Question

Integrate: sin(3t)^4 dt. Definite intregral from 0 to pi.

Answer 1

\[\int\limits_{0}^{\pi} \sin^{4}(3t) dt\]

Answer 2

you can use integration by parts, but its a lot of steps
there are tables that show integral sin^n(x)
Also wolfram provides steps as well

http://www.wolframalpha.com/input/?i=integrate+sin^4%283x%29dx

Answer 3

\[sin^2t = \frac{1 - cos2t}{2}\]
Use this to reduce the calculation and algebra and stuff...

Answer 4

\[\int_0^{\pi} \frac{1}{4}* (1 - cos6t )^2 dt\]

Answer 5

\[\frac{1}{4}\int _ 0^{\pi} (1 + cos^26t - 2cos6t)dt\]

Answer 6

Is there a way to do with without the reduction formula and just using trig identities and u-substitution?

Answer 7

\[cos^2t = \frac{1 + cos2t}{2}\]
\[\frac{1}{4} \int_0^{\pi} (1+ \frac{(1 + cos12t)}{4} -2cos6t)dt\]
\[\frac{1}{4} \int 1*dt -2cos6t*dt + \frac{1}{16}\int_0^{\pi} 1*dt + cos12t *dt \]

Answer 8

yeah, do what ishaan is showing you :)

Answer 9

\[\left| \frac{t}{4} -\frac{1}{12}sin6t +\frac{t}{16} + \frac{1}{16*12}sin12t\right|_0^{\pi}\]

Answer 10

Now Algebra ....you can do it ....I may messed up something so do check again the whole integration .... Good Luck

Answer 11

Ugh my browser was wigging and I couldn't ask questions. The 2nd time you used the half angle formula, why is there a 4 in the denominator?

Answer 12

ah my bad I told you I may have messed ...I messed it up sorry 2 should be there

Answer 13

That will make 16 into 8 as I multiplied an extra 2

Answer 14

So I have 3/8t - 1/12sin(6t) + 1/96sin(12t), but since anything with sine is going to be 0, it's pretty much 3pi/8 - 0, which is 3pi/8. OMG I kept trying to integrate before switching to the half angle the 2nd time and kept getting pi* some obscure fraction. THANK YOU THANK YOU THANK YOU!

Answer 15

\[\int\limits_{0}^{\pi}\sin^4{3t}\,\,dt\quad,\quad u=3t\quad,\quad dt=du/3\]
\[\frac{1}{3}\int\limits_{0}^{3\pi}\sin^4{u}\,\,du=\frac{6}{3}\int\limits_{0}^{\pi/2}\sin^4{u}\,\,du=2\int\limits_{0}^{\pi/2}\cos^0{u}\cdot\sin^4{u}\,\,du=\]\[=2\cdot\frac{1}{2}B\left(\frac{1}{2},\frac{5}{2}\right)=B\left(\frac{1}{2},\frac{5}{2}\right)=\frac{\Gamma\left(\frac{1}{2}\right)\cdot\Gamma\left(\frac{5}{2}\right)}{\Gamma\left(\frac{1}{2}+\frac{5}{2}\right)}=\frac{\Gamma\left(\frac{1}{2}\right)\cdot\Gamma\left(\frac{5}{2}\right)}{\Gamma\left(3\right)}=\]
\[=\frac{\sqrt{\pi}\cdot\frac{3}{4}\sqrt{\pi}}{2!}=\frac{3}{8}\pi\]

Answer 16

wow awesome LaTeX