U-Substitution of Indefinite Integrals Examples 1

Recall that if $f(x) = g(x)g'(x)$, then $\int f(x) \: dx$ can be evaluated using the U-Substitution method. We will now look at some examples of this technique.

Example 1

Evaluate the integral $\int (3 - x)^{10} \: dx$.

First let $u = 3 - x$, and therefore $du = -dx$. Making these substitutions we get that:

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\begin{align} \int (3 - x)^{10} \: dx = \int u^{10}(-1)du \\ \int (3 - x)^{10} \: dx = -\frac{u^{11}}{11} \\ \int (3 - x)^{10} \: dx = - \frac{(3 - x)^{11}}{11} + C \end{align}

Example 2

Evaluate the integral $\int \frac{x^3}{(1 + x^4)^{1/3}} \: dx$.

Let $u = 1 + x^4$ and therefore $du = 4x^3 \: dx$. Divide both sides by 4 to get $\frac{du}{4} = x^3 \: dx$, and making the appropriate substitution we obtain:

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\begin{align} \int \frac{x^3}{(1 + x^4)^{1/3}} \: dx = \int \frac{1}{4} \cdot \frac{1}{u^{1/3}} \: du \\ \int \frac{x^3}{(1 + x^4)^{1/3}} \: dx = \frac{1}{4} \int u^{-1/3} \: du \\ \int \frac{x^3}{(1 + x^4)^{1/3}} \: dx = \frac{1}{4} \frac{3u^{2/3}}{2} \\ \int \frac{x^3}{(1 + x^4)^{1/3}} \: dx = \frac{3(1 + x^4)^{2/3}}{8} + C \end{align}

Example 3

Evaluate the integral $\int 4\cos(3x) \: dx$.

Let $u = 3x$, and therefore $du = 3 \: dx$. Divide both sides by 3 to get $\frac{du}{3} = dx$ Making the appropriate substitutions we get:

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\begin{align} \int 4\cos(3x) \: dx = 4 \int \frac{1}{3} \cos u \: du \\ \int 4\cos(3x) \: dx = \frac{4}{3} \cdot \sin u \\ \int 4\cos(3x) \: dx = \frac{4\sin 3x}{3} + C \end{align}

Example 4

Evaluate the integral $\int \frac{\cos 5x}{e^{\sin 5x}} \: dx$.

Let $u = \sin 5x$, and therefore, $du = 5 \cos 5x \: dx$. Divide both sides by 5 and we get that $\frac{du}{5} = \cos 5x$. Making the appropriate substitutions:

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\begin{align} \int \frac{\cos 5x}{e^{\sin 5x}} \: dx = \int \frac{1}{5} \cdot \frac{1}{e^u} \: du \\ \int \frac{\cos 5x}{e^{\sin 5x}} \: dx = \frac{1}{5} \int e^{-u} \: du \\ \int \frac{\cos 5x}{e^{\sin 5x}} \: dx = \frac{-e^{-u}}{5} \\ \int \frac{\cos 5x}{e^{\sin 5x}} \: dx = \frac{-e^{-\sin 5x}}{5} + C \end{align}

Example 5

Evaluate the integral $\int (x + 2)(x - 1)^5 \: dx$.

First let $u = x - 1$ so that $du = dx$. We should also note that $u + 3 = x + 2$. Making the appropriate substitutions we get that:

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\begin{align} \int (x + 3)(x - 1)^5 \: dx = \int (u + 3)(u)^5 \: du \\ \int (x + 3)(x - 1)^5 \: dx = \int u^6 + 3u^5 \: du \\ \int (x + 3)(x - 1)^5 \: dx = \frac{u^7}{7} + 3\frac{u^6}{6} \\ \int (x + 3)(x - 1)^5 \: dx = \frac{(x - 1)^7}{7} + \frac{3(x-1)^6}{6} + C \end{align}