Tag: Method of Substitution

Evaluate the integral using substitution

by RoRi

October 17, 2015

Use the method of substitution to evaluate the following integral:

$\int x^2 (8x^3 + 27)^{\frac{2}{3}} \, dx.$

Let

$u = 8x^3 + 27 \quad \implies \quad du = 24x^2 \, dx.$

Then, we can evaluate the integral,

$\begin{align*} \int x^2 (8x^3 + 27)^{\frac{2}{3}} \, dx &= \frac{1}{24} \int u^{\frac{2}{3}} \, du \\ &= \frac{1}{40} u^{\frac{5}{3}} + C \\ &= \frac{1}{40} (8x^3 + 27)^{\frac{5}{3}} + C. \end{align*}$

Evaluate the integral using substitution

by RoRi

October 17, 2015

Use the method of substitution to evaluate the following integral:

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

Let

$\begin{align*} u &= (x^2+1)^{-\frac{1}{2}} \quad \implies \quad du = -x(x^2+1)^{-\frac{3}{2}} \\ x &= \frac{\sqrt{1-u^2}}{u}. \end{align*}$

Then, we can evaluate the integral,

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

Now, we make a second substitution. Let

$v = 1-u^2 \quad \implies \quad dv = -2u \, du.$

Then, we continue evaluating the integral,

$\begin{align*} \int \frac{-u}{\sqrt{1-u^2}} \, du &= \frac{1}{2} \int \frac{1}{\sqrt{v}} \, dv \\ &= \sqrt{v} + C \\ &= \sqrt{1-u^2} + C \\ &= \sqrt{1 - \frac{1}{x^2+1}} + C \\[9pt] &= \sqrt{\frac{x^2}{x^2+1}} + C \\[9pt] &= \frac{x}{\sqrt{x^2+1}} + C. \end{align*}$

Evaluate the integral using substitution

by RoRi

October 17, 2015

Use the method of substitution to evaluate the following integral:

$\int t (1+t)^{\frac{1}{4}} \, dt.$

Let

$u = 1 + t \quad \implies \quad du = dt.$

Then, we can evaluate the integral,

$\begin{align*} \int t(t+1)^{\frac{1}{4}} \, dt &= \int (u-1)u^{\frac{1}{4}} \, dt \\ &= \int u^{\frac{5}{4}} \, du - \int u^{\frac{1}{4}} \, du \\[9pt] &= \frac{4}{9} u^{\frac{9}{4}} - \frac{4}{5} u^{\frac{5}{4}} + C \\[9pt] &= \frac{4}{9} (1+t)^{\frac{9}{4}} - \frac{4}{5} (1+t)^{\frac{5}{4}} + C. \end{align*}$

Evaluate the integral using substitution

by RoRi

October 17, 2015

Use the method of substitution to evaluate the following integral:

$\int \frac{x^5 \, dx}{\sqrt{1-x^6}}.$

Let

$u = 1 - x^6 \quad \implies \quad du = -6x^5 \, dx.$

Then, we can evaluate the integral,

$\begin{align*} \int \frac{x^5 \, dx}{\sqrt{1-x^6}} &= -\frac{1}{6} \int u^{-\frac{1}{2}} \, du \\ &= -\frac{1}{3} u^{\frac{1}{2}} + C \\ &= -\frac{1}{3} (1 - x^6)^{\frac{1}{2}} + C. \end{align*}$

Evaluate the integral using substitution

by RoRi

October 17, 2015

Use the method of substitution to evaluate the following integral:

$\int x^{n-1} \sin (x^n) \, dx.$

Let

$u = x^n \quad \implies \quad du = nx^{n-1} \, dx.$

Then, we can evaluate the integral,

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

Evaluate the integral using substitution

by RoRi

October 17, 2015

Use the method of substitution to evaluate the following integral:

$\int_3^8 \frac{\sin \left( \sqrt{x + 1} \right)}{\sqrt{x+1}} \, dx.$

Let

$u = \sqrt{x+1} \quad \implies \quad du = \frac{1}{2} (x+1)^{-\frac{1}{2}} \, dx.$

For the limits of integration we have

$\begin{align*} u(3) &= 2 \\ u(8) &= 3 \end{align*}$

Then, we can evaluate the integral,

$\begin{align*} \int_3^8 \frac{\sin \left( \sqrt{x+1} \right)}{\sqrt{x+1}} \, dx &= 2 \int_2^3 \sin u \, du \\ &= 2 \left( -\cos u \Bigr \rvert_2^3 \right) \\ &= 2 (\cos 2 - \cos 3). \end{align*}$

Evaluate the integral using substitution

by RoRi

October 17, 2015

Use the method of substitution to evaluate the following integral:

$\int \frac{\sin x \, dx}{\sqrt{\cos^3 x}}.$

Let

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

Then, we can evaluate the integral,

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

Evaluate the integral using substitution

by RoRi

October 17, 2015

Use the method of substitution to evaluate the following integral:

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

Let

$u = 3 + \cos x \quad \implies \quad du = -\sin x \, dx.$

Then, we can evaluate the integral,

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

Evaluate the integral using substitution

by RoRi

October 17, 2015

Use the method of substitution to evaluate the following integral:

$\int_0^{\frac{\pi}{4}} \cos (2x) \sqrt{4-\sin (2x)} \, dx.$

Let

$u = 4 - \sin (2x) \quad \implies \quad du = -2\cos (2x) \, dx.$

For the limits of integration we have

$\begin{align*} u \left( \frac{\pi}{4} \right) &= 3 \\ u (0) &= 4 \end{align*}$

Then, we can evaluate the integral,

$\begin{align*} \int_0^{\frac{\pi}{4}} \cos (2x) \sqrt{4 - \sin (2x)} \, dx &= -\frac{1}{2} \int_{u(0)}^{u\left(\frac{\pi}{4}\right)} u^{\frac{1}{2}} \, du \\ &= -\frac{1}{2} \int_4^3 u^{\frac{1}{2}} \, du \\ &= -\frac{1}{3} u^{\frac{3}{2}} \Bigr \rvert_4^3 \\ &= -\frac{1}{3} \left( \sqrt{27} - 8 \right) \\ &= \frac{8}{3} - \frac{\sqrt{27}}{3} \\ &= \frac{8}{3} - \sqrt{3}. \end{align*}$