Question

$\underset{0}{\overset{\pi }{\int }}\frac{x^3{\cos }^{4}x{\sin }^{2}x}{({\pi }^2-3\pi x+3x^2)}dx$ is equal to

• A. $\frac{{\pi }^2}{8}$
• B. $\frac{{\pi }^2}{32}$
• C. $\frac{{\pi }^2}{16}$
• D. $\frac{{\pi }^2}{64}$

Answer 1

The correct option is B $\frac{{\pi }^2}{32}$

Let, $I=\underset{0}{\overset{\pi }{\int }}\frac{x^3{\cos }^{4}x{\sin }^{2}x}{({\pi }^2-3\pi x+3x^2)}dx\cdots \text{(i)}$
Using property $\underset{0}{\overset{a}{\int }}f\left(x\right)dx=\underset{0}{\overset{a}{\int }}f(a-x)dx$
$I=\underset{0}{\overset{\pi }{\int }}\frac{(\pi -x{)}^3{\cos }^4(\pi -x\left){\sin }^2\right(\pi -x)}{{\pi }^2-3\pi (\pi -x)+3(\pi -x{)}^2}dx$
$=\underset{0}{\overset{\pi }{\int }}\frac{({\pi }^3-x^3-3{\pi }^{2}x+3\pi x^2){\cos }^{4}x{\sin }^{2}x}{({\pi }^2-3\pi x+3x^2)}dx\cdots \text{(ii)}$
Adding $\left(i\right)$ and $\left(ii\right)$ we have
$2I=\underset{0}{\overset{\pi }{\int }}\frac{({\pi }^3-3{\pi }^{2}x+3\pi x^2){\cos }^{4}x{\sin }^{2}x}{({\pi }^2-3\pi x+3x^2)}dx$
$\Rightarrow 2I=\pi \underset{0}{\overset{\pi }{\int }}{\cos }^{4}x{\sin }^{2}xdx$
$\Rightarrow 2I=2\pi \underset{0}{\overset{\pi /2}{\int }}{\cos }^{4}x{\sin }^{2}xdx\therefore I=\pi \underset{0}{\overset{\pi /2}{\int }}{\cos }^{4}x{\sin }^{2}xdx$
Using walli's formula, we get
$I=\pi \frac{(3\cdot 1)\left(1\right)}{6\cdot 4\cdot 2}\cdot \frac{\pi }{2}=\frac{{\pi }^2}{32}\{\because I_{m,n}=\frac{(m-1)(m-3)(m-5)\cdots (n-1)(n-3)(n-5)}{(m+n)(m+n-2)(m+n-4)\cdots }\cdot \frac{\pi }{2}\text{, when both}m,n\text{is even}\}$