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OpenStudy (legomyego180):
What am I doing wrong in this integral?
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OpenStudy (legomyego180):
\[\int\limits_{\frac{ \pi }{ 3 }}^{\frac{ \pi }{ 2 }}\frac{ (3\sin(x)) }{ (4+2\cos(x))^2 }\]
OpenStudy (legomyego180):
\[u=2\cos(x)+4\]
\[du=-2\sin(x)\]
\[-\frac{ 2 }{ 3 }\int\limits_{\frac{ \pi }{ 3 }}^{\frac{ \pi }{ 2 }}\frac{ 1}{ u^2 }\]
OpenStudy (legomyego180):
that should be negative, sorry
myininaya (myininaya):
i think i see ...
\[du=-2 \sin(x) dx \\ \frac{3}{-2} du= 3 \sin(x) dx\]
you should have -3/2 out in front
OpenStudy (legomyego180):
Oh yea, you're right
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OpenStudy (legomyego180):
Im still getting the wrong answer though
myininaya (myininaya):
did you want to show the rest of your work
OpenStudy (holsteremission):
Check the limits. They don't stay the same once you change the variable.
OpenStudy (legomyego180):
\[-\frac{ 3 }{ 2 }\left[ -\frac{ 1 }{ 4 }+\frac{ 1 }{ 5 } \right]\]
OpenStudy (legomyego180):
Thats what I got after evaluating my bounds
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OpenStudy (holsteremission):
That's correct.
OpenStudy (legomyego180):
Oh I have to change the bounds by plugging into u?
OpenStudy (legomyego180):
Oh I typed it into wolfram wrong...
>.<
OpenStudy (legomyego180):
Thank you guys, I appreciate the help. I was afraid I had forgotten how to do integrals overnight...
OpenStudy (canimcan):
Just write it again and you'll solve it.
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