Use the Table of Integrals to evaluate the integral. \int \frac{\tan^{3}(1/z)}{z^{2}}dz

Question

Use the Table of Integrals to evaluate the integral. \int \frac{\tan^{3}(1/z)}{z^{2}}dz

b2sonicxh 2021-12-31 Answered

Use the Table of Integrals to evaluate the integral.

\int \frac{\tan^{3} (\frac{1}{z})}{z^{2}} d z

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Expert Answer

Karen Robbins

Answered 2022-01-01 Author has 49 answers

Step 1
To determine:
The value of the given integral

\int \frac{\tan^{3} (\frac{1}{z})}{z^{2}} d z

Step 2
Formula used:
The formula for the cube of the tangent function is given by

\int \tan^{3} (x) d x = \frac{1}{2} \tan^{2} x + \ln | \cos (x) | + C

Step 3
Calculation:
The given integral is

\int \frac{\tan^{3} (\frac{1}{z})}{z^{2}} d z

By using the substitution,
Put

u = \frac{1}{z}

Differentiate both sides of the above equation with respect to z,

\frac{d u}{d z} = \frac{- 1}{z^{2}}

Multiply both sides by dz,

d u = \frac{- 1}{z^{2}} d z

\Rightarrow d z = - z^{2} d u

Substitute the value of dz in the given integral

\int \frac{\tan^{3} (u)}{z^{2}} (- z^{2}) d u = \int - \tan^{3} (u) d u

Taking a negative sign outside the integral,

\int - \tan^{3} (u) d u = - \int \tan^{3} (u) d u

By using the formula for the cube of the tangent function,

- \int \tan^{3} (u) d u = - [\frac{1}{2} \tan^{2} u + \ln | \cos (u) |]

Distributing the negative sign,

- \int \tan^{3} (u) d u = - \frac{1}{2} \tan^{2} u - \ln | \cos (u) |

Resubstitute u = 1/z,
Thus,

\int \frac{\tan^{3} (\frac{1}{z})}{z^{2}} d z = - \frac{1}{2} \tan^{2} (\frac{1}{z}) - \ln | \cos (\frac{1}{z}) | + C

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vicki331g8

Answered 2022-01-02 Author has 37 answers

\int (\frac{{(\tan (\frac{1}{z}))}^{3}}{z^{2}} d z

Let us put the expression

- \frac{1}{z^{2}}

under the differential sign, i.e.:

(- \frac{1}{z^{2}}) d z = d (\frac{1}{z}), t = \frac{1}{z}

Then the initial integral can be written as follows:

\int (- {\tan (t)}^{3}) \cdot d t

\int - {\tan (z)}^{3} d z

We make a trigonometric substitution:

\tan (z) =

and then

d t = \frac{1}{1 + t^{2}}

\int - \frac{t^{3}}{t^{2} + 1} \cdot d t

Simplify the expression:

\int \frac{- z^{3}}{z^{2} + 1} d z

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 (- z) d z = - \frac{z^{2}}{2}

Integrating further, we get:

\int \frac{z}{z^{2} + 1} d z = \frac{\ln (z^{2} + 1)}{2}

Answer:

- \frac{z^{2}}{2} + \frac{\ln (z^{2} + 1)}{2} + C

or

- \frac{z^{2}}{2} + \ln (\sqrt{z^{2} + 1}) + C

Returning to the change of variables

(t = \tan (z))

, we get:

I = \frac{\ln ({\tan (x)}^{2} + 1)}{2} - \frac{{\tan (z)}^{2}}{2} + C

To write down the final answer, it remains to substitute 1 / z instead of t.

\frac{\ln ({\tan (\frac{1}{z})}^{2} + 1)}{2} - \frac{{\tan (\frac{1}{z})}^{2}}{2} + C

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Vasquez

Answered 2022-01-07 Author has 460 answers

$\begin{matrix} \int \frac{\tan (\frac{1}{z})^{3}}{z^{2}} d z \\ \int - \tan (t)^{3} d t \\ - \int \tan (t)^{3} d t \\ - (\frac{1}{2} \times \tan (t)^{2} - \int \tan (t) d t) \\ - (\frac{1}{2} \times \tan (t)^{2} - \int \frac{\sin t}{\cos t} d t) \\ - (\frac{1}{2} \times \tan (t)^{2} - \int - \frac{1}{u} d u) \\ - (\frac{1}{2} \times \tan (t)^{2} + \int \frac{1}{u} d u) \\ - (\frac{1}{2} \times \tan (t)^{2} + \ln (| u |)) \\ - (\frac{1}{2} \times \tan (t)^{2} + \ln (| \cos (t) |)) \\ - (\frac{1}{2} \times \tan (\frac{1}{z})^{2} + \ln (| \cos (\frac{1}{z}) |)) \\ - \frac{\tan (\frac{1}{z})^{2}}{2} - \ln (| \cos (\frac{1}{z}) |) \\ - \frac{\tan (\frac{1}{z})^{2}}{2} - \ln (| \cos (\frac{1}{z}) |) + C \end{matrix}$