How do you integrate $\int {tan}^{3} (\frac{π x}{2}) {sec}^{2} (\frac{π x}{2})$ $int tan^3((pix)/2)sec^2((pix)/2)$ ?

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How do you integrate $\int {tan}^{3} (\frac{π x}{2}) {sec}^{2} (\frac{π x}{2})$ $int tan^3((pix)/2)sec^2((pix)/2)$ ?

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Answer 1 · 2016-11-29T05:01:14

$\int {tan}^{3} (\frac{π x}{2}) {sec}^{2} (\frac{π x}{2}) d x = \frac{{tan}^{4} (\frac{π x}{2})}{2 π} + C$ $inttan^3((pix)/2)sec^2((pix)/2)dx=tan^4((pix)/2)/(2pi)+C$

Explanation:

$I = \int {tan}^{3} (\frac{π x}{2}) {sec}^{2} (\frac{π x}{2}) d x$ $I=inttan^3((pix)/2)sec^2((pix)/2)dx$

We can first eliminate some ugliness by letting $u = \frac{π x}{2}$ $u=(pix)/2$ . Differentiating this gives $d u = \frac{π}{2} d x$ $du=pi/2dx$ , so rearrange the integral as follows:

$I = \frac{2}{π} \int {tan}^{3} (\frac{π x}{2}) {sec}^{2} (\frac{π x}{2}) (\frac{π}{2} d x)$ $I=2/piinttan^3((pix)/2)sec^2((pix)/2)(pi/2dx)$

$I = \frac{2}{π} \int {tan}^{3} (u) {sec}^{2} (u) d u$ $I=2/piinttan^3(u)sec^2(u)du$

Notice that ${sec}^{2}$ $sec^2$ is the derivative of $tan$ $tan$ , so let $v = tan (u)$ $v=tan(u)$ so $d v = {sec}^{2} (u) d u$ $dv=sec^2(u)du$ :

$I = \frac{2}{π} \int v^{3} d v$ $I=2/piintv^3dv$

$I = \frac{2}{π} \frac{v^{4}}{4}$ $I=2/piv^4/4$

$I = \frac{v^{4}}{2 π}$ $I=(v^4)/(2pi)$

$I = \frac{{tan}^{4} (u)}{2 π}$ $I=tan^4(u)/(2pi)$

$I = \frac{{tan}^{4} (\frac{π x}{2})}{2 π} + C$ $I=tan^4((pix)/2)/(2pi)+C$

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How do you integrate $\int {tan}^{3} (\frac{π x}{2}) {sec}^{2} (\frac{π x}{2})$ $int tan^3((pix)/2)sec^2((pix)/2)$ ?

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Explanation:

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How do you integrate $\int {tan}^{3} (\frac{π x}{2}) {sec}^{2} (\frac{π x}{2})$ $int tan^3((pix)/2)sec^2((pix)/2)$ ?