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

  • Thread starter cooltee13
  • Start date
1. The problem statement, all variables and given/known data

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

2. Relevant equations
just standard integral equations


3. 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?
 
well it seems trivial to me... lol
This question seems to be just tricky, nothing else.
 
101
0
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
 
[tex]\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=?????[/tex]

What does this equal to??
 
ok, thanks guys lol. I feel dumb now
 
ok, thanks guys lol. I feel dumb now
Ok, integrate the following:

[tex]\int_0^{ln(1)} sin(x)e^{-x^{2}}dx[/tex]

This question was somewhere i dunno where though. Give a shot to it.

HINT: THis is also tricky.
 
Last edited:
[tex]\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=?????[/tex]

What does this equal to??


Isnt that just equal to 1?
 
1,750
1
Look at your limits of Integration

From 0 to Ln(1)
 

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