Integral-Calculus Question 504
Question: The value of $ \int{\frac{\sin x}{\sin 4x}dx} $ is
Options:
A) $ \frac{1}{4}\log | \frac{\sin x-1}{\sin x+1} |-\frac{1}{\sqrt{2}}\log | \frac{\sqrt{2}\sin x-1}{\sqrt{2}\sin x+1} |+C $
B) $ \frac{1}{8}\log | \frac{\cos x-1}{\cos x+1} |-\frac{1}{2\sqrt{2}}\log | \frac{\sqrt{2}\cos x-1}{\sqrt{2}\cos x+1} |+C $
C) $ \frac{1}{8}\log | \frac{\sin x-1}{sinx+1} |-\frac{1}{4\sqrt{2}}\log | \frac{\sqrt{2}\sin x-1}{\sqrt{2}\sin x+1} |+C $
D) None of these.
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ I=\int{\frac{\sin xdx}{4\sin x\cos x\cos 2x}} $ $ =\frac{1}{4}\int{\frac{\cos x,dx}{(1-{{\sin }^{2}}x)(1-2,{{\sin }^{2}}x)}} $ $ =\frac{1}{4}\int{\frac{dt}{(1-t^{2})(1-2t^{2})}} $ $ [t=sin,x] $ $ =\frac{1}{4}\int{( \frac{2}{1-2t^{2}}-\frac{1}{1-t^{2}} )dt} $ $ =\frac{1}{4}{ \frac{2}{2\sqrt{2}}\log | \frac{1+\sqrt{2}t}{1-\sqrt{2}t} |-\frac{1}{2}\log | \frac{1+t}{1-t} | }+C $ $ =\frac{1}{8}\log | \frac{\sin x-1}{\sin x+1} |-\frac{1}{4\sqrt{2}}\log | \frac{\sqrt{2}\sin x-1}{\sqrt{2}\sin x+1} |+C $
BETA
