Question

Integrate...again.
See below.

Answer 1

\[\int\limits \sin \pi t \cos \pi t dt\]

Answer 2

π is a constant.
so it would be the same as
cos(2x)

Answer 3

I think you want to use u substitution.

Answer 4

i know that u = pi*t
so du = pi*dt
so then pi integral (sinu)(cosu) du
but i don't know how to integrate sin u cos u ...

Answer 5

You have sin and cos. you know that
f'(t) of sin(t) = cos(t). That's a hint that you can use u-substitution to replace one.

Answer 6

Try \(v = \sin(u) \quad dv = \cos(u)du\).

Answer 7

ohhhh thanks @abb0t

Answer 8

yw! integrals are my favorite.

Answer 9

Did you figure out the answer?

Answer 10

i did it a slightly different way but i got \[-\frac{ 1 }{ 4 \pi } \cos(2 \pi t) +C\]

Answer 11

Wait, where did the (2πt) and 4π come from?

Answer 12

@abb0t chain rule

Answer 13

i didn't really use u-sub. i used the double angle formula sin2x = 2cosxsinx and played with it and used 2pi*t as my u.

Answer 14

You can substitute for either trigonometric function.$$u=\sin\pi t\\du=\pi\cos\pi t\ dt\\\frac1\pi du=\cos\pi t\ dt\\\int\sin\pi t\cos\pi t\ dt=\frac1\pi \int u\ du=\frac1{2\pi}u^2+C=\frac1{2\pi}\sin^2\pi t+C$$... or$$u=$$

Answer 15

@oldrin.bataku
I understand the chain rule, but that's wasn't in the original problem stated. Unless she made a mistake.

Answer 16

\[
\int \sin (\pi t) \cos (\pi t) dt
\]\(u = \pi t \quad du = \pi dt\) \[
\int \sin (u) \cos (u) \frac{du}{\pi}
\] \(v = \sin (u) \quad dv = \cos(u)du\) \[

\int v \frac{dv}{\pi}
\]\[= \frac{v^2}{2\pi} +C = \frac{[\sin(u)]^2}{2\pi}+C = \frac{[\sin(\pi x)]^2}{2\pi}+C
\]

Answer 17

jenny, you should look @ oldrin's explanation, it's a bit simpler, step-wise.

Answer 18

Should be \(t\) not \(x\).

Answer 19

sin2x = 2sinxcosx
1/2 sin2x = sinxcosx
\[\int\limits \sin(2\pi t)dt\] is what i did

Answer 20

http://www.wolframalpha.com/input/?i=-1%2F4pi+%28cos%282pi*t%29+%3D+%5B-cos%5E2%28pi*t%29%5D%2F2pi
it's the same thing...

Answer 21

here's what i did.

Answer 22

@jennychan12 $$\int\sin\pi t\cos\pi t\ dt\\\sin\pi t\cos\pi t=\frac12\sin2\pi t\\u=2\pi t\\du=2\pi\ dt\\\frac1{2\pi}du=dt\\\frac12\int\sin2\pi t\ dt=\frac1{4\pi}\int\sin u\ du=-\frac1{4\pi}\cos u+C=-\frac1{4\pi}\cos2\pi t+C$$

Answer 23

yeah that's what i did.

Answer 24

by parts would work also... lots of ways to do this problem

Answer 25

The integrals are the same.
http://www.wolframalpha.com/input/?i=1%2F%282pi%29+sin%5E2+%28pi+t%29+%3D+-1%2F%284pi%29+cos+%282pi+t%29

... the only difference is that they will use a different constant of integration. this happens a lot in calculus; depending on what technique you use, the results may look different.