 b2sonicxh

2021-12-31

Use the Table of Integrals to evaluate the integral.
$\int \frac{{\mathrm{tan}}^{3}\left(\frac{1}{z}\right)}{{z}^{2}}dz$ Karen Robbins

Step 1
To determine:
The value of the given integral
$\int \frac{{\mathrm{tan}}^{3}\left(\frac{1}{z}\right)}{{z}^{2}}dz$
Step 2
Formula used:
The formula for the cube of the tangent function is given by
$\int {\mathrm{tan}}^{3}\left(x\right)dx=\frac{1}{2}{\mathrm{tan}}^{2}x+\mathrm{ln}|\mathrm{cos}\left(x\right)|+C$
Step 3
Calculation:
The given integral is
$\int \frac{{\mathrm{tan}}^{3}\left(\frac{1}{z}\right)}{{z}^{2}}dz$
By using the substitution,
Put $u=\frac{1}{z}$
Differentiate both sides of the above equation with respect to z,
$\frac{du}{dz}=\frac{-1}{{z}^{2}}$
Multiply both sides by dz,
$du=\frac{-1}{{z}^{2}}dz$
$⇒dz=-{z}^{2}du$
Substitute the value of dz in the given integral
$\int \frac{{\mathrm{tan}}^{3}\left(u\right)}{{z}^{2}}\left(-{z}^{2}\right)du=\int -{\mathrm{tan}}^{3}\left(u\right)du$
Taking a negative sign outside the integral,
$\int -{\mathrm{tan}}^{3}\left(u\right)du=-\int {\mathrm{tan}}^{3}\left(u\right)du$
By using the formula for the cube of the tangent function,
$-\int {\mathrm{tan}}^{3}\left(u\right)du=-\left[\frac{1}{2}{\mathrm{tan}}^{2}u+\mathrm{ln}|\mathrm{cos}\left(u\right)|\right]$
Distributing the negative sign,
$-\int {\mathrm{tan}}^{3}\left(u\right)du=-\frac{1}{2}{\mathrm{tan}}^{2}u-\mathrm{ln}|\mathrm{cos}\left(u\right)|$
Resubstitute u = 1/z,
Thus, $\int \frac{{\mathrm{tan}}^{3}\left(\frac{1}{z}\right)}{{z}^{2}}dz=-\frac{1}{2}{\mathrm{tan}}^{2}\left(\frac{1}{z}\right)-\mathrm{ln}|\mathrm{cos}\left(\frac{1}{z}\right)|+C$ vicki331g8

$\int \left(\frac{{\left(\mathrm{tan}\left(\frac{1}{z}\right)\right)}^{3}}{{z}^{2}}dz$
Let us put the expression $-\frac{1}{{z}^{2}}$ under the differential sign, i.e.:
$\left(-\frac{1}{{z}^{2}}\right)dz=d\left(\frac{1}{z}\right),t=\frac{1}{z}$
Then the initial integral can be written as follows:
$\int \left(-{\mathrm{tan}\left(t\right)}^{3}\right)\cdot dt$
$\int -{\mathrm{tan}\left(z\right)}^{3}dz$
We make a trigonometric substitution: $\mathrm{tan}\left(z\right)=$ and then $dt=\frac{1}{1+{t}^{2}}$
$\int -\frac{{t}^{3}}{{t}^{2}+1}\cdot dt$
Simplify the expression:
$\int \frac{-{z}^{3}}{{z}^{2}+1}dz$
Degree the numerator P (x) is greater than or equal to the degree of the denominator Q (x), so we divide the polynomials.
$\frac{-{z}^{3}}{{z}^{2}+1}=-z+\frac{z}{{z}^{2}+1}$
Integrating the integer part, we get:
$\int \left(-z\right)dz=-\frac{{z}^{2}}{2}$
Integrating further, we get:
$\int \frac{z}{{z}^{2}+1}dz=\frac{\mathrm{ln}\left({z}^{2}+1\right)}{2}$
$-\frac{{z}^{2}}{2}+\frac{\mathrm{ln}\left({z}^{2}+1\right)}{2}+C$
or
$-\frac{{z}^{2}}{2}+\mathrm{ln}\left(\sqrt{{z}^{2}+1}\right)+C$
Returning to the change of variables $\left(t=\mathrm{tan}\left(z\right)\right)$, we get:
$I=\frac{\mathrm{ln}\left({\mathrm{tan}\left(x\right)}^{2}+1\right)}{2}-\frac{{\mathrm{tan}\left(z\right)}^{2}}{2}+C$
To write down the final answer, it remains to substitute 1 / z instead of t.
$\frac{\mathrm{ln}\left({\mathrm{tan}\left(\frac{1}{z}\right)}^{2}+1\right)}{2}-\frac{{\mathrm{tan}\left(\frac{1}{z}\right)}^{2}}{2}+C$ Vasquez

$\begin{array}{}\int \frac{\mathrm{tan}\left(\frac{1}{z}{\right)}^{3}}{{z}^{2}}dz\\ \int -\mathrm{tan}\left(t{\right)}^{3}dt\\ -\int \mathrm{tan}\left(t{\right)}^{3}dt\\ -\left(\frac{1}{2}×\mathrm{tan}\left(t{\right)}^{2}-\int \mathrm{tan}\left(t\right)dt\right)\\ -\left(\frac{1}{2}×\mathrm{tan}\left(t{\right)}^{2}-\int \frac{\mathrm{sin}t}{\mathrm{cos}t}dt\right)\\ -\left(\frac{1}{2}×\mathrm{tan}\left(t{\right)}^{2}-\int -\frac{1}{u}du\right)\\ -\left(\frac{1}{2}×\mathrm{tan}\left(t{\right)}^{2}+\int \frac{1}{u}du\right)\\ -\left(\frac{1}{2}×\mathrm{tan}\left(t{\right)}^{2}+\mathrm{ln}\left(|u|\right)\right)\\ -\left(\frac{1}{2}×\mathrm{tan}\left(t{\right)}^{2}+\mathrm{ln}\left(|\mathrm{cos}\left(t\right)|\right)\right)\\ -\left(\frac{1}{2}×\mathrm{tan}\left(\frac{1}{z}{\right)}^{2}+\mathrm{ln}\left(|\mathrm{cos}\left(\frac{1}{z}\right)|\right)\right)\\ -\frac{\mathrm{tan}\left(\frac{1}{z}{\right)}^{2}}{2}-\mathrm{ln}\left(|\mathrm{cos}\left(\frac{1}{z}\right)|\right)\\ -\frac{\mathrm{tan}\left(\frac{1}{z}{\right)}^{2}}{2}-\mathrm{ln}\left(|\mathrm{cos}\left(\frac{1}{z}\right)|\right)+C\end{array}$