A) \[\tan \frac{x}{2}+c\]
B) \[x+\tan \frac{x}{2}+c\]
C) \[x-\frac{1}{2}\tan \frac{x}{2}+c\]
D) \[\frac{x-\tan \frac{x}{2}}{2}+c\]
E) \[x-\tan \frac{x}{2}+c\]
Correct Answer: E
Solution :
\[f(x)=\cos x-{{\cos }^{2}}x+{{\cos }^{3}}x-\infty \] \[=\frac{\cos x}{1+\cos x}\] \[\therefore \]\[\int{f(x)}dx=\int{\frac{1+\cos x}{1+\cos x}}dx\] \[-\int{\frac{1}{1+\cos x}}dx\] \[=x-\frac{1}{2}\tan \frac{x}{2}.2+c\] \[=x-\tan \frac{x}{2}.2+c\]You need to login to perform this action.
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