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Mathematics 13 Online
OpenStudy (anonymous):

Find definite integral of cos0/1+2sin0 d0 on the interval pi/6, 0

OpenStudy (anonymous):

\[\int\limits_{0}^{ \frac{ \pi }{ 6 }}\frac{ \cos \theta }{ 1+2\sin \theta } d \theta\]

OpenStudy (anonymous):

@ganeshie8

OpenStudy (asnaseer):

HINT: Use the substitution \(u=1+2\sin(\theta)\)

OpenStudy (anonymous):

u sub. let u = 1 + 2sin\(\theta\)

OpenStudy (anonymous):

ok, so the it equals \[\frac{ 1 }{ 2 } \int\limits \frac{ 1 }{ u }du\]

OpenStudy (asnaseer):

correct - remember to adjust the limits of the integral accordingly

OpenStudy (anonymous):

how does that work? I never learned about adjusting the limits

OpenStudy (asnaseer):

well you have used:\[u=1+2\sin(\theta)\]so you need the value of u for the lower bound of theta and for the upper bound of theta

OpenStudy (asnaseer):

lower bound is zero, so what is u when theta=0?

OpenStudy (anonymous):

oh is it just solving the integral? plugging in the limits?

OpenStudy (asnaseer):

yes :)

OpenStudy (anonymous):

can i make the function\[\frac{ 1 }{ 2 }\ln \left| u \right|\]

OpenStudy (anonymous):

then sub in u as 1+2sin theta then solve for each of the limits and subtract them?

OpenStudy (asnaseer):

yes you can also do that

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

i get a really weird number for the first limit

OpenStudy (anonymous):

0.34657

OpenStudy (anonymous):

@asnaseer

OpenStudy (anonymous):

@ganeshie8

OpenStudy (asnaseer):

leave it in terms of the ln function - what answer do you get?

OpenStudy (anonymous):

ln sqrt(2), thank you!

OpenStudy (asnaseer):

are you sure???

OpenStudy (anonymous):

i think so....

OpenStudy (asnaseer):

actually yes - its correct - I got:\[\frac{\ln(2)}{2}\]which is indeed the same as yours. :)

OpenStudy (anonymous):

:)

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