Find the integral for the following. a. \int ( \sqrt[3]{x}-\frac{1}{\sqrt[3]{x}} )^2 dx b. \int...

Question:

Find the integral for the following.

{eq}\displaystyle a. \int \left ( \sqrt[3]{x}-\frac{1}{\sqrt[3]{x}} \right )^2\ dx\\ \displaystyle b. \int \frac{x^3}{\sqrt{x^2-1}} \ dx\\ \displaystyle c. \int_0^{\frac{x}{12}} \tan 3x \ dx {/eq}

Integration


{eq}\displaystyle{\text{Power Rule:}\int x^n \ dx = \frac{x^{n + 1}}{n +1} +C\\[10pt] \text{ We will use the substitution method to solve the intergals presented in the question. We will add a constant C after performing the required integration.}\\[10pt] \text{The required steps to evaluate}\int f(x)\quad dx \quad\text{by substitution are:}\\[10pt] \text{1. Put x=g(y), dx = g'(y) dy in the integrand }\\[10pt] \text{2. Evaluate the resulting integral in y.}\\[10pt] \text{3. Express the result obtained in terms of x.}\\} {/eq}

Answer and Explanation:


{eq}\displaystyle{a. \int \left ( \sqrt[3]{x}-\frac{1}{\sqrt[3]{x}} \right )^2\ dx\\[10pt] \int \left(x^\frac{1}{3} - x^\frac{-1}{3}\right)^2 \ dx\\[10pt] \text{ On expanding by using}\quad(a - b)^2 = a^2 -2ab +b^2\\[10pt] \int \left(x^\frac{2}{3} -2x^\frac{1}{3}\cdot x^\frac{-1}{3} +x^\frac{-2}{3} \right) \ dx\\[10pt] \int \left(x^\frac{2}{3} -2x^0 + x^\frac{-2}{3} \right) \ dx\\[10pt] \int \left(x^\frac{2}{3} + x^\frac{-2}{3} - 2 \right) \ dx\hspace{100pt}(\because x^0 = 1)\\[10pt] \frac{3x^\frac{5}{3}}{5} +\frac{ 3x^\frac{1}{3}}{1} -2x +C\\[10pt] \boxed{\frac{3x^\frac{5}{3}}{5} + 3x^\frac{1}{3} -2x +C}\\[10pt] b. \int \frac{x^3}{\sqrt{x^2-1}} \ dx\hspace{60pt}(1)\\ \text{ Substitute u = }x^2 - 1\\[10pt] \frac{du}{dx} = 2x\\[10pt] dx = \frac{1}{2} du\\[10pt] \text{Also,} x^2 = u + 1\\[10pt] \text{(1) becomes,}\\[10pt] = \frac{1}{2}\left(\int \frac{u + 1}{\sqrt{u}}\ du\right)\\[10pt] = \frac{1}{2}\left(\int u^\frac{1}{2} \ du + \int ^\frac{-1}{2} \ du\right)\\[10pt] = \frac{1}{2}\left(\frac{2u^\frac{3}{2}}{3} + \frac{2\sqrt{u}}{1}\right) +C\hspace{50pt}\left(\text{By applying}\int x^n \ dx = \frac{x^{n + 1}}{n +1} +C\right)\\[10pt] = \left(\frac{u^\frac{3}{2}}{3} + \frac{\sqrt{u}}{1}\right) +C\\[10pt] = \frac{(x^2 -1)^\frac{3}{2}}{3} + \sqrt{x^2 -1} +C\\[10pt] = \frac{\sqrt{x^2 -1}(x^2 - 1+3)}{3} +C\\[10pt] \boxed{= \frac{\sqrt{x^2 -1}(x^2 +2)}{3} +C}\hspace{50pt}\text{( By taking common out)}\\[10pt] c. \int_0^{\frac{x}{12}} \tan 3x \ dx\\[10pt] \int_0^{\frac{x}{12}}\frac{ \sin 3x}{\cos 3x} \ dx\quad\quad(1)\hspace{50pt}\left(\because\tan x = \frac{\sin x}{\cos x}\right)\\[10pt] \text{Substitute} u = \cos 3x\\[10pt] \frac{du}{dx} = -3\sin 3x\\[10pt] dx = \frac{du}{-3 \sin 3x}\\[10pt] \text{(1) becomes,}\\[10pt] \frac{-1}{3}\int_0^{\frac{x}{12}} \frac{1}{u} \ du\\[10pt] \frac{-1}{3}\begin{vmatrix}\ln u\end{vmatrix}_0^{\frac{x}{12}}\\[10pt] \frac{-1}{3}\begin{vmatrix}\ln \cos 3x\end{vmatrix}_0^{\frac{x}{12}}\\[10pt] \frac{-1}{3}\left(\ln \cos \frac{x}{4} - \ln \cos 0\right)\\[10pt] \frac{-1}{3}\left(\ln \cos \frac{x}{4} - \ln 1\right)\hspace{50pt}(\because \cos 0 = 1)\\[10pt] \frac{-1}{3}\left(\ln \cos \frac{x}{4} -0\right)\hspace{50pt}(\because \ln 1 = 0)\\[10pt] \boxed{\frac{-\ln \cos \frac{x}{4}}{3}}\\} {/eq}


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How to Solve Integrals Using Substitution

from Math 104: Calculus

Chapter 13 / Lesson 5
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