Question

The value of $\underset{0}{\overset{\pi }{\int }}\frac{5x^3\sin x{\cos }^{6}x}{({\pi }^3-3{\pi }^{2}x+3\pi x^2)}dx$ is

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

Answer 1

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

$I={\int }_0^{\pi }\frac{5x^{3}co{s}^{6}xsinx}{{\pi }^3-3{\pi }^{3}x+3\pi x^2-x^3+x^3}$
$I={\int }_0^{\pi }\frac{5x^{3}co{s}^{6}xsinx}{(\pi -x{)}^3+x^3}dx$ ...(1)
${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx.$
$I={\int }_0^{\pi }\frac{5(\pi -x{)}^{3}co{s}^{6}xsinx}{x^3+(\pi -x)3}$ ...(2)
(1) + (2)
$2I={\int }_0^{\pi }5co{s}^{6}xsinx\times \left(\frac{x^3+(\pi -x{)}^3}{x^3+(\pi -x{)}^3}\right)dx$
$\Rightarrow 2I={\int }_0^{\pi }5co{s}^{6}xsinxdx$
Let $cosx=t\Rightarrow sinx=-dt$
$2I=-{\int }_1^{-1}5t^{2}dt={\int }_{-1}^{1}5t^{6}dt$
${\begin{array}{c}2I=5\frac{t^7}{7}\end{array}|}_{-1}^1=5(\frac{1}{7}+\frac{1}{7})$
$\Rightarrow 2I=2\times \frac{5}{7}\Rightarrow I=\frac{5}{7}.$