integration techniques

[intervade]

Hello, I'm working on a problem that I'm not sure where to start, maybe someone could point me in the right direction.

 

[galactus]

One way to do it is good ol' trig sub.

Let $\displaystyle x=\frac{3}{2}tan{\theta}, \;\ \frac{3}{2}sec^{2}{\theta}d{\theta}$

Making the subs and simplifying gives:

$\displaystyle \frac{3}{16}\int\frac{sin^{3}{\theta}}{cos^{2}{\theta}}d{\theta}$

Let $\displaystyle u=cos{\theta}, \;\ -du=sin{\theta}d{\theta}$

Make the subs and remember the identity $\displaystyle sin^{2}{\theta}=1-cos^{2}{\theta}$

$\displaystyle \frac{3}{16}\int du-\frac{3}{16}\int \frac{1}{u^{2}}du$

Integrate and resub

 

[Deleted member 4993]

You would not have to re-sub, if you change your limits along with your variables.

 

[galactus]

You would not have to re-sub, if you change your limits along with your variables.

Yes, of course. That is the best thing to do.

 

[BigGlenntheHeavy]

$\displaystyle Another \ way: \ Stealing \ galactus's \ thunder, \ we \ have:$

$\displaystyle \frac{3}{16}\int_{0}^{\pi/3}\frac{sin^{3}(\theta)}{cos^{2}(\theta)}d\theta \ = \ \frac{3}{16}\int_{0}^{\pi/3}\frac{sin(\theta)sin^{2}(\theta)}{cos^{2}(\theta)}d\theta$

$\displaystyle = \ \frac{3}{16}\int_{0}^{\pi/3}\frac{sin(\theta)[1-cos^{2}(\theta)]}{cos^{2}(\theta)}d\theta \ = \ \frac{3}{16}\int_{0}^{\pi/3}\frac{sin(\theta)-sin(\theta)cos^{2}(\theta)}{cos^{2}(\theta)}d\theta$

$\displaystyle = \ \frac{3}{16}\int_{0}^{\pi/3}\frac{sin(\theta)}{cos^{2}(\theta)}d\theta \ - \ \frac{3}{16}\int_{0}^{\pi/3}sin(\theta)d\theta$

$\displaystyle I \ trust \ that \ whoever \ is \ interested, \ can \ take \ it \ from \ here.$