Integral Calculus Question 156
Question: $ \int{\frac{dx}{\sin x(3+{{\cos }^{2}}x)}} $ is equal to
Options:
A) $ \log | y^{2}-1 |-{{\tan }^{-1}}y+C $
B) $ {{\tan }^{-1}}\frac{y}{\sqrt{3}}+C $
C) $ \log | \frac{y-1}{y+1} |+C $
D) $ \frac{1}{4}\log | \frac{y-1}{y+1} |-\frac{1}{4\sqrt{3}}{{\tan }^{-1}}\frac{y}{\sqrt{3}}+C $
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Answer:
Correct Answer: D
Solution:
[d] $ \int{\frac{dx}{\sin x(3+{{\cos }^{2}}x)}} $ $ =\int{\frac{\sin x,dx}{{{\sin }^{2}}x(3+{{\cos }^{2}}x)}} $ $ =\int{\frac{\sin x,dx}{(1-{{\cos }^{2}}x)(3+{{\cos }^{2}}x)}} $ $ =\int{\frac{dy}{(y^{2}-1)(y^{2}+3)}} $ (Putting $ \cos x=y $ ) $ =\frac{1}{4}\int{[ \frac{1}{y^{2}-1}-\frac{1}{y^{2}+3} ]dy} $ $ =\frac{1}{4}\log | \frac{y-1}{y+1} |-\frac{1}{4\sqrt{3}}{{\tan }^{-1}}\frac{y}{\sqrt{3}}+C $
BETA
