Question

If $\int {\sin }^{3}x{\cos }^{5}xdx=A{\sin }^{4}x+B{\sin }^{6}x+C{\sin }^{8}x+D$. Then

• 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

Answer 1

Answer: (A)

$I=\int {\sin }^{3}x\cdot {\cos }^{5}xdx$
Put $\sin x=t\Rightarrow \cos xdx=dt$
$I=\int {\sin }^{3}x\cdot {\cos }^{4}x\cdot \cos xdx=\int t^3{\left(1-t^2\right)}^{2}dt$
$=\int \left(t^3-2t^5+t^7\right)dt=\frac{1}{4}{t}^4-\frac{2}{6}{t}^6+\frac{1}{8}{t}^8+D$
$=\frac{1}{4}{\sin }^{4}x-\frac{1}{3}{\sin }^{6}x+\frac{1}{8}{\sin }^{8}x+D$