A) \[{{e}^{x}}{{\sec }^{2}}\frac{x}{2}+C\]
B) \[{{e}^{x}}\tan \frac{x}{2}+C\]
C) \[{{e}^{x}}\sec \frac{x}{2}+C\]
D) \[{{e}^{x}}+\tan x+C\]
Correct Answer: B
Solution :
We have \[\int{{{e}^{x}}\left( \frac{1+\sin x}{1+\cos x} \right)d\,x}\] \[=\int{{{e}^{x}}\left( \frac{(1+2\sin \frac{x}{2}\cos \frac{\pi }{2})}{2+{{\cos }^{2}}\frac{x}{2}} \right)dx}\] \[[\sin 2x=2\sin x\cos x,\,1+\cos 2x=2{{\cos }^{2}}x]\] \[=\int{{{e}^{x}}\left[ \left( \frac{1}{2{{\cos }^{2}}x/2} \right)+\left( \frac{2\sin x/2\cos x/2}{2{{\cos }^{2}}x/2} \right) \right]dx}\]\[\therefore \] using \[\left[ \int{{{e}^{x}}(f(x)+f(x),\,dx={{e}^{x}}f(x)+C} \right]\] Here \[f(x)=\tan \frac{x}{2}\] and \[f(x)=\frac{1}{2}{{\sec }^{2}}\frac{x}{2}\] \[\Rightarrow \] \[{{e}^{x}}\left( \tan \frac{x}{2} \right)+C\] \[\therefore \] \[\int{{{e}^{x}}\left( \frac{1+\sin x}{1+\cos x} \right)dx={{e}^{x}}=\tan \frac{x}{2}+C}\]
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