1. ## Integration of (x-1)^99

Hi there,

(x-1)^99 ∂x

Also, cos^3 (ø) sin (ø) ∂ø

t*e^-t^2 ∂t

many thanks

Bryn

2. Originally Posted by Bryn
Hi there,

(x-1)^99

many thanks

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

3. 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

4. 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 ?

5. Originally Posted by skeeter
$\int (x-1)^{99} \, dx = \frac{(x-1)^{100}}{100} + C$
let $u=x-1$, then $du=dx$

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

6. 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 ...?