Integral Calculus Question 422
Question: $ \int_{{}}^{{}}{{{\cos }^{5}}x\ dx=} $
Options:
A) $ \sin x-\frac{2}{3}{{\sin }^{3}}x+\frac{1}{5}{{\sin }^{5}}x+c $
B) $ \sin x+\frac{2}{3}{{\sin }^{3}}x+\frac{1}{5}{{\sin }^{5}}x+c $
C) $ \sin x-\frac{2}{3}{{\sin }^{3}}x-\frac{1}{5}{{\sin }^{5}}x+c $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{{{\cos }^{5}}x,dx}=\int_{{}}^{{}}{{{\cos }^{4}}x\cos x,dx}=\int_{{}}^{{}}{{{(1-{{\sin }^{2}}x)}^{2}}\cos xdx} $ Put $ \sin x=t\Rightarrow \cos x\ dx=dt $ , then it reduces to $ \int_{{}}^{{}}{{{(1-t^{2})}^{2}}dt=\int_{{}}^{{}}{(1+t^{4}-2t^{2})\ dt=\frac{t^{5}}{5}-\frac{2t^{3}}{3}+t+c}} $ $ =\frac{{{\sin }^{5}}x}{5}-\frac{2{{\sin }^{3}}x}{3}+\sin x+c $ .
BETA
