A) \[\sin \frac{x}{2}\]
B) \[\cos \frac{x}{2}\]
C) \[\tan \frac{x}{2}\]
D) \[\log \frac{x}{2}\]
Correct Answer: C
Solution :
\[I=\int_{{}}^{{}}{{{e}^{x}}\left( \frac{1+\sin x}{1+\cos x} \right)\,dx}=\int_{{}}^{{}}{{{e}^{x}}\left[ \frac{1+2\sin (x/2)\,\cos (x/2)}{2{{\cos }^{2}}(x/2)} \right]dx}\] \[I=\int_{{}}^{{}}{{{e}^{x}}\left[ \frac{1}{2}{{\sec }^{2}}(x/2)+\tan (x/2) \right]\,dx}={{e}^{x}}.\tan (x/2)+c\] \[\{\because \,\,\,\int_{{}}^{{}}{{{e}^{x}}[f(x)+{f}'(x)\,]dx={{e}^{x}}.\,f(x)+c\}}\]You need to login to perform this action.
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