Integral Calculus Question 43
Question: $ \int_{{}}^{{}}{{{\sin }^{5}}x{{\cos }^{4}}x\ dx=} $
Options:
A) $ -\frac{1}{5}{{\cos }^{5}}x+\frac{2}{7}{{\cos }^{7}}x-\frac{1}{9}{{\cos }^{9}}x+c $
B) $ \frac{1}{5}{{\cos }^{5}}x+\frac{2}{7}{{\cos }^{7}}x-\frac{1}{9}{{\cos }^{9}}x+c $
C) $ \frac{1}{5}{{\cos }^{5}}x+\frac{2}{7}{{\cos }^{7}}x+\frac{1}{9}{{\cos }^{9}}x+c $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Put $ \cos x=t\Rightarrow -\sin x,dx=dt, $ then $ \int_{{}}^{{}}{{{(1-{{\cos }^{2}}x)}^{2}}.{{\cos }^{4}}x\sin x,dx}=-\int_{{}}^{{}}{{{(1-t^{2})}^{2}}.,t^{4}dt} $ $ =-\frac{t^{5}}{5}+\frac{2}{7}t^{7}-\frac{1}{9}t^{9}+c=-\frac{{{\cos }^{5}}x}{5}+\frac{2}{7}{{\cos }^{7}}x-\frac{1}{9}{{\cos }^{9}}x+c $ . Aliter : By reduction formula.
BETA
