# 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?

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.

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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?