdx\Rightarrow I= \frac{1}{4}\int\frac{1}{cosx cos2x }dx =\frac{1}{4} \int\frac{cosx}{cos^{2}x cos 2x}dx ⇒I=14∫cosx(1−sin2x)(1−2sin2x)dx\Rightarrow I =\frac{1}{4} \int\frac{cosx}{\left(1-sin^{2}x\right)\left(1-2sin^{2}x\right)}dx puttingsinx=t⇒cosxdx=dt sinx = t \Rightarrow cosx dx = dt, we get I=14∫dt(1−t2)(1−2t2) I= \frac{1}{4}\int\frac{dt}{\left(1-t^{2}\right)\left(1-2t^{2}\right)} Lett2=yLet t^{2}=y. Then, 1(1−t2)(1−2t2)=1(1−y)(1−2y) \frac{1}{\left(1-t^{2}\right)\left(1-2t^{2}\right)}=\frac{1}{\left(1-y\right)\left(1-2y\right)} and 1(1−y)(1−2y)=A1−y+B1−2y\frac{1}{\left(1-y\right)\left(1-2y\right)}= \frac{A}{1-y} +\frac{B}{1-2y}. ⇒1=A(1−2y)+B(1−y).......(i)\Rightarrow 1 =A\left(1-2y\right)+B\left(1-y\right) .......\left(i\right) Putting y=1andy=12respectivelyin(i),wegetA=−1andB=2y=1 and y=\frac{1}{2} respectively in \left(i\right), we get A = -1 and B = 2 ∴1(1−y)(1−2y)=−11−y+21−2y\therefore\frac{1}{\left(1-y\right)\left(1-2y\right)}= -\frac{1}{1-y} +\frac{2}{1-2y} ⇒1(1−t2)(1−2t2)=−11−t2+21−2t2\Rightarrow \frac{1}{\left(1-t^{2}\right)\left(1-2t^{2}\right)}=-\frac{1}{1-t^{2}}+\frac{2}{1-2t^{2}} ⇒I=14∫(−11−t2+21−2t2)dt\Rightarrow I= \frac{1}{4}\int \left(-\frac{1}{1-t^{2}}+\frac{2}{1-2t^{2}}\right)dt ⇒I=−14⋅12log∣1+t1−t∣+12⋅122log∣1+2t1−2t∣+C\Rightarrow I =-\frac{1}{4}\cdot\frac{1}{2}log\left|\frac{1+t}{1-t}\right| +\frac{1}{2}\cdot\frac{1}{2\sqrt{2}}log\left|\frac{1+\sqrt{2t}}{1-\sqrt{2t}}\right|+ C ⇒I=−18∣1+sinx1−sinx∣+142log∣1+2sinx1−2sinx∣+C\Rightarrow I = -\frac{1}{8}\left|\frac{1+sinx}{1-sinx }\right| +\frac{1}{4\sqrt{2}}log\left|\frac{1+\sqrt{2 sinx}}{1-\sqrt{2 sinx}}\right| +C