# Thread: What is wrong with my Integration?

1. ## What is wrong (Integration)?Thanks,for help but is there a simpler/other way?

The question is: Give an exakt value to P so that:
$\displaystyle IntP-sin^3v):dv = 1$ Int: limits are from $\displaystyle 0->pi$

$\displaystyle sin3v = 3sinv-4sin^3v -> 4sin^3v = 3sinv-sin3v -> sin^3v = 3/4sinv-1/4sin3v$

$\displaystyle Int(P-3/4sinv+1/4sin3v):dv$ from $\displaystyle 0;pi$
$\displaystyle Int(P-3/4sinv+1/4sin3v):dv$ = $\displaystyle (Pv+3/4cosv-3/4cos3v)$

$\displaystyle P(pi)+3/4+3/4-3/4-3/4 = 1$ -> $\displaystyle P(pi) = 1$ but the answer is $\displaystyle P = 7/3pi$

Anyway, Here is the solution:
$\displaystyle \int_{0}^{\pi} ( P - sin^3(v) ) dv = 1$
$\displaystyle P\int_{0}^{\pi} dv - \int_{0}^{\pi} sin^3(v) dv = 1$
$\displaystyle (P)(\pi) - \int_{0}^{\pi} ( 1-cos^2(v) ) sin(v) dv = 1$
For the integral substitute
$\displaystyle t=cos(v)$
$\displaystyle dt=-sin(v) dv$
$\displaystyle t(\pi)=-1$
$\displaystyle t(0)=1$
$\displaystyle (P)(\pi) + \int_{1}^{-1} ( 1 - t^2 ) dt = 1$
$\displaystyle (P)(\pi) - \int_{-1}^{1} ( 1 - t^2 ) dt = 1$
since $\displaystyle f(t)=1-t^2$ is an even function, the last equation can be writte as follows:
$\displaystyle (P)(\pi) - 2 \int_{0}^{1} ( 1 - t^2 ) dt = 1$
Integrate and evaluate, to get:
$\displaystyle (P)(\pi) - 2 [ 1 - \frac{1}{3} ] = 1$
$\displaystyle (P)(\pi) - 2 (\frac{2}{3}) = 1$
$\displaystyle (P)(\pi) - \frac{4}{3} = 1$
$\displaystyle (P)(\pi) = \frac{7}{3}$
$\displaystyle P = (\frac{7}{3}) (\frac{1}{\pi})$
$\displaystyle P=\frac{7}{3\pi}$

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Use \pi to write pi`s symbol in LaTeX.

3. ## Thanks

I´ll give you Thanks for sure.

But Oh My God is that what I have to do to solve it?
Haven´t learned a thing about Integration by substitution yet, I checked the wikipedia for info on the subject, but I think it will take some time to melt. You wouldn´t happen to have a e-book/book to recommend on the subject? Or know anywhere I can find some info on how to integrate by substitution?