Indefinite Integral Help - Sin[x]Cos[x]Cos[x] & e^(Sqrt[x+1])

[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.

 

[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.

 

[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

 

[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

 

[huan.conchito]

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

 

[Data]

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