## Problem Statement

Calculate the indefinite integral:

$$
\int x \cdot \sin ^{2} x d x
$$
## Solution

$$
\int x \cdot \sin ^{2} x d x=
$$

Let's denote:

$$
\begin{aligned}
& u=x \cdot \sin x ; d u=(\sin x+x \cdot \cos x) d x \\
& d v=\sin x d x ; v=-\cos x
\end{aligned}
$$

We will use the integration by parts formula $\int u d v=u v-\int v d u$. We get:

$$
\begin{aligned}
& =-\cos x \cdot x \cdot \sin x-\int(-\cos x) \cdot(\sin x+x \cdot \cos x) d x= \\
& =-\frac{x}{2} \cdot \sin 2 x+\int \cos x \cdot \sin x \cdot d x+\int x \cdot \cos ^{2} x \cdot d x= \\
& =-\frac{x}{2} \cdot \sin 2 x+\int \frac{1}{2} \cdot \sin 2 x \cdot d x+\int x \cdot\left(1-\sin ^{2} x\right) d x= \\
& =-\frac{x}{2} \cdot \sin 2 x+\frac{1}{2} \cdot\left(-\frac{1}{2} \cdot \cos 2 x\right)+\int x d x-\int x \cdot \sin ^{2} x d x= \\
& =-\frac{x}{2} \cdot \sin 2 x-\frac{1}{4} \cdot \cos 2 x+\frac{x^{2}}{2}+C_{1}-\int x \cdot \sin ^{2} x d x
\end{aligned}
$$

Thus, we get:

$$
\begin{aligned}
& \int x \cdot \sin ^{2} x d x=-\frac{x}{2} \cdot \sin 2 x-\frac{1}{4} \cdot \cos 2 x+\frac{x^{2}}{2}+C_{1}-\int x \cdot \sin ^{2} x d x \\
& 2 \cdot \int x \cdot \sin ^{2} x d x=-\frac{x}{2} \cdot \sin 2 x-\frac{1}{4} \cdot \cos 2 x+\frac{x^{2}}{2}+C_{1} \\
& \int x \cdot \sin ^{2} x d x=-\frac{x}{4} \cdot \sin 2 x-\frac{1}{8} \cdot \cos 2 x+\frac{x^{2}}{4}+\frac{C_{1}}{2}
\end{aligned}
$$

Without loss of generality, let $C=\frac{C_{1}}{2}:$

$$
\int x \cdot \sin ^{2} x d x=-\frac{x}{4} \cdot \sin 2 x-\frac{1}{8} \cdot \cos 2 x+\frac{x^{2}}{4}+C
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+1-30$ »

Categories: Kuznetsov's Problem Book Integrals Problem 1 | Integrals

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## Problem Kuznetsov Integrals 1-31

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