Integral Calculus Question 386
Question: If $ \int_{{}}^{{}}{\frac{e^{x}(1+\sin x)dx}{1+\cos x}=e^{x}f(x)+c} $ , then $ f(x)= $
[RPET 1997; Karnataka CET 2003, 05; Orissa JEE 2004]
Options:
A) $ \sin \frac{x}{2} $
B) $ \cos \frac{x}{2} $
C) $ \tan \frac{x}{2} $
D) $ \log \frac{x}{2} $
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Answer:
Correct Answer: C
Solution:
$ I=\int_{{}}^{{}}{e^{x}( \frac{1+\sin x}{1+\cos x} ),dx}=\int_{{}}^{{}}{e^{x}[ \frac{1+2\sin (x/2),\cos (x/2)}{2{{\cos }^{2}}(x/2)} ]dx} $ $ I=\int_{{}}^{{}}{e^{x}[ \frac{1}{2}{{\sec }^{2}}(x/2)+\tan (x/2) ],dx}=e^{x}.\tan (x/2)+c $ $ {\because ,\int_{{}}^{{}}{e^{x}[f(x)+{f}’(x),]dx=e^{x}.,f(x)+c}} $
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