Question

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

Answer 1

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

Answer 2

@ganeshie8

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

correct - remember to adjust the limits of the integral accordingly

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

yes :)

Answer 12

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

Answer 13

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

Answer 14

yes you can also do that

Answer 15

ok

Answer 16

i get a really weird number for the first limit

Answer 17

0.34657

Answer 18

@asnaseer

Answer 19

@ganeshie8

Answer 20

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

Answer 21

ln sqrt(2), thank you!

Answer 22

are you sure???

Answer 23

i think so....

Answer 24

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

Answer 25

:)