Integral Solutions to Complex Equations

[-EquinoX-]

Homework Statement

Find the integral of
$$\int{9cos(t)e^{9sin(t)} - e^{cos(t)}sin(t)} dt$$
$$\int{ - t^2sin(t) + 2tcos(t)} dt$$
$$\int{-8(sin(t))^2 - 2cos(t)sin(2sin((t)))} dt$$

Homework Equations

The Attempt at a Solution

I got -2tsin(2t) - t^2cos(t) + 2 cos(t) - 2t*sin(t)

is this correct?

 

[xepma]

You calculated the derivative... Not the primitive.

(It's not even possible to write down the primitive in terms of elementary functions)

 

[-EquinoX-]

oops.. that's right, so how do I get the integrals of it

 

[n!kofeyn]

For the first listed integral, break the integral into two integrals and then perform a separate u-substitution for both integrals. I.e., let $u$ be something in the first integral, and let $v$ be something in the second integral.

For the second listed integral, again break the integral into two integrals and then use integration by parts. You will need to do integration by parts twice for the first of the two integrals.

Let's hold off on the third integral for now. Further hints can be given if you are still stuck.

 

[-EquinoX-]

are you saying u = 9cos(t)e^{9sin(t)} or are you asking to break this up to a u and v

 

[n!kofeyn]

are you saying u = 9cos(t)e^{9sin(t)} or are you asking to break this up to a u and v

No. That isn't how you apply u-substitution. Break the integral up as:
$$\int \left( 9\cos t e^{9\sin t} - \sin t e^{\cos t} \right) \,dt = \int 9\cos t e^{9\sin t} \,dt + \int (-\sin t) e^{\cos t} \,dt$$

For the integral on the left. Let $u = 9\sin t$. Then $du = 9\cos t \,dt$. Then the integral on the left becomes
$$\int \underbrace{e^{9\sin t}}_{e^u} \, \underbrace{9\cos t \,dt}_{du} = \int e^u \,du = e^u + C = e^{9\sin t} + C$$

Now try the integral on the right by letting v be something, similar to the one I just did.

 

[-EquinoX-]

thanks for the help