Question

The value of ${\int }_0^{\pi }\frac{5x^3\sin x{\cos }^{6}x}{({\pi }^3-3{\pi }^{2}x+3\pi x^2)}dx$ is

• A. $\frac{5}{7}$
• B. $\frac{2}{7}$
• C. $\frac{12}{14}$
• D. None

Answer 1

The correct option is A $\frac{5}{7}$

$I={\int }_0^{\pi }\frac{5x^3{\cos }^{6}x\sin x}{{\pi }^3-3{\pi }^{3}x+3\pi x^2-x^3+x^3}$
$I={\int }_0^{\pi }\frac{5x^3{\cos }^{6}x\sin x}{(\pi -x{)}^3+x^3}dx\cdots \left(1\right)$
${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx.$
$I={\int }_0^{\pi }\frac{5(\pi -x{)}^3{\cos }^{6}x\sin x}{x^3+(\pi -x)3}\cdots \left(2\right)$
Add $\left(1\right)+\left(2\right)$
$2I={\int }_0^{\pi }5{\cos }^{6}x\sin x\times \left(\frac{x^3+(\pi -x{)}^3}{x^3+(\pi -x{)}^3}\right)dx$
$\Rightarrow 2I={\int }_0^{\pi }5{\cos }^{6}x\sin xdx$
Let $\cos x=t\Rightarrow \sin xdx=-dt$
$2I=-{\int }_1^{-1}5t^{6}dt={\int }_{-1}^{1}5t^{6}dt$
${\begin{array}{c}2I=5\times \frac{t^7}{7}\end{array}|}_{-1}^1=5(\frac{1}{7}+\frac{1}{7})$
$\Rightarrow 2I=2\times \frac{5}{7}\Rightarrow I=\frac{5}{7}$