Integration by Parts: Verify Formula for $\int x^{n} sin x dx$

[clairez93]

Homework Statement

\int\frac{t^{2}}{\sqrt{2+3t}}

Use integration by parts to verify the formula:
\int x^{n} sin x dx = -x^{n} cos x + n\int x^{n-1} cos x dx

Homework Equations

The Attempt at a Solution

For the first one, I attached the picture of my work on paper, as it would take me forever to type out in latex code, I think. For the second one:

u = sin x
du = cos x
dV = x^{n}
V = \frac{x^{n+1}}{n+1}

\int x^{n} sin x dx = -x^{n} cos x + n\int x^{n-1} cos x dx =
sin x (\frac{x^{n+1}}{n+1}) - \int \frac{x^{n+1}}{n+1} cos x dx

That doesn't really look like the formula to me. Am I supposed to use an identity of some sorts?

 

[mathie.girl]

Homework Statement

\int\frac{t^{2}}{\sqrt{2+3t}}

Use integration by parts to verify the formula:
\int x^{n} sin x dx = -x^{n} cos x + n\int x^{n-1} cos x dx

Homework Equations

The Attempt at a Solution

For the first one, I attached the picture of my work on paper, as it would take me forever to type out in latex code, I think. For the second one:

u = sin x
du = cos x
dV = x^{n}
V = \frac{x^{n+1}}{n+1}

\int x^{n} sin x dx = -x^{n} cos x + n\int x^{n-1} cos x dx =
sin x (\frac{x^{n+1}}{n+1}) - \int \frac{x^{n+1}}{n+1} cos x dx

That doesn't really look like the formula to me. Am I supposed to use an identity of some sorts?

I can't see the work for the first one, so I can't tell if that's right or not.

For the second one, you differentiated \sin{x} to get \cos{x}and integrated x^n to get \frac{x^{n+1}}{n+1}, is that right? However, since the formula you're supposed to end up with has an x^{n-1}, I would differentiate the x^n and integrate the \sin{x} and see what you get.

 

[n!kofeyn]

mathie.girl is right. A good thing to remember when doing integration by parts is that you let u be the term such that when you differentiate it, du is "simpler" than u. For example, you let u=sinx so that du=cosx. That really doesn't do any simplifying. That means you'll let dv be the term that doesn't really simplify when taking the derivative.