# Integrate int (sin^2(t) + cos^2(t) -1)dt from e^x to e^(2x)

cooltee13

## Homework Statement

Integrate from e^x to e^2x: (sin^2(t) + cos^2(t) -1)dt

## Homework Equations

just standard integral equations

## The Attempt at a Solution

I know how to do most of it, my only question is: is (sin^2(e^2x) + cos^2(e^x) -1) a special trig identity? or would i just solve it like a normal interation in parts problem?

## Answers and Replies

sutupidmath
well it seems trivial to me... lol
This question seems to be just tricky, nothing else.

regor60
Aren't the sin and cos together adding to 1 the way you've written it, thus 1-1=0 ? You may want to double check you have it correct

sutupidmath
$$\int_{e^{x}}^{e^{2x}}(sin^{2}(t)+cos^{2}(t)-1)dt=\int_{e^{x}}^{e^{2x}}(1-1)dt=\int_{e^{x}}^{e^{2x}}(0)dt=?$$

What does this equal to??

cooltee13
ok, thanks guys lol. I feel dumb now

sutupidmath
ok, thanks guys lol. I feel dumb now

Ok, integrate the following:

$$\int_0^{ln(1)} sin(x)e^{-x^{2}}dx$$

This question was somewhere i don't know where though. Give a shot to it.

HINT: THis is also tricky.

Last edited:
cooltee13
$$\int_{e^{x}}^{e^{2x}}(sin^{2}(t)+cos^{2}(t)-1)dt=\int_{e^{x}}^{e^{2x}}(1-1)dt=\int_{e^{x}}^{e^{2x}}(0)dt=?$$

What does this equal to??

Isnt that just equal to 1?

rocomath
Look at your limits of Integration

From 0 to Ln(1)

sutupidmath
Isnt that just equal to 1?

Why on Earth do u think it is equal to 1?

sutupidmath
Look at your limits of Integration

From 0 to Ln(1)
How did u change your display name?