Integral Calculus Question 165
Question: If $ \int{{{\sin }^{3}}x{{\cos }^{5}}xdx} $ $ =A{{\sin }^{4}}x+B{{\sin }^{6}}x+C{{\sin }^{8}}x+D $ Then
Options:
A) $ A=\frac{1}{4},B=-\frac{1}{3},C=\frac{1}{8},D\in R $
B) $ A=\frac{1}{8},B=\frac{1}{4},C=\frac{1}{3},D\in R $
C) $ A=0,B=-\frac{1}{6},C=\frac{1}{8},D\in R $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ I=\int{{{\sin }^{3}}x.{{\cos }^{5}}xdx} $ Put $ \sin x=t\Rightarrow cosxdx=dt $ $ I=\int{{{\sin }^{3}}x.{{\cos }^{4}}x.\cos xdx=\int{t^{3}{{( 1-t^{2} )}^{2}}dt}} $ $ =\int{( t^{3}-2t^{5}+t^{7} )dt=\frac{1}{4}t^{4}-\frac{2}{6}t^{6}+\frac{1}{8}t^{8}+D} $ $ =\frac{1}{4}{{\sin }^{4}}x-\frac{1}{3}{{\sin }^{6}}x+\frac{1}{8}{{\sin }^{8}}x+D $
BETA
