# Integration by Parts

posted by Ashley

integral from 0 to 2pi of isin(t)e^(it)dt. I know my answer should be -pi.
**I pull i out because it is a constant.
My work:
let u=e^(it) du=ie^(it)dt dv=sin(t) v=-cos(t)
i integral sin(t)e^(it)dt= -e^(it)cos(t)+i*integral cost(t)e^(it)dt

do integration by parts again, then I get
=-e^(it)cos(t)+i[e^(it)sin(t)-i*integral e^(it)sin(t)dt.
then i add i*integral e^(it)sin(t)dt to both sides
2i*integral e^(it)sin(t)dt= -e^(it)(cos(t)-i(sin(t)))
divide by 2 and because e^(-it)=cos(t)-isin(t)

i*integral e^(it)sin(t)dt=-1/2

So my question is how do I get -pi from this? I have not plugged in 0 to 2pi interval and I should have -t/2 as a function to plug the interval in.

1. Steve

since sin(t) = (e^-t - e^-it)/2,
isin(t) = i/2 (e^it - e^-it)
i sint e^it = i/2 (e^2it - 1)

so, the integral is just

i/2 ((1/2i) e^2it - t)
= -1/4 e^2it - t/2
do that from 0 to 2π and you have
[1/4 e^4πi - π]-[1/4]
= 1/4 - π - 1/4
= -π

I haven't checked your integration by parts, but I get

-t/2 + 1/4 (sin 2t - i cos 2t)

## Similar Questions

1. ### calc

d/dx integral from o to x of function cos(2*pi*x) du is first i do the integral and i find the derivative right. by the fundamental theorem of calculus, if there is an integral from o to x, don't i just plug the x in the function. …
2. ### trig integration

s- integral endpoints are 0 and pi/2 i need to find the integral of sin^2 (2x) dx. i know that the answer is pi/4, but im not sure how to get to it. i know: s sin^2(2x)dx= 1/2 [1-cos (4x)] dx, but then i'm confused. The indefinite …
3. ### trig integration

i'm having trouble evaluating the integral at pi/2 and 0. i know: s (at pi/2 and 0) sin^2 (2x)dx= s 1/2[1-cos(2x)]dx= s 1/2(x-sin(4x))dx= (x/2)- 1/8[sin (4x)] i don't understand how you get pi/4 You made a few mistakes, check again. …
4. ### Integral

That's the same as the integral of sin^2 x dx. Use integration by parts. Let sin x = u and sin x dx = dv v = -cos x du = cos x dx The integral is u v - integral of v du = -sinx cosx + integral of cos^2 dx which can be rewritten integral …
5. ### Math/Calculus

How would I integrate the following by parts: Integral of: (x^2)(sin (ax))dx, where a is any constant. Just like you did x^2 exp(x) below. Also partial integration is not the easiest way to do this integral. You can also use this method. …
6. ### Calculus

I need help with this integral. w= the integral from 0 to 5 24e^-6t cos(2t) dt. i found the the integration in the integral table. (e^ax/a^2 + b^2) (a cos bx + b sin bx) im having trouble finishing the problem from here.
7. ### math

I'm trying to find the convolution f*g where f(t)=g(t)=sin(t). I set up the integral and proceed to do integration by parts twice, but it keeps working out to 0=0 or sin(t)=sin(t). How am I supposed to approach it?
8. ### Integral Help

I need to find the integral of (sin x)/ cos^3 x I let u= cos x, then got -du= sin x (Is this right correct?
9. ### Calculus help??

I'm not sure how to solve this and help would be great! d/dx [definite integral from 0 to x of (2pi*u) du] is: a. 0 b. 1/2pi sin x c. sin(2pi x) d. cos (2pi x) e. 2pi cos (2pi x) This is the fundamental theorem, right?
10. ### Math

Evaluate the integral of (e^2x)*sin^3 x dx I let u = e^2x, du = (1/2)e^2x dx v= (-1/3)cos^3 x , dv =sin^3 x dx When I used integration by parts and solved it all out I got: (37/36)intgral of (e^2x)*sin^3 x dx = (-1/3)(e^2x)*cos^3 x …

More Similar Questions