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.
tan3(1z)z2dz
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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
tan3(1z)z2dz
Step 2
Formula used:
The formula for the cube of the tangent function is given by
tan3(x)dx=12tan2x+ln|cos(x)|+C
Step 3
Calculation:
The given integral is
tan3(1z)z2dz
By using the substitution,
Put u=1z
Differentiate both sides of the above equation with respect to z,
dudz=1z2
Multiply both sides by dz,
du=1z2dz
dz=z2du
Substitute the value of dz in the given integral
tan3(u)z2(z2)du=tan3(u)du
Taking a negative sign outside the integral,
tan3(u)du=tan3(u)du
By using the formula for the cube of the tangent function,
tan3(u)du=[12tan2u+ln|cos(u)|]
Distributing the negative sign,
tan3(u)du=12tan2uln|cos(u)|
Resubstitute u = 1/z,
Thus, tan3(1z)z2dz=12tan2(1z)ln|cos(1z)|+C
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vicki331g8
Answered 2022-01-02 Author has 37 answers
((tan(1z))3z2dz
Let us put the expression 1z2 under the differential sign, i.e.:
(1z2)dz=d(1z),t=1z
Then the initial integral can be written as follows:
(tan(t)3)dt
tan(z)3dz
We make a trigonometric substitution: tan(z)= and then dt=11+t2
t3t2+1dt
Simplify the expression:
z3z2+1dz
Degree the numerator P (x) is greater than or equal to the degree of the denominator Q (x), so we divide the polynomials.
z3z2+1=z+zz2+1
Integrating the integer part, we get:
(z)dz=z22
Integrating further, we get:
zz2+1dz=ln(z2+1)2
Answer:
z22+ln(z2+1)2+C
or
z22+ln(z2+1)+C
Returning to the change of variables (t=tan(z)), we get:
I=ln(tan(x)2+1)2tan(z)22+C
To write down the final answer, it remains to substitute 1 / z instead of t.
ln(tan(1z)2+1)2tan(1z)22+C
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Vasquez
Answered 2022-01-07 Author has 460 answers

tan(1z)3z2dztan(t)3dttan(t)3dt(12×tan(t)2tan(t)dt)(12×tan(t)2sintcostdt)(12×tan(t)21udu)(12×tan(t)2+1udu)(12×tan(t)2+ln(|u|))(12×tan(t)2+ln(|cos(t)|))(12×tan(1z)2+ln(|cos(1z)|))tan(1z)22ln(|cos(1z)|)tan(1z)22ln(|cos(1z)|)+C

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