Integral Calculus Question 36
Question: $ \int_{{}}^{{}}{{{\sin }^{3}}x{{\cos }^{2}}x\ dx=} $
Options:
A) $ \frac{{{\cos }^{5}}x}{5}-\frac{{{\cos }^{3}}x}{3}+c $
B) $ \frac{{{\cos }^{5}}x}{5}+\frac{{{\cos }^{3}}x}{3}+c $
C) $ \frac{{{\sin }^{5}}x}{5}-\frac{{{\sin }^{3}}x}{3}+c $
D) $ \frac{{{\sin }^{5}}x}{5}+\frac{{{\sin }^{3}}x}{3}+c $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \int_{{}}^{{}}{{{\sin }^{3}}x{{\cos }^{2}}x,dx}=\int_{{}}^{{}}{(1-{{\cos }^{2}}x){{\cos }^{2}}x,.,\sin x,dx} $ Put $ \cos x=t\Rightarrow -\sin x,dx=dt, $ then it reduces to $ -\int_{{}}^{{}}{(t^{2}-t^{4})dt}=\frac{t^{5}}{5}-\frac{t^{3}}{3}+c=\frac{{{(\cos x)}^{5}}}{5}-\frac{{{(\cos x)}^{3}}}{3}+c $ .
BETA
