#### Provide Solution For R.D.Sharma Maths Class 12 Chapter 19 definite Integrals Exercise 19.1 Question 29 Maths Textbook Solution.

Answer: $\sqrt{2}+\frac{\pi }{2\sqrt{2}}-\frac{\pi ^{2}}{16\sqrt{2}}-2$

Hint: Use indefinite formula then put the limit to solve this integral

Given: $\int_{0}^{\frac{\pi }{4}}x^{2}\sin xdx$

Solution:

$\int_{0}^{\frac{\pi }{4}}x^{2}\sin xdx$

Integrating by parts then, $x^{2}$ and sin x be the 2 parts

\begin{aligned} =&\left[x^{2} \int \sin x d x\right]_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}}\left(\frac{d\left(x^{2}\right)}{d x} \int \sin x d x\right) d x \\ =&\left[x^{2}(-\cos x)\right]_{0}^{\frac{\pi}{4}}-\int_{0}^{\frac{\pi}{4}}[2 x(-\cos x) d x] \end{aligned} \quad\left[\int \sin x d x=-\cos x\right]

$=-\left[x^{2} \cos x\right]_{0}^{\frac{\pi}{4}}+2 \int_{0}^{\frac{\pi}{4}} x \cos x d x$

(Again using integration by parts method x and cos x be the 2 terms)

$=-\left[\frac{\pi^{2}}{4^{2}} \cos \frac{\pi}{4}-0 \times \cos 0\right]+2\left[x \int \cos x d x\right]_{0}^{\frac{\pi}{4}}-2 \int_{0}^{\frac{\pi}{4}}\left(\frac{d(x)}{d x} \int \cos x d x\right) d x$

$=-\left[\frac{\pi^{2}}{16} \frac{1}{\sqrt{2}}-0\right]+2[x \sin x]_{0}^{\frac{\pi}{4}}-2 \int_{0}^{\frac{\pi}{4}} 1 \cdot \sin x d \quad\left[\begin{array}{l} \int \cos x d x=\sin x \\ \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}, \cos 0=1 \end{array}\right]$

\begin{aligned} &=\frac{-\pi^{2}}{16 \sqrt{2}}+2\left[\frac{\pi}{4} \sin \frac{\pi}{4}-0 \times \sin 0\right]-2 \int_{0}^{\frac{\pi}{4}} \sin x d x \\ &=\frac{-\pi^{2}}{16 \sqrt{2}}+2\left[\frac{\pi}{4} \frac{1}{\sqrt{2}}-0\right]-2[-\cos x]_{0}^{\frac{\pi}{4}} \end{aligned}\left[\begin{array}{l} \int \sin x d x=\cos x \\ \sin \frac{\pi}{4}=\frac{1}{\sqrt{2}}, \sin 0=0 \end{array}\right]

\begin{aligned} &=\frac{-\pi^{2}}{16 \sqrt{2}}+2\left[\frac{\pi}{4 \sqrt{2}}\right]+2[\cos x]_{0}^{\frac{\pi}{4}} \\ &=\frac{-\pi^{2}}{16 \sqrt{2}}+\frac{2 \pi}{4 \sqrt{2}}+2\left[\cos \frac{\pi}{4}-\cos 0\right] \quad\left[\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}, \cos 0=1\right] \end{aligned}

\begin{aligned} &=\frac{-\pi^{2}}{16 \sqrt{2}}+\frac{2 \pi}{4 \sqrt{2}}+2\left[\frac{1}{\sqrt{2}}-1\right] \\ &=\frac{-\pi^{2}}{16 \sqrt{2}}+\frac{2 \pi}{4 \sqrt{2}}+\frac{\sqrt{2} \times \sqrt{2}}{\sqrt{2}}-2 \\ &=\sqrt{2}+\frac{\pi}{2 \sqrt{2}}-\frac{\pi^{2}}{16 \sqrt{2}}-2 \end{aligned}