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Stuck on integration

I'm confused how the integral of cosysin3y dy is (1/4)sin4y... can someone please explain it through? Thanks in advance :biggrin:
it is a substitution; let u = siny

then find du/dy

make dy the subject

replace dy with the result of previous line (call it ?)

write the integral as cosy.u4.?

simplify

integrate with respect to u

rewrite the answer in terms of y

don't forget +c
Original post by chloe17
I'm confused how the integral of cosysin3y dy is (1/4)sin4y... can someone please explain it through? Thanks in advance :biggrin:


If u=sin⁡yu=\sin y then dudy=cos⁡y⇒du=cos⁡ydy\frac{du}{dy} = \cos y \Rightarrow du = \cos y dy.

Hence ∫cos⁡ysin⁡3ydy=∫sin⁡3ycos⁡ydy=∫u3du=⋯\int \cos y \sin^3 y dy = \int \sin^3 y \cos y dy = \int u^3 du = \cdots
Original post by the bear
x


Original post by atsruser
x


Is substitution really necessary here? This can be solved pretty simply through recognition.
Original post by StrangeBanana
Is substitution really necessary here? This can be solved pretty simply through recognition.


most students find the step by step method easier to use
Original post by StrangeBanana
Is substitution really necessary here? This can be solved pretty simply through recognition.


What can be done "pretty simply" by you cannot be done "pretty simply" by the OP.
Original post by atsruser
What can be done "pretty simply" by you cannot be done "pretty simply" by the OP.


exactly.... "Why doesn't OP use a Laplace Transform... it's really simple ?"

:rofl:
Original post by atsruser
What can be done "pretty simply" by you cannot be done "pretty simply" by the OP.


I'm not going to have a dick-swinging contest on here; the OP should learn integration by recognition if they don't know how to do it. It is much faster, and much simpler, than substitution.

Original post by the bear
exactly.... "Why doesn't OP use a Laplace Transform... it's really simple ?"

:rofl:


What a ridiculous comparison. :curious: Laplace transforms aren't in the A-level syllabus, and are significantly more complex than both recognition and substitution. Recognition is taught at A-level, and is the simplest method of integration once learnt, but if the OP wants to ignore the fastest method because they are too lazy to learn it, fine by me.
(edited 9 years ago)
I am stuck on this question if some one could help me please ?
integrate [dt (e^-1/2t) fn(t)] from 0 to infinity , I have tried to solve it by integrating by part taking u =fn(t) and dv=e^-1/2t and teh result is -1/2 L[F(t)] not sure if its right ? could some one help help help

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