# I need help with integration, !

1. Apr 2, 2005

### huan.conchito

I need help with integration, urgent!

Can someone please show me the steps in calculating the indefinite integrals of:

Sin[x]Cos[x]Cos[x]
and
e^(Sqrt[x+1])

Please show all the steps, and not just the answer, which I can do on my calculator.

Thank you.

Last edited: Apr 2, 2005
2. Apr 2, 2005

### Data

For the first one, just substitute $u = \cos{x} \Longrightarrow du = -\sin{x} \ dx$.

For the second, substitute $u = \sqrt{x+1} \Longrightarrow du = \frac{dx}{2\sqrt{x+1}} \Longrightarrow 2u \ du = dx$, and then integrate by parts.

Last edited: Apr 2, 2005
3. Apr 2, 2005

### huan.conchito

thats exactly what I did, but I can't get the right answer. Could you please do it, i'll give you 10 gmail invites, hehe

4. Apr 2, 2005

### Data

Using the substitutions I posted above,

$$\int \sin{x} \cos^2{x} \ dx = -\int u^2 \ du = -\frac{u^3}{3} + C = -\frac{\cos^3{x}}{3} + C$$

and

$$\int e^{\sqrt{x+1}} \ dx = 2\int ue^u \ du = 2\left[ue^u - \int e^u \ du\right] = 2\left[ue^u - e^u\right] + C= 2e^u(u-1) + C = 2e^{\sqrt{x+1}}\left(\sqrt{x+1}-1\right) + C$$

Last edited: Apr 2, 2005
5. Apr 2, 2005

### huan.conchito

thanks, in the first one, i forgot the sin[x] dissapears, when you integrate, 2nd one still trying myself

6. Apr 2, 2005

### Data

Well, it doesn't dissapear. It's part of the $du$.