## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{x \cdot \cos x}{\sin ^{3} x} d x
$$
## Solution

$$
\int \frac{x \cdot \cos x}{\sin ^{3} x} d x=
$$

Let's denote:

$$
\begin{aligned}
& u=\frac{x \cdot \cos x}{\sin x}=x \cdot \operatorname{ctg} x ; d u=\left(\operatorname{ctg} x-\frac{x}{\sin ^{2} x}\right) d x \\
& d v=\frac{1}{\sin ^{2} x} d x ; v=-\operatorname{ctg} x
\end{aligned}
$$

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

$$
\begin{aligned}
& =-\operatorname{ctg} x \cdot x \cdot \operatorname{ctg} x-\int(-\operatorname{ctg} x) \cdot\left(\operatorname{ctg} x-\frac{x}{\sin ^{2} x}\right) d x= \\
& =-\operatorname{ctg}^{2} x \cdot x+\int \operatorname{ctg}^{2} x \cdot d x-\int \frac{x \cdot \operatorname{ctg} x}{\sin ^{2} x} d x= \\
& =-\operatorname{ctg}^{2} x \cdot x+\int \frac{\cos ^{2} x}{\sin ^{2} x} d x-\int \frac{x \cdot \cos x}{\sin ^{3} x} d x= \\
& =-\operatorname{ctg}^{2} x \cdot x+\int \frac{1-\sin ^{2} x}{\sin ^{2} x} d x-\int \frac{x \cdot \cos x}{\sin ^{3} x} d x= \\
& =-\operatorname{ctg}^{2} x \cdot x+\int \frac{1}{\sin ^{2} x} d x-\int d x-\int \frac{x \cdot \cos x}{\sin ^{3} x} d x= \\
& =-\operatorname{ctg}^{2} x \cdot x-\operatorname{ctg} x-x+C_{1}-\int \frac{x \cdot \cos x}{\sin ^{3} x} d x
\end{aligned}
$$

Thus, we have:

$$
\begin{aligned}
& \int \frac{x \cdot \cos x}{\sin ^{3} x} d x=-\operatorname{ctg}^{2} x \cdot x-\operatorname{ctg} x-x+C_{1}-\int \frac{x \cdot \cos x}{\sin ^{3} x} d x \\
& 2 \cdot \int \frac{x \cdot \cos x}{\sin ^{3} x} d x=-\operatorname{ctg}^{2} x \cdot x-\operatorname{ctg} x-x+C_{1}
\end{aligned}
$$

$$
\begin{aligned}
& 2 \cdot \int \frac{x \cdot \cos x}{\sin ^{3} x} d x=-x \cdot\left(\operatorname{ctg}^{2} x+1\right)-\operatorname{ctg} x+C_{1} \\
& 2 \cdot \int \frac{x \cdot \cos x}{\sin ^{3} x} d x=-x \cdot \frac{1}{\sin ^{2} x}-\frac{\cos x}{\sin x}+C_{1} \\
& 2 \cdot \int \frac{x \cdot \cos x}{\sin ^{3} x} d x=-\frac{x+\cos x \cdot \sin x}{\sin ^{2} x}+C_{1} \\
& \int \frac{x \cdot \cos x}{\sin ^{3} x} d x=-\frac{x+\cos x \cdot \sin x}{2 \sin ^{2} x}+\frac{C_{1}}{2}
\end{aligned}
$$

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

$$
\int \frac{x \cdot \cos x}{\sin ^{3} x} d x=-\frac{x+\cos x \cdot \sin x}{2 \sin ^{2} x}+C
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_1-31$ » Categories: Kuznetsov's Problem Book Integrals Problem 1 | Integrals

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- Last edited: 00:34, 23 February 2009.
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