Thread: 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.

that should be

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

Originally Posted by harish21

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

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?

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.

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