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ksaimouli
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3siny sqtcosy dy
ksaimouli
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\[\int\limits_{}^{}3\sin y \sqrt{cosy} \]
Jemurray3
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substitute u = cos(y), du = -sin(y) dy
ksaimouli
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yup i did i am getting \[-3\sqrt{cosy}\]
Jemurray3
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but cos(y) is u.
Jemurray3
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\[ \int -3\sqrt{u} du \]is a pretty easy integral.
ksaimouli
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|dw:1357088383865:dw|
ksaimouli
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see i am getting <-3 {sqrtcosy}/>
ksaimouli
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\[-3 {\sqrt cosy}\]
Jemurray3
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I reiterate that cos(y) is just u.
Jemurray3
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so that is -3 sqrt(u)
ksaimouli
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yes thats what i did -3 sqrt(u) and before we said u=cosx so i rewrote it back
ksaimouli
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is that right answer for this question 3siny sqtcosy dy
Jemurray3
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No. After your substitution the integral becomes
\[ \int -3 \sqrt{u} du\]
you need to integrate this, which you should be able to do in two seconds. Then, you can replace all the u's with cos(y)'s and then you'll be done.
MrDoe
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jemurray3 is correct, just think of the square root as u^1/2 i think thats what your having trouble with
ksaimouli
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-2u^(3/2)
ksaimouli
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thx