- #1
tangibleLime
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Homework Statement
Integrate from 0 to sqrt(pi/2). I can't figure out how to make the LaTeX work correctly for it, so it just looks like an indefinite integral.
[tex]\int 7\theta^3cos(\theta^2) d\theta[/tex]
Homework Equations
[tex]\int udv = uv - \int vdu[/tex]
The Attempt at a Solution
First I took the 7 out of the integral and pulled a theta out of theta cubed in order to match it with the theta squared.
[tex]7 \int \theta^2cos(\theta^2)\theta d\theta[/tex]
Then I substituted [tex]w = \theta^2, dw = 2\theta d\theta, \frac{dw}{2}=\theta d\theta[/tex].
To make the bounds of integration to work with this substitution...
[tex]w = 0^2 = 0[/tex]
[tex]w = \sqrt{pi/2} = \frac{pi}{2}[/tex]
[tex]w = \sqrt{pi/2} = \frac{pi}{2}[/tex]
After pulling out the 2 that comes along with the dw, this is what my integral is looking like.
[tex]7\frac{1}{2}\int w*cos(w) dw[/tex]
Now, using integration by parts...
[tex]u = w[/tex]
[tex]du = dw[/tex]
[tex]v = sinw[/tex]
[tex]dv = cosw dw[/tex]
[tex]uv - \int vdu[/tex]
[tex]7\frac{1}{2}(w*sinw - \int sinw dw)[/tex]
[tex]7\frac{1}{2}(w*sinw + cosw)[/tex]
[tex]du = dw[/tex]
[tex]v = sinw[/tex]
[tex]dv = cosw dw[/tex]
[tex]uv - \int vdu[/tex]
[tex]7\frac{1}{2}(w*sinw - \int sinw dw)[/tex]
[tex]7\frac{1}{2}(w*sinw + cosw)[/tex]
Substituting back the original value of w...
[tex]7\frac{1}{2}(\theta^2*sin\theta^2 + cos\theta^2)[/tex]
Now using the Fundamental Theorem...
[tex]7\frac{1}{2}(\frac{pi}{2}^2*sin\frac{pi}{2}^2 + cos\frac{pi}{2}^2) - 7\frac{1}{2}(0^2*sin0^2 + cos0^2)[/tex]
This gives some ridiculous answer:
[tex]3.5 ((pi/2)^2 sin((pi/2)^2)+cos((pi/2)^2))-3.5 (0^2 sin(0^2)+cos(0^2)) = -3.5 cosh(0)+3.5 (cosh(-i (pi/2)^2)-cos(pi/2+(pi/2)^2) (pi/2)^2)[/tex]
...which is obviously incorrect. Where did I go wrong?
Thanks.