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Finding an integral using trig. substitution

  • Thread starter adam199
  • Start date
  • #1
17
0
The integral from 0 to pi/2 of:

cos(t)/sqrt(1+sin^2(t)) dt

I'm supposed to use trig. substitution to find the solution. I started by using the formula a^2+x^2 to get x=atanx. In this case, sin(t)=(1)tan(θ), and so cos(t)dt=sec^2(θ)dθ and so I substituted this into the equation and got:

sec^2(θ)/sqrt(1+tan^2(θ)) -> sec^2(θ)/sqrt(sec^2(θ)) -> sec(θ)

Now, I have the integral of sec(θ)dθ, which equals ln abs(secθ+tanθ). When I take this integral at 0, it turns out to be 0, and so I'm left with the answer being solely the integral at pi/2. The problem is that the secant at pi/2 is infinity and so is the tangent at pi/2, and so the answer ends up being infinity, and this is apparently wrong.

Any help would be greatly appreciated.
 

Answers and Replies

  • #2
3,812
92
It would be better if you substitute sin(t)=x.
 
  • #3
33,507
5,192
A trig substitution can be used after you use the substitution that Pranav-Arora recommends.
 
  • #4
17
0
It would be better if you substitute sin(t)=x.
I substituted sin(t)=x and got:

1/sqrt(1+x^2)dx

but when I use the trig. substitution with this integral, I come up with the same thing I had before. I don't think I understand how to do it substituting sin(t)=x and then using a trig. substitution, like Mark said.
 
  • #5
17
0
Never mind. I eventually figured out how to translate the upper and lower limits of the integrand over to the final versions. Thanks.
 

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