Integration question (1 Viewer)

Shuuya · Jan 15, 2016

I need some help with evaluating this... I think I've got the integration bit right, but I don't know what to do with the 'n' when evaluating

Thanks!

DatAtarLyfe · Jan 15, 2016

Deal with it like any other exponent in an integral, raise the power, divided by the raised power, sub in limits

InteGrand · Jan 15, 2016

Shuuya said:
I need some help with evaluating this... I think I've got the integration bit right, but I don't know what to do with the 'n' when evaluating

Thanks!

$\noindent Let $u=\cos x$, $\mathrm{d}u=-\sin x \text{ d}x$. When $x=0,u=1$ and when $x=\frac{\pi}{2},u=0$, so we have$

$\begin{align*}\int _0 ^{\frac{\pi}{2}} \sin x \cos ^n x \text{ d}x &= \int _1 ^ 0 - u^n \text{ d}u \\ &= \int_0 ^1 u^n \text{ d}u \\ &= \frac{1}{n+1}.\end{align*}$

Shuuya · Jan 15, 2016

InteGrand said:
$\noindent Let $u=\cos x$, $\mathrm{d}u=-\sin x \text{ d}x$. When $x=0,u=1$ and when $x=\frac{\pi}{2},u=0$, so we have$

$\begin{align*}\int _0 ^{\frac{\pi}{2}} \sin x \cos ^n x \text{ d}x &= \int _1 ^ 0 - u^n \text{ d}u \\ &= \int_0 ^1 u^n \text{ d}u \\ &= \frac{1}{n+1}.\end{align*}$

This might be a dumb question but how did you get from the second line to the last line?

... Wait don't worry, I got it

Thanks!

InteGrand · Jan 15, 2016

Shuuya said:
This might be a dumb question but how did you get from the second line to the last line?

$\noindent We have$

$\begin{align*}\int _0 ^1 u^n \text{ d}u &= \left[\frac{u^{n+1}}{n+1} \right]_0 ^1 \quad (\text{using the power rule for integration, assuming }n\neq -1) \\ &= \frac{1^{n+1}}{n+1} - \frac{0^{n+1}}{n+1} \\ &= \frac{1}{n+1} - 0 \\ &= \frac{1}{n+1}. \end{align*}$

InteGrand · Jan 15, 2016

Shuuya · Jan 15, 2016

InteGrand said:
$\noindent We have$

$\begin{align*}\int _0 ^1 u^n \text{ d}u &= \left[\frac{u^{n+1}}{n+1} \right]_0 ^1 \quad (\text{using the power rule for integration, assuming }n\neq -1) \\ &= \frac{1^{n+1}}{n+1} - \frac{0^{n+1}}{n+1} \\ &= \frac{1}{n+1} - 0 \\ &= \frac{1}{n+1}. \end{align*}$

I was trying to sub in cosx for u so I was getting confused, but makes sense now

DatAtarLyfe · Jan 15, 2016

Shuuya said:
I was trying to sub in cosx for u so I was getting confused, but makes sense now

you don't sub the function for u after you've already changed the limits in a definite integral

Integration question (1 Viewer)

Shuuya

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InteGrand

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Shuuya

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InteGrand

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Shuuya

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