Integral Calculus Question 493
Question: $ \int_{{}}^{{}}{\frac{\sin x}{\sin x-\cos x}}\ dx= $
[Roorkee 1988]
Options:
A) $ \frac{1}{2}\log (\sin x-\cos x)+x+c $
B) $ \frac{1}{2}[\log (\sin x-\cos x)+x]+c $
C) $ \frac{1}{2}\log (\cos x-\sin x)+x+c $
D) $ \frac{1}{2}[\log (\cos x-\sin x)+x]+c $
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Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{\frac{\sin x,dx}{\sin x-\cos x}}=\frac{1}{2}\int_{{}}^{{}}{\frac{2\sin x}{\sin x-\cos x},dx} $ $ =\frac{1}{2}\int_{{}}^{{}}{\frac{(\sin x-\cos x+\sin x+\cos x)}{\sin x-\cos x},dx} $ $ =\frac{1}{2}\int_{{}}^{{}}{( 1+\frac{\sin x+\cos x}{\sin x-\cos x} ),dx}=\frac{1}{2}[x+\log (\sin x-\cos x)]+c $ .
BETA
