Question
Evaluate the following definite integrals:$\int_{0}^\limits{\frac{\pi}{4}}\text{x}^2\sin\text{x}\text{ dx}$
✓
Answer
Let $\int_{0}^\limits{\frac{\pi}{4}}\text{x}^2\sin\text{x}\text{ dx}$ Then, Integrating by parts.$\text{I}=\big[-\text{x}^2\cos\text{x}\big]^{\frac{\pi}{4}}_0-\int_{0}^\limits{\frac{\pi}{4}}-2\text{x}\cos\text{x}\text{ dx}$
$\Rightarrow\text{I}=\big[-\text{x}^2\cos\text{x}\big]^{\frac{\pi}{4}}_0+\big[2\text{x }\sin\text{x}\big]^{\frac{\pi}{4}}_0-\int_{0}^\limits{\frac{\pi}{4}}2\sin\text{x dx}$
$\Rightarrow\text{I}=\big[-\text{x}^2\cos\text{x}\big]^{\frac{\pi}{4}}_0+\big[2\text{x }\sin\text{x}\big]^{\frac{\pi}{4}}_0+\big[2\cos\text{x}\big]^{\frac{\pi}{4}}_0$
$\Rightarrow\text{I}=\frac{-\pi^2}{16\sqrt{2}}+\frac{\pi}{2\sqrt{2}}+\frac{2}{\sqrt{1}}-2$
$\Rightarrow\text{I}=\sqrt{2}+\frac{\pi}{2\sqrt{2}}-\frac{\pi^2}{16\sqrt{2}}-2$
$\Rightarrow\text{I}=\big[-\text{x}^2\cos\text{x}\big]^{\frac{\pi}{4}}_0+\big[2\text{x }\sin\text{x}\big]^{\frac{\pi}{4}}_0-\int_{0}^\limits{\frac{\pi}{4}}2\sin\text{x dx}$
$\Rightarrow\text{I}=\big[-\text{x}^2\cos\text{x}\big]^{\frac{\pi}{4}}_0+\big[2\text{x }\sin\text{x}\big]^{\frac{\pi}{4}}_0+\big[2\cos\text{x}\big]^{\frac{\pi}{4}}_0$
$\Rightarrow\text{I}=\frac{-\pi^2}{16\sqrt{2}}+\frac{\pi}{2\sqrt{2}}+\frac{2}{\sqrt{1}}-2$
$\Rightarrow\text{I}=\sqrt{2}+\frac{\pi}{2\sqrt{2}}-\frac{\pi^2}{16\sqrt{2}}-2$
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