Author Topic: Integration help! (Read 615 times) Tweet Share

samsiexD · « **on:** May 16, 2013, 06:23:41 pm »

I'm having trouble with integration and using substituition for these two questions
1. Evaluate ∫(cosx)/(root(1+2sinx)) using an appropriate substituition method
This is what ive done
∫Cosx(1+2sinx)^-1/2
Let u= 1+2sinx
du/dx = 2cosx
..and then i got it wrong
The answer is root(1+2sinx))

And
2a. Find an antiderivative of x(x-2)^8
I got ((x-2)^9)/9 + c but thats also wrong

B. hence write down the answer to d/dx[(9x+2)(x-2)^9]

If anyone could help me out please and tell me what im doing wrong, i'd be so grateful

:

b^3 · « **Reply #1 on:** May 16, 2013, 06:35:40 pm »

'...and then I got it wrong'. What did you do after this?

Spoiler

$\begin{alignedat}{1}\int\frac{\cos\left(x\right)}{\sqrt{1+\sin\left(x\right)}}dx \\ \text{Let }u=1+2\sin\left(x\right) \\ \frac{du}{dx}=2\cos\left(x\right) \\ dx=\frac{1}{2\cos\left(x\right)}du \\ \int\frac{\cos\left(x\right)}{\sqrt{1+\sin\left(x\right)}}dx & =\frac{1}{2}\int\frac{1}{u^{\frac{1}{2}}}du \\ & =\frac{1}{2}\int u^{-\frac{1}{2}}du \\ & =\frac{1}{2}\times2u^{\frac{1}{2}}+C \\ & =u^{\frac{1}{2}}+C \\ & =\sqrt{1+2\sin\left(x\right)}+C \end{alignedat} $

$\begin{alignedat}{1}\int x\left(x-2\right)^{8}dx \\ \text{Let }u=x-2 \\ \frac{du}{dx}=1 \\ dx=du \\ x=u+2 \\ \int x\left(x-2\right)^{8}dx & =\int\left(u+2\right)u^{8}du \\ & =\int u^{9}+2u^{8}du \end{alignedat} $
You should be able to go from there

As for part B, your answer from part a will be a multiple of what you need from part b, so you should be able to work from there.

Hope that helps.

samsiexD · « **Reply #2 on:** May 16, 2013, 07:24:54 pm »

Thank you so much, very helpful

Search the forums now!

We have moved!

Author Topic: Integration help! (Read 615 times) Tweet Share

samsiexD

Integration help!

b^3

Re: Integration help!

samsiexD

Re: Integration help!

Recent Posts

User:
Password: