Respuesta :
[tex]\displaystyle\int\frac{x^3}{\sqrt{25+9x^2}}\,\mathrm dx[/tex]
Substitute [tex]x=\dfrac53\tan y[/tex] so that [tex]\mathrm dx=\dfrac53\sec^2y\,\mathrm dy[/tex].
Then you have
[tex]\displaystyle\int\frac{\frac53\tan y\times \frac53\sec^2y}{\sqrt{25+9\left(\frac53\tan y\right)^2}}\,\mathrm dy[/tex]
Expand the denominator and simplify.
[tex]\sqrt{25+9\left(\dfrac53\tan y\right)^2}=\sqrt{25+25\tan^2y}=5\sqrt{\sec^2y}=5\sec y[/tex]
where the last simplification only holds for certain intervals.
Then the integral is
[tex]\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[/tex]
To compute the remaining integral, use the Pythagorean expansion for [tex]\tan^2y[/tex]:
[tex]\tan^3y\sec y=\tan^2 y\tan y\sec y=(\sec^2y-1)\tan y\sec y[/tex]
Now substituting [tex]z=\sec y[/tex] so that [tex]\mathrm dz=\sec y\tan y\,\mathrm dy[/tex] allows you to rewrite the integral as
[tex]\displaystyle\frac{5^3}{3^4}\int(z^2-1)\,\mathrm dz=\frac{5^3}{3^4}\left(\frac13z^3-z\right)+C[/tex]
Back-substitute:
[tex]\displaystyle\frac{5^3}{3^4}\left(\frac13z^3-z\right)+C[/tex]
[tex]=\displaystyle\frac{5^3}{3^5}z^3-\frac{5^3}{3^4}z+C[/tex]
[tex]=\displaystyle\frac{5^3}{3^5}\sec^3y-\frac{5^3}{3^4}\sec y+C[/tex]
[tex]=\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[/tex]
Simplify the trigonometric terms. Your reference triangle involves an angle with the side opposite the angle having length [tex]3x[/tex] and the side adjacent to it having length [tex]5[/tex]. This means the missing length is [tex]\sqrt{25+9x^2}[/tex].
The secant ratio for this angle is then [tex]\dfrac{\sqrt{25+9x^2}}{5}[/tex].
So the antiderivative is equivalent to
[tex]\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[/tex]
[tex]=\displaystyle\frac{1}{3^5}(25+9x^2)^{3/2}-\frac{5^2}{3^4}(25+9x^2)^{1/2}+C[/tex]
[tex]=\displaystyle\frac1{3^5}(25+9x^2)^{1/2}(9x^2-50)+C[/tex]
Substitute [tex]x=\dfrac53\tan y[/tex] so that [tex]\mathrm dx=\dfrac53\sec^2y\,\mathrm dy[/tex].
Then you have
[tex]\displaystyle\int\frac{\frac53\tan y\times \frac53\sec^2y}{\sqrt{25+9\left(\frac53\tan y\right)^2}}\,\mathrm dy[/tex]
Expand the denominator and simplify.
[tex]\sqrt{25+9\left(\dfrac53\tan y\right)^2}=\sqrt{25+25\tan^2y}=5\sqrt{\sec^2y}=5\sec y[/tex]
where the last simplification only holds for certain intervals.
Then the integral is
[tex]\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[/tex]
To compute the remaining integral, use the Pythagorean expansion for [tex]\tan^2y[/tex]:
[tex]\tan^3y\sec y=\tan^2 y\tan y\sec y=(\sec^2y-1)\tan y\sec y[/tex]
Now substituting [tex]z=\sec y[/tex] so that [tex]\mathrm dz=\sec y\tan y\,\mathrm dy[/tex] allows you to rewrite the integral as
[tex]\displaystyle\frac{5^3}{3^4}\int(z^2-1)\,\mathrm dz=\frac{5^3}{3^4}\left(\frac13z^3-z\right)+C[/tex]
Back-substitute:
[tex]\displaystyle\frac{5^3}{3^4}\left(\frac13z^3-z\right)+C[/tex]
[tex]=\displaystyle\frac{5^3}{3^5}z^3-\frac{5^3}{3^4}z+C[/tex]
[tex]=\displaystyle\frac{5^3}{3^5}\sec^3y-\frac{5^3}{3^4}\sec y+C[/tex]
[tex]=\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[/tex]
Simplify the trigonometric terms. Your reference triangle involves an angle with the side opposite the angle having length [tex]3x[/tex] and the side adjacent to it having length [tex]5[/tex]. This means the missing length is [tex]\sqrt{25+9x^2}[/tex].
The secant ratio for this angle is then [tex]\dfrac{\sqrt{25+9x^2}}{5}[/tex].
So the antiderivative is equivalent to
[tex]\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[/tex]
[tex]=\displaystyle\frac{1}{3^5}(25+9x^2)^{3/2}-\frac{5^2}{3^4}(25+9x^2)^{1/2}+C[/tex]
[tex]=\displaystyle\frac1{3^5}(25+9x^2)^{1/2}(9x^2-50)+C[/tex]
The integration is [tex]\boxed{\dfrac{1}{{{3^5}}}{{\left( {25 + 9{x^2}} \right)}^{\dfrac{1}{2}}}\left( {9{x^2} - 50} \right) + C}.[/tex]
Further explanation:
Given:
The integral is [tex]\int {\dfrac{{{x^3}}}{{\sqrt {25 + 9{x^2}} }}dx}.[/tex]
Explanation:
The given integral is [tex]\int {\dfrac{{{x^3}}}{{\sqrt {25 + 9{x^2}} }}dx}.[/tex]
Plug [tex]x = \dfrac{5}{3}\tan y.[/tex]
Differentiate [tex]x = \dfrac{5}{3}\tan y[/tex] with respect to y.
[tex]dx = \dfrac{5}{3}{\sec ^2}ydy[/tex]
The integral can be expressed as,
[tex]\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}[/tex]
Further solve the above equation,
[tex]\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}[/tex]
Substitute [tex]t[/tex] for [tex]{\sec ^2}y.[/tex]
[tex]\begin{aligned}t&= \sec y\\dt&= \sec y\tan ydy\\\end{aligned}[/tex]
The integral can be written as follows,
[tex]\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}[/tex]
Substitute [tex]\sec y[/tex] for t.
[tex]\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}[/tex]
The secant ratio of the angle is given by [tex]\dfrac{{\sqrt {25 + 9{x^2}} }}{5}.[/tex]
The integral can be expressed as follows,
[tex]\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}[/tex]
The integration is [tex]\boxed{\dfrac{1}{{{3^5}}}{{\left( {25 + 9{x^2}} \right)}^{\dfrac{1}{2}}}\left( {9{x^2} - 50} \right) + C}.[/tex]
Learn more:
- Learn more about inverse of the functionhttps://brainly.com/question/1632445.
- Learn more about equation of circle brainly.com/question/1506955.
- Learn more about range and domain of the function https://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.
