Tag: Method of Substitution

Evaluate the integral using substitution

by RoRi

October 17, 2015

Use the method of substitution to evaluate the following integral:

$\int z(z-1)^{\frac{1}{3}} \, dz.$

Let

$u = z - 1 \quad \implies \quad du = dz.$

Then, we can evaluate the integral,

$\begin{align*} \int z(z-1)^{\frac{1}{3}} &= \int (u+1)u^{\frac{1}{3}} \, du \\ &= \int u^{\frac{4}{3}} \, du + \int u^{\frac{1}{3}} \, du \\ &= \frac{3}{7} u^{\frac{7}{3}} + \frac{3}{4} u^{\frac{4}{3}} + C\\ &= \frac{3}{7} (z-1)^{\frac{7}{3}} + \frac{3}{4} (z-1)^{\frac{4}{3}} + C. \end{align*}$

Evaluate the integral using substitution

by RoRi

October 17, 2015

Use the method of substitution to evaluate the following integral:

$\int \sin^3 x \, dx.$

First, we rearrange the integral to get something easier to work with,

$\begin{align*} \int \sin^3 x \, dx &= \int ( \sin x \sin^2 x ) \, dx \\ &= \int \sin x (1 - \cos^2 x) \, dx \\ &= \int \sin x \, dx - \int \sin x \cos^2 x \, dx \\ \end{align*}$

Now, let

$u = \cos x \quad \implies \quad du = - \sin x \, dx.$

Then, we can evaluate the integral,

$\begin{align*} \int \sin^3 x \, dx &= \int \sin x \, dx - \int \sin x \cos^2 x \, dx \\ &= -\cos x + \int u^2 \, du \\ &= -\cos x + \frac{1}{3}u^3 + C \\ &= \frac{1}{3} \cos^3 x - \cos x + C. \end{align*}$

Evaluate the integral using substitution

by RoRi

October 17, 2015

Use the method of substitution to evaluate the following integral:

$\int \frac{(x+1)\, dx}{(x^2+2x+3)^3}.$

Let

$u = x^2 + 2x + 3 \quad \implies \quad du = 2(x + 1) \, dx.$

Then, we can evaluate the integral,

$\begin{align*} \int \frac{(x+1) \, dx}{(x^2+ 2x + 3)^3} &= \frac{1}{2} \int u^{-3} \, du \\ &= -\frac{1}{4} u^{-2} + C \\ &= -\frac{1}{4} (x^2 + 2x + 3)^{-2} + C. \end{align*}$

Evaluate the integral using substitution

by RoRi

October 17, 2015

Use the method of substitution to evaluate the following integral:

$\int_{-\frac{2}{3}}^{\frac{1}{3}} \frac{x \, dx}{\sqrt{2-3x}}.$

Let

$\begin{align*} u &= 2 - 3x &\implies && x &= \frac{2-u}{3} \\ du &= -3 \, dx & \implies && dx &= -\frac{1}{3} \, du. \end{align*}$

For the limits of integration under this substitution we have

$\begin{align*} u \left( -\frac{2}{3} \right) &= 4 \\ u \left( \frac{1}{3} \right) &= 1 \end{align*}$

Then, we can evaluate the integral,

$\begin{align*} \int_{-\frac{2}{3}}^{\frac{1}{3}} \frac{x \, dx}{\sqrt{2 - 3x}} &= -\frac{1}{3} \int_{u \left(-\frac{2}{3} \right)}^{u \left( \frac{1}{3} \right)} \left( \frac{1}{3}(2 - u)u^{-\frac{1}{2}} \right) \, du \\ &= -\frac{1}{9} \int_4^1 \left( 2u^{-\frac{1}{2}} - u^{\frac{1}{2}}\right) \, du \\ &= -\frac{1}{9} \left. \left( 4u^{\frac{1}{2}} - \frac{2}{3}u^{\frac{3}{2}} \right) \right|_4^1 \\ &= -\frac{1}{9} \left( 4 - \frac{2}{3} - 8 + \frac{16}{3} \right) \\ &= -\frac{2}{27}. \end{align*}$

Evaluate the integral using substitution

by RoRi

October 16, 2015

Use the method of substitution to evaluate the following integral:

$\int x^2 \sqrt{x+1} \, dx.$

Let

$u = x + 1 \qquad \implies \qquad du = dx.$

Then,

$\begin{align*} \int x^2 \sqrt{x+1} \, dx &= \int (u-1)^2 \sqrt{u} \, du \\ &= \int \left(u^{\frac{5}{2}} - 2u^{\frac{3}{2}} + u^{\frac{1}{2}} \right) \, du \\[9pt] &= \frac{2}{7} u^{\frac{7}{2}} - \frac{4}{5} u^{\frac{5}{2}} + \frac{2}{3} u^{\frac{3}{2}} + C \\[9pt] &= \frac{2}{7} (x+1)^{\frac{7}{2}} - \frac{4}{5} (x+1)^{\frac{5}{2}} + \frac{2}{3} (x+1)^{\frac{3}{2}} + C. \end{align*}$

Evaluate the integral using substitution

by RoRi

October 16, 2015

Use the method of substitution to evaluate the following integral:

$\int x \sqrt{1 + 3x} \, dx.$

Let

$\begin{align*} u &= 1 + 3x & \implies && x = \frac{u-1}{3} \\ du &= 3 \, dx & \implies && dx = \frac{1}{3} \, du. \end{align*}$

Then, we have

$\begin{align*} \int x \sqrt{1+3x} \, dx &= \int \left( \frac{u-1}{3} \sqrt{u} \right) \frac{du}{3} \\ &= \frac{1}{9} \int \left( u^{\frac{3}{2}} - u^{\frac{1}{2}} \right) \, du \\ &= \frac{1}{9} \left( \int u^{\frac{3}{2}} \, du - \int u^{\frac{1}{2}} \, du \right)\\ &= \frac{1}{9} \left( \frac{2}{5} u^{\frac{5}{2}} - \frac{2}{3} u^{\frac{3}{2}} \right) + C \\ &= \frac{2}{45} (1+3x)^{\frac{5}{2}} - \frac{2}{27} (1+3x)^{\frac{3}{2}} + C. \end{align*}$

Evaluate the integral using substitution

by RoRi

October 16, 2015

Use the method of substitution to evaluate the following integral:

$\int \sqrt{2x+1} \, dx.$

Let

$u = 2x + 1 \qquad \implies \qquad du = 2 \, dx.$

So, using substitution

$\begin{align*} \int \sqrt{2x+1} \, dx &= \int \frac{\sqrt{u}}{2} \, du \\ &= \left(\frac{1}{2} \right) \left( \frac{2}{3} \right) u^{\frac{3}{2}} + C \\ &= \frac{1}{3}(2x+1)^{\frac{3}{2}} + C. \end{align*}$