Evaluate the following definite integrals: _ 0 ^ 2 x^2 ^2x dx

Question

Evaluate the following definite integrals:
$\int\limits_{0}^{\frac{\pi}{2}}\text{x}^2\cos^2\text{x}\text{ dx}$

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Answer

We have,
$\int\text{x}^2\cos^2\text{x dx}=\int\text{x}^2\Big(\frac{1+\cos2\text{x}}{2}\Big)\text{dx}\\=\frac{1}{2}\int(\text{x}^2+\text{x}^2\cos2\text{x})\text{dx}=\frac{1}{2}\big[\int\text{x}^2\text{dx}+\int\text{x}^2\cos2\text{x dx}\big]\ ...(\text{A})$
Now,
$\int_{0}^\limits{\frac{\pi}{2}}\text{x}^2\text{ dx}=\Big[\frac{\text{x}^3}{3}\Big]^{\frac{\pi}{2}}_0=\frac{\pi^3}{24}\ ....(\text{B})$
$\int\text{x}^2\cos2\text{x dx}=\text{x}^2\int\cos2\text{x dx}-\int2\text{x}\big(\int\cos2\text{x dx}\big)$
$=\frac{\text{x}^2\sin2\text{x}}{2}-\int\frac{\sin2\text{x}}{2}2\text{x dx}$
$=\frac{\text{x}^2\sin2\text{x}}{2}-\Big[\text{x}\int\sin2\text{x}-\int\big(\int\sin2\text{x dx}\big)\text{dx}\Big]$
$=\frac{\text{x}^2\sin2\text{x}}{2}+\frac{\text{x}\cos2\text{x}}{2}-\int\frac{\cos2\text{x}}{2}\text{ dx}$
$\therefore\ \int_{0}^\limits{\frac{\pi}{2}}\text{x}^2\cos2\text{x dx}=\Big[\frac{\text{x}^2\sin2\text{x}}{2}+\frac{\text{x}\cos2\text{x}}{2}-\frac{\sin2\text{x}}{4}\Big]^{\frac{\pi}{2}}_0=\frac{-\pi}{4}\ ...(\text{C})$
Now, put (B) & (C) in (A), we get
$\int_{0}^\limits{\frac{\pi}{2}}\text{x}^2\cos^2\text{x}\text{ dx}=\int_{0}^\limits{\frac{\pi}{2}}\text{x}^2\text{ dx}+\int_{0}^\limits{\frac{\pi}{2}}\text{x}^2\cos2\text{x dx}\\=\frac{1}{2}\Big[\frac{\pi^2}{24}-\frac{\pi}{4}\Big]=\frac{\pi^3}{48}-\frac{\pi}{8}$

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