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Find the general indefinite integral: (a) \int (3x-2)^{2}dr (b)\int \frac{8x-5}{\sqrt[3]{x}}dr ...

Question:

Find the general indefinite integral:

(a) {eq}\int (3x-2)^{2}dx {/eq}

(b) {eq}\int \frac{8x-5}{\sqrt[3]{x}}dx {/eq}

(c) {eq}\int {/eq} cos {eq}(3x)\sqrt[3]{sin(3x)}dx {/eq}

Integration with the use of Substitution:

One integration technique is known as the u-substitution. This method makes a substitution on the given integrand which changes the differential variable. An appropriate substitution relates to an integration process that is easier and finishes with a fewer number of steps.

Answer and Explanation:

For the first problem, we allow {eq}\displaystyle u = 3x+2 {/eq} and {eq}\displaystyle du = 3dx {/eq} or {eq}\displaystyle \frac{du}{3} = dx {/eq}. We perform the substitution and then proceed with the integration.

{eq}\begin{align} \displaystyle \int (3x-2)^2 dx &= \int u^2 \frac{du}{3}\\ &= \frac{1}{3}\int u^2du\\ &= \frac{u^3}{9} +C\\ \text{We revert the substitution.}\\ &= \frac{(3x-2)^3}{9} +C\\ \end{align} {/eq}

For the second problem, we express the fraction as a difference between two terms and then simplify the fractions before performing the integral.

{eq}\begin{align} \displaystyle \int \frac{8x-5}{\sqrt[3]{x}}dx &= \int \left( \frac{8x}{\sqrt[3]x}- \frac{5}{\sqrt[3]x}\right)dx\\ &= \int \left( 8x^{1-\frac{1}{3}}- 5x^{- \frac{1}{3}}\right)dx\\ &= \int \left( 8x^{\frac{2}{3}}- 5x^{- \frac{1}{3}}\right)dx\\ &= 8\frac{x^\frac{5}{3}}{\frac{5}{3}} - 5 \frac{x^ \frac{2}{3}}{\frac{2}{3}} +C\\ &= \frac{24}{5}x^ \frac{5}{3} - \frac{15}{2}x^ \frac{2}{3} +C \end{align} {/eq}

For the third problem, we allow {eq}\displaystyle u = \sin(3x) {/eq} such that {eq}\displaystyle du = 3\cos(3x)dx {/eq} or {eq}\displaystyle \frac{du}{3} = \cos(3x)dx {/eq}. Again, we perform the substitution and then proceed with the integration.

{eq}\begin{align} \displaystyle \int \cos(3x)\sqrt[3]{\sin(3x)}dx &= \int u^ \frac{1}{3}\frac{du}{3}\\ &= \frac{1}{3}\int u^ \frac{1}{3}du\\ &= \frac{1}{3}\frac{u^ \frac{4}{3}}{\frac{4}{3}}+C\\ &= 4u^ \frac{4}{3} +C\\ &= 4\sin^ \frac{4}{3}(3x) +C \end{align} {/eq}


Learn more about this topic:

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How to Solve Integrals Using Substitution

from Math 104: Calculus

Chapter 13 / Lesson 5
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