# How do you integrate int tan^3((pix)/2)sec^2((pix)/2)?

Nov 29, 2016

$\int {\tan}^{3} \left(\frac{\pi x}{2}\right) {\sec}^{2} \left(\frac{\pi x}{2}\right) \mathrm{dx} = {\tan}^{4} \frac{\frac{\pi x}{2}}{2 \pi} + C$

#### Explanation:

$I = \int {\tan}^{3} \left(\frac{\pi x}{2}\right) {\sec}^{2} \left(\frac{\pi x}{2}\right) \mathrm{dx}$

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

$I = \frac{2}{\pi} \int {\tan}^{3} \left(\frac{\pi x}{2}\right) {\sec}^{2} \left(\frac{\pi x}{2}\right) \left(\frac{\pi}{2} \mathrm{dx}\right)$

$I = \frac{2}{\pi} \int {\tan}^{3} \left(u\right) {\sec}^{2} \left(u\right) \mathrm{du}$

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

$I = \frac{2}{\pi} \int {v}^{3} \mathrm{dv}$

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

$I = \frac{{v}^{4}}{2 \pi}$

$I = {\tan}^{4} \frac{u}{2 \pi}$

$I = {\tan}^{4} \frac{\frac{\pi x}{2}}{2 \pi} + C$