Integration of (x-1)^99

**Bryn** · November 27th 2009, 05:36 AM

Hi there,

(x-1)^99 ∂x

Also, cos^3 (ø) sin (ø) ∂ø

t*e^-t^2 ∂t

many thanks

Bryn

**skeeter** · November 27th 2009, 05:42 AM

Originally Posted by Bryn

Hi there,

(x-1)^99

many thanks

Bryn

$\int (x-1)^{99} \, dx = \frac{(x-1)^{100}}{100} + C$

**Bryn** · November 27th 2009, 05:57 AM

But why is this, why doesn't the x-1 need integrating or differentiating

I can't see how you can just integrate x-1 as if it was q for example

Thanks

**skeeter** · November 27th 2009, 06:09 AM

Originally Posted by Bryn

But why is this, why doesn't the x-1 need integrating or differentiating

I can't see how you can just integrate x-1 as if it was q for example

Thanks

because (x-1) is a function

q is ?

**dedust** · November 27th 2009, 07:28 AM

Originally Posted by skeeter

$\int (x-1)^{99} \, dx = \frac{(x-1)^{100}}{100} + C$

let $u=x-1$ , then $du=dx$

and $\int (x-1)^{99} \, dx =\int (u)^{99} \, du =\frac{u^{100}}{100} + C= \frac{(x-1)^{100}}{100} + C$

**Grandad** · November 27th 2009, 08:18 AM

Hello Bryn

Originally Posted by Bryn

Hi there,

(x-1)^99 ∂x

Also, cos^3 (ø) sin (ø) ∂ø

t*e^-t^2 ∂t

many thanks

Bryn

All these integrals can be solved by substitution. The first has already been done. I'll start you off for numbers 2 & 3.

2) Put $\cos\phi = u$

Then $-\sin\phi\; d\phi=du$

So $\int\cos^3\phi\sin\phi\;d\phi = -\int u^3\;du=...$

3) Put $e^{-t^2}=u$

Then $e^{-t^2}(-2t)\;dt=du$

So ...?

Grandad

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