Question

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

Answer 1

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

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

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

Answer 2

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

Learn more:

1. Learn more about inverse of the functionhttps://brainly.com/question/1632445.
2. Learn more about equation of circle brainly.com/question/1506955.
3. Learn more about range and domain of the function brainly.com/question/3412497

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.