Integral Calculus Question 63
Question: $ \int_{{}}^{{}}{\frac{dx}{(\sin x+\sin 2x)}=} $
[IIT 1984]
Options:
A) $ \frac{1}{6}\log (1-\cos x)+\frac{1}{2}\log (1+\cos x)-\frac{2}{3}\log (1+2\cos x) $
B) $ 6\log (1-\cos x)+2\log (1+\cos x)-\frac{2}{3}\log (1+2\cos x) $
C) $ 6\log (1-\cos x)+\frac{1}{2}\log (1+\cos x)+\frac{2}{3}\log (1+2\cos x) $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ I=\int_{{}}^{{}}{\frac{dx}{\sin x(1+2\cos x)}}=\int_{{}}^{{}}{\frac{\sin x,dx}{{{\sin }^{2}}x(1+2\cos x)}} $
$ =\int_{{}}^{{}}{\frac{\sin x,dx}{(1-\cos x)(1+\cos x)(1+2\cos x)}} $
Now differential coefficient of $ \cos x $ is $ -\sin x $ which is given in numerator and hence we make the substitution $ \cos x=t\Rightarrow -\sin x,dx=dt $
$ \therefore ,I=-\int_{{}}^{{}}{\frac{dt}{(1-t)(1+t)(1+2t)}} $
We split the integrand into partial fractions
\ $ I=-\int{[ \frac{1}{6(1-t)}-\frac{1}{2(1+t)}+\frac{4}{3(1+2t)} ]},dt $ etc.
BETA
