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For trigonometric substitution to solve the above integral, fill in the blanks below using the picture of the triangle given.

integral (x^3)/(sqrt(25+9x^2)) dx / l / l / l / l / l / l / l / l / l side A / l / l --------- side B *** sorry hard to draw on the computer side a=3x side b=? side c=? 3x/5=? (3/5) dx=? (sqrt(25+9x^2))/(5)=?

Verified Answer :

LammettHash

\(\displaystyle\int\frac{x^3}{\sqrt{25+9x^2}}\,\mathrm dx\)

Substitute \(x=\dfrac53\tan y\) so that \(\mathrm dx=\dfrac53\sec^2y\,\mathrm dy\).

Then you have

\(\displaystyle\int\frac{\frac53\tan y\times \frac53\sec^2y}{\sqrt{25+9\left(\frac53\tan y\right)^2}}\,\mathrm dy\)

Expand the denominator and simplify.

\(\sqrt{25+9\left(\dfrac53\tan y\right)^2}=\sqrt{25+25\tan^2y}=5\sqrt{\sec^2y}=5\sec y\)

where the last simplification only holds for certain intervals.

Then the integral is

\(\displaystyle\int\frac{\left(\frac53\tan y\right)^3\times \frac53\sec^2y}{5\sec y}\,\mathrm dy=\frac{5^3}{3^4}\int \tan^3y\sec y\,\mathrm dy\)

To compute the remaining integral, use the Pythagorean expansion for \(\tan^2y\):

\(\tan^3y\sec y=\tan^2 y\tan y\sec y=(\sec^2y-1)\tan y\sec y\)

Now substituting \(z=\sec y\) so that \(\mathrm dz=\sec y\tan y\,\mathrm dy\) allows you to rewrite the integral as

\(\displaystyle\frac{5^3}{3^4}\int(z^2-1)\,\mathrm dz=\frac{5^3}{3^4}\left(\frac13z^3-z\right)+C\)

Back-substitute:

\(\displaystyle\frac{5^3}{3^4}\left(\frac13z^3-z\right)+C\)
\(=\displaystyle\frac{5^3}{3^5}z^3-\frac{5^3}{3^4}z+C\)
\(=\displaystyle\frac{5^3}{3^5}\sec^3y-\frac{5^3}{3^4}\sec y+C\)
\(=\displaystyle\frac{5^3}{3^5}\sec^3\left(\arctan\frac{3x}5\right)-\frac{5^3}{3^4}\sec\left(\arctan\frac{3x}5\right)+C\)

Simplify the trigonometric terms. Your reference triangle involves an angle with the side opposite the angle having length \(3x\) and the side adjacent to it having length \(5\). This means the missing length is \(\sqrt{25+9x^2}\).

The secant ratio for this angle is then \(\dfrac{\sqrt{25+9x^2}}{5}\).

So the antiderivative is equivalent to

\(\displaystyle\frac{5^3}{3^5}\left(\frac{\sqrt{25+9x^2}}{5}\right)^3-\frac{5^3}{3^4}\left(\frac{\sqrt{25+9x^2}}{5}\right)+C\)
\(=\displaystyle\frac{1}{3^5}(25+9x^2)^{3/2}-\frac{5^2}{3^4}(25+9x^2)^{1/2}+C\)
\(=\displaystyle\frac1{3^5}(25+9x^2)^{1/2}(9x^2-50)+C\)

The integration is \(\boxed{\dfrac{1}{{{3^5}}}{{\left( {25 + 9{x^2}} \right)}^{\dfrac{1}{2}}}\left( {9{x^2} - 50} \right) + C}.\)

Further explanation:

Given:

The integral is \(\int {\dfrac{{{x^3}}}{{\sqrt {25 + 9{x^2}} }}dx}.\)

Explanation:

The given integral is \(\int {\dfrac{{{x^3}}}{{\sqrt {25 + 9{x^2}} }}dx}.\)

Plug \(x = \dfrac{5}{3}\tan y.\)

Differentiate \(x = \dfrac{5}{3}\tan y\) with respect to y.

\(dx = \dfrac{5}{3}{\sec ^2}ydy\)

The integral can be expressed as,

\(\begin{aligned}I&= \int {\frac{{{{\left( {\frac{5}{3}\tan y} \right)}^3} \times \frac{5}{3}{{\sec }^2}y}}{{\sqrt {25 + 9{{\left( {\frac{5}{3}\tan y} \right)}^2}} }}dy}\\&= \int {\frac{{{{\left( {\frac{5}{3}} \right)}^4}{{\tan }^3}y \times {{\sec }^2}y}}{{\sqrt {25\left( {1 + {{\tan }^2}y} \right)} }}dy}\\&= {\left( {\frac{5}{3}} \right)^4}\int {\frac{{{{\tan }^3}y \times {{\sec }^2}y}}{{5\sec y}}dy} \\\end{aligned}\)

Further solve the above equation,

\(\begin{aligned}I&= {\left( {\frac{5}{3}} \right)^4}\int {{{\tan }^3}y\sec ydy} \\&= {\left( {\frac{5}{3}} \right)^4}\int {\left( {{{\tan }^2}y} \right)\tan y\sec ydy}\\ &= {\left( {\frac{5}{3}} \right)^4}\int {\left( {{{\sec }^2}y - 1} \right)\tan y\sec ydy}  \\\end{aligned}\)

Substitute \(t\) for \({\sec ^2}y.\)

\(\begin{aligned}t&= \sec y\\dt&= \sec y\tan ydy\\\end{aligned}\)

The integral can be written as follows,

\(\begin{aligned}I&= \frac{{125}}{{81}}\int {\left( {{t^2} - 1} \right)} dt\\&= \frac{{125}}{{81}}\left( {\frac{1}{3}{t^3} - t} \right) + C\\&= \frac{{125}}{{243}}{t^3} - \frac{{125}}{{81}}t + C\\\end{aligned}\)

Substitute \(\sec y\) for t.

\(\begin{aligned}I&= \frac{{125}}{{243}}{\sec ^3}y - \frac{{125}}{{81}}\sec y + C\\&= \frac{{125}}{{243}}{\sec ^3}\left( {{{\tan }^{ - 1}}\frac{{3x}}{5}} \right) - \frac{{125}}{{81}}\sec \left( {{{\tan }^{ - 1}}\frac{{3x}}{5}} \right) + C\\\end{aligned}\)

The secant ratio of the angle is given by \(\dfrac{{\sqrt {25 + 9{x^2}} }}{5}.\)

The integral can be expressed as follows,

\(\begin{aligned}I&= \frac{{125}}{{243}}{\left( {\frac{{\sqrt {25 + 9{x^2}} }}{5}} \right)^3} - \frac{{125}}{{81}}\left( {\frac{{\sqrt {25 + 9{x^2}} }}{5}} \right) + C\\&= \frac{1}{{243}}{\left( {\sqrt {25 + 9{x^2}} } \right)^3} - \frac{{25}}{{81}}\left( {\sqrt {25 + 9{x^2}} } \right) + C\\&= \frac{1}{{243}}\left( {\sqrt {25 + 9{x^2}} } \right)\left( {9{x^2} - 50} \right) + C \\\end{aligned}\)

The integration is \(\boxed{\dfrac{1}{{{3^5}}}{{\left( {25 + 9{x^2}} \right)}^{\dfrac{1}{2}}}\left( {9{x^2} - 50} \right) + C}.\)

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Answer details:

Grade: College

Subject: Mathematics

Chapter: Integrals

Keywords: function, decreasing, increasing, monotonic, integral, monotone, data is not monotone, table, sketch, possible, graph, integration, area under the curve.

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