Integration by Parts of Indefinite Integrals Examples 2

We will now look at some more examples of Integration by Parts of Indefinite Integrals. Please check our the Integration by Parts Examples 1 first before reviewing these examples.

(1)

\begin{align} \int u \: dv = uv - \int v \: du \end{align}

Example 1

Evaluate the integral $\int \ln x \: dx$.

At first this example may not look that suitable for integration by parts, however, if we let $u = \ln x$, while when differentiated is simpler, and let $dv = dx$, then we can easily apply integration by parts. Therefore $du = \frac{1}{x} \: dx$ and $v = x$. Substituting this into our integration by parts formula we get:

(2)

\begin{align} \int \ln x \: dx = x \ln x - \int x \frac{1}{x} \: dx \\ \int \ln x \: dx = x \ln x - x \end{align}

Example 2

Evaluate the integral $\int e^x \cos x \: dx$.

We note that choosing either $u = e^x$ or $u = \cos x$ for differentiation with both result in functions that are relatively the same in terms of difficulty. Let's choose $u = e^x$ though, and therefore $dv = \cos x \: dx$. Then $du = e^x \: dx$ and $v = \sin x \: dx$. Substituting this into the integration by parts formula we obtain that:

(3)

\begin{align} \int e^x \cos x \: dx = e^x \sin x - \int e^x \sin x \: dx \\ \end{align}

Now we will apply integration by parts again. Let $u = e^x$ and $dv = \sin x \: dx$. Therefore, $du = e^x \: dx$ and $v = -\cos x$. Substituting this back in we get:

(4)

\begin{align} \int e^x \cos x \: dx = e^x \sin x - \left ( \int e^x \sin x \: dx \right ) \\ \: \int e^x \cos x \: dx = e^x \sin x - \left ( -e^x \cos x - \int -e^x \cos x \: dx \right ) \\ \: \int e^x \cos x \: dx = e^x \sin x + e^x \cos x - \int e^x \cos x \: dx \\ 2 \int e^x \cos x \: dx = e^x \sin x + e^x \cos x \\ \int e^x \cos x \: dx = \frac{1}{2} \left ( e^x \sin x + e^x \cos x \right ) + C \end{align}

Example 3

Evaluate the integral $\int \sin ^{-1} (3x) \: dx$.

We will choose $u = \sin ^{-1} (3x)$ since when differentiated, becomes simpler. Therefore, $dv = dx$. Therefore, $du = \frac{3}{\sqrt{1 - (3x)^2}}$ and let $v = x$. Making the appropriate substitutions we get that:

(5)

\begin{align} \int \sin^{-1} (3x) \: dx = x\sin ^{-1} (3x) - \int \frac{3x}{\sqrt{1 - 9x^2}} \: dx \end{align}

We will now use the technique of U-Substitution of Indefinite Integrals. Let $u = 1 - 9x^2$ so that $du = -18x \: dx$ and $\frac{-1}{6} du = 3x \: dx$. Therefore:

(6)

\begin{align} \int \sin^{-1} (3x) \: dx = x\sin ^{-1} (3x) - \int \frac{-1}{6} \frac{1}{\sqrt{u}} \: du \\ \int \sin^{-1} (3x) \: dx = x\sin ^{-1} (3x) + \frac{1}{6} \int \frac{1}{\sqrt{u}} \: du \\ \int \sin^{-1} (3x) \: dx = x\sin ^{-1} (3x) + \frac{1}{6} \int u^{-1/2} \: du \\ \int \sin^{-1} (3x) \: dx = x\sin ^{-1} (3x) + \frac{1}{6} \frac{u^{1/2}}{(1/2)} \: du \\ \int \sin^{-1} (3x) \: dx = x\sin ^{-1} (3x) + \frac{1}{3} (1 - 9x^2)^{1/2} + C \end{align}

Example 4

Evaluate the integral $\int 2x \tan ^{-1} x \: dx$.

We will choose $u = \tan^{-1} x$ since when differentiated it becomes simpler. Therefore $dv = 2x \: dx$. Thus, $du = \frac{1}{1 + x^2}$ and $v = x^2$. Substituting into the integration by parts formula we obtain that:

(7)

\begin{align} \int 2x \tan^{-1} x \: dx = x^2\tan^{-1} x - \int \frac{x^2}{1 + x^2} \: dx \end{align}

Now note that $\frac{x^2}{1 + x^2} = \frac{x^2 + 1 - 1}{1 + x^2} = \frac{(x^2 + 1) - 1}{(x^2 + 1)} = 1 - \frac{1}{1 + x^2}$. Therefore:

(8)

\begin{align} \int 2x \tan^{-1} x \: dx = x^2\tan^{-1} x - \int 1 - \frac{1}{1 + x^2} \: dx \\ \int 2x \tan^{-1} x \: dx = x^2\tan^{-1} x - \left ( \int 1 \: dx - \int \frac{1}{1 + x^2} \: dx \right )\\ \int 2x \tan^{-1} x \: dx = x^2\tan^{-1} x - x + \tan ^{-1} x + C \end{align}