Couple of Problems - simple integrals

1. Oct 9, 2006

sapiental

1) definite integral 0 to 1 [1/1+x^2 + 9e^x - 1/sqrt(x)]
= tan^-1(x) + 9e^x - 2x^1/2
I know that to evaluate it I just plug in b for x - a for x

2) indefinte integral sin(3t)cos(3t)

= 1/3 (-cos(3t)sin(3t))dt

2. Oct 9, 2006

dextercioby

2) is definitely wrong. As for 1), i think you mean the simpler version, but from what you've written one understands

$$\int_{0}^{1}\frac{dx}{1+x^{2}+9e^{x}-\frac{1}{\sqrt{x}}}$$

Daniel.

3. Oct 9, 2006

sapiental

hi, for #1 I meant the integral from 0 to 1 [(1/1+x^2)+ 9e^x + (1/sqrt(x))]dx

sorry, I'm not sure how to do edit it as well as you did.

Where exactly did I mess up on #2 and how should I approach it?

Thanks!

4. Oct 9, 2006

dextercioby

HINT: $(\sin 3x)'=3\cos 3x$. So i guess a substitution will do.

Daniel.

5. Oct 9, 2006

sapiental

thanks alot for the hint it was very helpful

now I get: 1/6(sin3x)^2+C

please let me know if this is correct.

Thanks again.

6. Oct 9, 2006

dextercioby

Surely it is. You can always check the result by differentiation.

Daniel.