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integration techniques
- Thread starter intervade
- Start date
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
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
D
Deleted member 4993
Guest
You would not have to re-sub, if you change your limits along with your variables.
Subhotosh Khan said: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
Senior Member
- Joined
- Mar 8, 2009
- Messages
- 1,577
\(\displaystyle Another \ way: \ Stealing \ galactus's \ thunder, \ we \ have:\)
\(\displaystyle \frac{3}{16}\int_{0}^{\pi/3}\frac{sin^{3}(\theta)}{cos^{2}(\theta)}d\theta \ = \ \frac{3}{16}\int_{0}^{\pi/3}\frac{sin(\theta)sin^{2}(\theta)}{cos^{2}(\theta)}d\theta\)
\(\displaystyle = \ \frac{3}{16}\int_{0}^{\pi/3}\frac{sin(\theta)[1-cos^{2}(\theta)]}{cos^{2}(\theta)}d\theta \ = \ \frac{3}{16}\int_{0}^{\pi/3}\frac{sin(\theta)-sin(\theta)cos^{2}(\theta)}{cos^{2}(\theta)}d\theta\)
\(\displaystyle = \ \frac{3}{16}\int_{0}^{\pi/3}\frac{sin(\theta)}{cos^{2}(\theta)}d\theta \ - \ \frac{3}{16}\int_{0}^{\pi/3}sin(\theta)d\theta\)
\(\displaystyle I \ trust \ that \ whoever \ is \ interested, \ can \ take \ it \ from \ here.\)
\(\displaystyle \frac{3}{16}\int_{0}^{\pi/3}\frac{sin^{3}(\theta)}{cos^{2}(\theta)}d\theta \ = \ \frac{3}{16}\int_{0}^{\pi/3}\frac{sin(\theta)sin^{2}(\theta)}{cos^{2}(\theta)}d\theta\)
\(\displaystyle = \ \frac{3}{16}\int_{0}^{\pi/3}\frac{sin(\theta)[1-cos^{2}(\theta)]}{cos^{2}(\theta)}d\theta \ = \ \frac{3}{16}\int_{0}^{\pi/3}\frac{sin(\theta)-sin(\theta)cos^{2}(\theta)}{cos^{2}(\theta)}d\theta\)
\(\displaystyle = \ \frac{3}{16}\int_{0}^{\pi/3}\frac{sin(\theta)}{cos^{2}(\theta)}d\theta \ - \ \frac{3}{16}\int_{0}^{\pi/3}sin(\theta)d\theta\)
\(\displaystyle I \ trust \ that \ whoever \ is \ interested, \ can \ take \ it \ from \ here.\)