Let \(u\) and \(v\) be differentiable functions of \(x\). Then the differential of the product \(uv\) is found by the product rule

$$d(uv)=udv+vdu$$

Whence, by integration, we have

$$uv=\int{udv}+\int{vdu}$$

Re-arranging leads to the formula

$$\bbox[5pt, border:2px solid blue]
{
\int{udv}=uv-\int{vdu}\qquad{ (1) }
}
$$

This formula is called integration by parts. It is often used to integrate expressions, that may be represented as a product of two factors \(u\) and \(dv\), in such a way that the finding of the function \(v\) from its differential \(dv\) and the evaluation of the integral \(\int{vdu}\) taken together, be a simpler problem than the direct evaluation of the integral \(\int{udv}\). To become skilled at breaking up a given element of integration into factors \(u\) and \(dv\), one has to solve problems.

Example 1. \(\begin{aligned}
\int{x\sin {x}dx}
\end{aligned}\)

Solution. Let \(u=x\) and \(dv=\sin {x}\). Then, \(du=1\) and \(v=-\cos {x}\).

\(\begin{aligned}
I&=-x\cos {x}+\int{\cos {x}dx}\\
&=-x\cos {x}+\sin {x}+C\\
\end{aligned}\)

Example 2. \(\begin{aligned}
\int{x\cos {x}dx}
\end{aligned}\)

Solution. Let \(u=x\) and \(dv=\cos {x}\). Then, \(du=1\) and \(v=\sin {x}\).

\(\begin{aligned}
I&=x\sin {x}-\int{\sin {x}dx}\\
&=x\sin {x}+\cos {x}+C\\
\end{aligned}\)

The rule of integration by parts is widely used. For example, integrals of the form

$$\int{x^{k}\sin {ax}}, \int{x^{k}\cos {ax}}$$

$$\int{x^{k}e^{ax}}, \int{x^{k}\ln{ax}}$$

and certain integrals containing inverse trignometric functions are evaluated by means of integration by parts.

Example 3. \(\begin{aligned}
\int{x^{3}\sin {x}dx}
\end{aligned}\)

Solution. In the table below, the first column represents \(x^{3}\) and its derivatives and the second column represents \(\sin {x}\) and its integrals.

$$
\begin{array}{ccc}
x^{3} & & \sin x \\
\hline
\\
3x^{2} & & -\cos x \\
& + &\\
6x & & -\sin x \\
& – &\\
6 & & \cos x \\
& + &\\
0 & & \sin x
\end{array}
$$

This table shows the repeated application of integration by parts. Following the table, the final answer is given by:

\(\begin{aligned}
I = -3x^{2}\sin x -6x\cos x + 6\sin x
\end{aligned}\)

Example 4. \(\begin{aligned}
\int{x^{4}\cos {x}dx}
\end{aligned}\)

Solution. In the table below, the first column represents \(x^{4}\) and its derivatives and the second column represents \(\cos {x}\) and its integrals.

$$
\begin{array}{ccc}
x^{4} & & \cos x \\
\hline
\\
4x^{3} & & \sin x \\
& + &\\
12x^{2} & & -\cos x \\
& – &\\
24x & & -\sin x \\
& + &\\
24 & & \cos x\\
& – &\\
0 & & \sin x\\
\end{array}
$$

Following the table, the final answer is given by:

\(\begin{aligned}
I = -4x^{3}\cos x +12x^{2}\sin x + 24\cos x – 24\sin x
\end{aligned}\)

Example 5. \(\begin{aligned}
\int{x^{2}e^{x}dx}
\end{aligned}\)

Solution. In the table below, the first column represents \(x^{2}\) and its derivatives and the second column represents \(e^{x}\) and its integrals.

$$
\begin{array}{ccc}
x^{2} & & e^{x} \\
\hline
\\
2x & & e^{x} \\
& + &\\
2 & & e^{x} \\
& – &\\
0 & & e^{x}
\end{array}
$$

Following the table, the final answer is given by:

\(\begin{aligned}
I = 2xe^{x}-2e^{x}
\end{aligned}\)

Example 6. \(\begin{aligned}
\int{x^{n}\ln^{x}dx}
\end{aligned}\)

Solution. Let \(u=\ln x\) and \(dv=x^{n}\). Then, \(du=\frac{1}{x}\) and \(v=\frac{x^{n+1}}{n+1}\).

\(\begin{aligned}
I&=\ln x\cdot\frac{x^{n+1}}{n+1}-\frac{1}{n+1}\int{x^{n}dx}\\
&=\ln x\cdot\frac{x^{n+1}}{n+1}-\frac{x^{n+1}}{(n+1)^{2}}+C\\
\end{aligned}\)

Example 7. \(\begin{aligned}
\int{\arcsin x dx}
\end{aligned}\)

Solution. Let \(u=\arcsin x\) and \(dv=1\). Then, \(du=\frac{1}{\sqrt{1-x^{2}}}\) and \(v=x\).

\(\begin{aligned}
I&=x\arcsin {x} – \int{\frac{x}{\sqrt{1-x^{2}}}dx}\\
&=x\arcsin x + \sqrt{1-x^{2}}+C\\
\end{aligned}\)

Example 7. \(\begin{aligned}
\int{\arccos x dx}
\end{aligned}\)

Solution. Let \(u=\arccos x\) and \(dv=1\). Then, \(du=-\frac{1}{\sqrt{1-x^{2}}}\) and \(v=x\).

\(\begin{aligned}
I&=x\arccos {x} + \int{\frac{x}{\sqrt{1-x^{2}}}dx}\\
&=x\arccos {x} – \sqrt{1-x^{2}}+C\\
\end{aligned}\)

Example 8. \(\begin{aligned}
\int{\arctan x dx}
\end{aligned}\)

Solution. Let \(u=\arctan x\) and \(dv=1\). Then, \(du=\frac{1}{1+x^{2}}\) and \(v=x\).

\(\begin{aligned}
I&=x\arctan {x} – \int{\frac{x}{1+x^{2}}dx}\\
&=x\arctan {x} – \sqrt{1+x^{2}}+C\\
\end{aligned}\)

Example 9. \(\begin{aligned}
\int{\ln x dx}
\end{aligned}\)

Solution. Let \(u=\ln x\) and \(dv=1\). Then, \(du=\frac{1}{x}\) and \(v=x\).

\(\begin{aligned}
I&=x\ln{x} – \int{dx}\\
&=x\ln{x} – x + C
\end{aligned}\)

Example 10. \(\begin{aligned}
\int{\ln (1-x) dx}
\end{aligned}\)

Solution. Let \(u=\ln (1-x)\) and \(dv=1\). Then, \(du=-\frac{1}{1-x}\) and \(v=x\).

\(\begin{aligned}
I&=x\ln{(1-x)} + \int{\frac{xdx}{1-x}}\\
&=x\ln{(1-x)} – \int{\frac{(1-x-1)dx}{(1-x)}}\\
&=x\ln{(1-x)} – \int{dx} + \int{\frac{dx}{1-x}}\\
&=x\ln{(1-x)} – x – \ln(1-x)\\
\end{aligned}\)

Example 11. \(\begin{aligned}
\int{\frac{x^{2}}{\sqrt{1-x^{2}}} dx}
\end{aligned}\)

Solution. Let \(u=x\) and \(dv=\frac{x}{\sqrt{1-x^{2}}}\). Then, \(du=1\) and \(v=-\sqrt{1-x^{2}}\).

\(\begin{aligned}
I&=-x\sqrt{1-x^{2}} + \int{\sqrt{1-x^{2}}dx}\\
&=-x\sqrt{1-x^{2}} + \int{\sqrt{1-x^{2}}\cdot\frac{\sqrt{1-x^{2}}}{\sqrt{1-x^{2}}}dx}\\
&=-x\sqrt{1-x^{2}} + \int{\frac{1-x^{2}}{\sqrt{1-x^{2}}}dx}\\
&=-x\sqrt{1-x^{2}} + \int{\frac{dx}{\sqrt{1-x^{2}}}} – \int{\frac{x^{2}dx}{\sqrt{1-x^{2}}}}\\
&=-x\sqrt{1-x^{2}} + \arcsin {x} – I\\
2I&=-x\sqrt{1-x^{2}} + \arcsin {x}\\
I&=-\frac{x}{2}\sqrt{1-x^{2}} + \frac{1}{2}\arcsin {x}
\end{aligned}\)

Example 12. \(\begin{aligned}
\int{\sqrt{r^{2}-x^{2}} dx}
\end{aligned}\)

Solution. We multiply the numerator and denominator by \(\sqrt{r^{2}-x^{2}}\). Consequently, we have :

\(\begin{aligned}
I&=\int{\sqrt{r^{2}-x^{2}}\times\frac{\sqrt{r^{2}-x^{2}}}{\sqrt{r^{2}-x^{2}}}dx}\\
&=\int{\frac{r^{2}-x^{2}}{\sqrt{r^{2}-x^{2}}}dx}\\
&=r^{2}\int{\frac{dx}{\sqrt{r^{2}-x^{2}}}}-\int{\frac{x^{2}dx}{\sqrt{r^{2}-x^{2}}}}\\
&=r^{2}\arcsin {x} – \int{\frac{x^{2}dx}{\sqrt{r^{2}-x^{2}}}}\\
\end{aligned}\)

In the latter integral, let \(u=x\) and \(dv=\frac{xdx}{\sqrt{r^{2}-x^{2}}}\). Then, \(du=1\) and \(v=-\sqrt{r^{2}-x^{2}}\). Therefore, we have

\(\begin{aligned}
I&=r^{2}\arcsin {x} – \int{x\cdot\frac{x}{\sqrt{r^{2}-x^{2}}}dx}\\
&=r^{2}\arcsin {x} + x\sqrt{r^{2}-x^{2}}- \int{\sqrt{r^{2}-x^{2}}dx}\\
&=r^{2}\arcsin {x} + x\sqrt{r^{2}-x^{2}} – I\\
2I&=r^{2}\arcsin {x} + x\sqrt{r^{2}-x^{2}}\\
I&=\frac{r^{2}}{2}\arcsin {x}+\frac{x}{2}\sqrt{r^{2}-x^{2}}+C
\end{aligned}\)

Example 13. \(\begin{aligned}
\int{x \arcsin {x} dx}
\end{aligned}\)

Solution. Let \(u=\arcsin {x}\) and \(dv=x\). Then, \(du=\frac{1}{\sqrt{1-x^{2}}}\) and \(v=\frac{x^{2}}{2}\).

\(\begin{aligned}
I&=\frac{x^{2}}{2}\arcsin {x}-\frac{1}{2}\left[\int{\frac{x^{2}dx}{\sqrt{1-x^{2}}}}\right]\\
&=\frac{x^{2}}{2}\arcsin {x}-\frac{1}{2}I_{1}\\
\end{aligned}\)

In the latter integral \(I_{1}\), we let \(u=x\) and \(dv=\frac{xdx}{\sqrt{1-x^{2}}}\). Then, \(du=1\) and \(v=-\sqrt{1-x^{2}}\). Consequently, we have

\(\begin{aligned}
I_{1}&=\int{x\cdot\frac{xdx}{\sqrt{1-x^{2}}}}\\
&=-x\sqrt{1-x^{2}}+\int{\sqrt{1-x^{2}}dx}\\
&=-x\sqrt{1-x^{2}}+\int{\sqrt{1-x^{2}}\times\frac{\sqrt{1-x^{2}}}{\sqrt{1-x^{2}}}dx}\\
&=-x\sqrt{1-x^{2}}+\int{\frac{1-x^{2}}{\sqrt{1-x^{2}}}dx}\\
&=-x\sqrt{1-x^{2}}+\int{\frac{dx}{\sqrt{1-x^{2}}}}-\int{\frac{x^{2}}{\sqrt{1-x^{2}}}}\\
&=-x\sqrt{1-x^{2}}+\arcsin {x} – I_{1}\\
2I_{1}&=-x\sqrt{1-x^{2}}+\arcsin {x}\\
I_{1}&=-\frac{x}{2}\sqrt{1-x^{2}}+\frac{1}{2}\arcsin {x}+C
\end{aligned}\)

The final answer is given by,

\(\begin{aligned}
I&=\frac{x^{2}}{2}\arcsin {x}-\frac{1}{2}\left[-\frac{x}{2}\sqrt{1-x^{2}}+\frac{1}{2}\arcsin {x}\right]+C\\
&=\frac{x^{2}}{2}\arcsin {x}+\frac{x}{4}\sqrt{1-x^{2}}-\frac{1}{4}\arcsin {x}+C\\
\end{aligned}\)