Use the Table of Integrals to evaluate the integral. ∫tan3(1z)z2dz

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Use the Table of Integrals to evaluate the integral.  ∫tan3(1z)z2dz

Karen Robbins · Accepted Answer

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=−12tan2u−ln|cos⁡(u)|  Resubstitute u = 1/z,  Thus, ∫tan3(1z)z2dz=−12tan2(1z)−ln|cos⁡(1z)|+C

vicki331g8 · Accepted Answer

∫((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+1⋅dt  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)2−tan⁡(z)22+C  To write down the final answer, it remains to substitute 1 / z instead of t.  ln⁡(tan⁡(1z)2+1)2−tan⁡(1z)22+C

Vasquez · Accepted Answer

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

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