Mathematics

The value of $$2\displaystyle \int \sin x \text{ cosec } 4x\ dx$$ is equal to:


ANSWER

$$\dfrac{1}{2\sqrt2}\ln \left|\dfrac{1+\sqrt2\sin x}{1-\sqrt2\sin x}\right|- \dfrac{1}{4}\ln \left|\dfrac{1+\sin x}{1-\sin x}\right|+c$$


SOLUTION
$$ Q \displaystyle \int 2\sin x\, cosec\, 4 x dx $$

$$  \displaystyle = \int \frac{2 \sin x}{\sin 4x}dx = \int\frac{2\sin x}{2\sin 2x\cos 2x}dx $$

$$ \displaystyle = \int \frac{2 \sin x}{2 \sin x \cos x \cos 2x}dx $$

$$ \displaystyle = \frac{1}{2} \int \frac{1}{\cos x \cos 2 x}dx = \frac{1}{2} \int \frac{\cos x}{(\cos^2 x)(1-2 \sin^2 x)}dx $$

$$ \displaystyle = \frac{1}{2} \int \frac{\cos x}{(1-\sin^2 x)(1-2 \sin^2 x)}dx $$

put $$ \sin x = t $$
     $$ \cos x\, dx  = dt $$
$$  \displaystyle =  \frac{1}{2} \int\frac{1}{(1-t^2)(1-2t^2)}dt $$

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

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

$$ \displaystyle = \frac{1}{2}  \frac{1}{2\sqrt {2}} ln \left| \frac{1+\sqrt{2}t}{1-\sqrt{2}t}\right| -\frac{1}{2}. \frac{1}{2} ln \left| \frac{1+t}{1-t}\right| +c $$

$$ \displaystyle = \frac{1}{4 \sqrt{2}}ln \left| \frac{1+\sqrt{2}\sin x}{1- \sqrt{2}\sin x}\right| -\frac{1}{4} ln \left| \frac{1+\sin x}{1-\sin x}\right| +c $$
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Single Correct Hard Published on 17th 09, 2020
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