# Thread: How to solve definite integrals?

1. ## How to solve definite integrals?

Example: the integral from 0 to 0.5pi of 2cosx with respect to x. (Sorry for writing it like this but I don't know any other ways) What I got was sin(0.5pi)^2 - (sin(0)^2)=1 However, the back of the book says that the answer is 2. Can someone explain what I am doing wrong? Thanks.

2. ## Re: How to solve definite integrals?

that should be $\int_0^{0.5\pi} 2 \cos(x)\;dx = [2\sin(x)]_0^{0.5 \pi}=...$

how did you get sin(0.5pi)^2 - (sin(0)^2)=1

3. ## Re: How to solve definite integrals?

Originally Posted by harish21
that should be $\int_0^{0.5\pi} 2 \cos(x)\;dx = [2\sin(x)]_0^{0.5 \pi}=...$
how did you get sin(0.5pi)^2 - (sin(0)^2)=1
$2\sin^2 \left( {\frac{\pi }{2}} \right) = 2$

4. ## Re: How to solve definite integrals?

Well, I treated cosx just like an x. If you integrate 2x you would get x^2 so I figured that I have to square the sin and divide it by 2.
Plato, why do you keep 2sin? Don't you divide by the new power which would leave just a sin?

5. ## Re: How to solve definite integrals?

Originally Posted by Emanresu
Well, I treated cosx just like an x. If you integrate 2x you would get x^2 so I figured that I have to square the sin and divide it by 2.
Plato, why do you keep 2sin? Don't you divide by the new power which would leave just a sin?
???

I think you need to look at your textbook. Integrating 2cos(x) has nothing to do with integrating 2x.. at all. Pull the 2 out of the integrand. What would you take the derivative of to get cos(x)? That is the integral. This has nothing to do with increasing the exponent and dividing by the new exponent because this is not 2x.

6. ## Re: How to solve definite integrals?

I really have problems, don't I? lol
Anyway, thanks to everyone who replied.