Ocr c4 integration help

given that y=pi/4 when x=1 solve the differential equation

dy/dx = x sec y cosec^3y


....i've rearranged to get an integral but i'm struggling now

i have cos y sin^3y dy = x dx

help please?

i'm thinking integration by parts? ..the answer i'm supposed to get is

sin^4y = 2x^2 + k

wheni get that i'll be able to do it

$\frac{\mathrm{d}y}{\mathrm{d}x} = x \sec y \, \mathrm{cosec}^3 y[br][br]\Rightarrow \; \cos y \sin^3 y \; \mathrm{d}y = x \; \mathrm{d} x [br][br]\Rightarrow \; \int \cos y \sin^3 y \; \mathrm{d}y = \int x \; \mathrm{d} x$

The RHS should be very simple to integrate; but look at the LHS. And think of what differentiates to give it - this sort of simple 'recognition' of what differentiates to give what saves you a nasty Integration by Parts job:

$\Rightarrow \; \sin^4\ y = \frac{x^2}{2} + c$

And then you sub in your numbers to get $c = \frac{-1}{4}$.

Would it not be $\frac{1}{4} sin^4 y$ on the LHS?

Yeah it would be, he's was just saying try and differentiate $sin^4y = \frac{1}{2}x^2 + C$ and see what adjustments have to be made e.g. coefficients.

Would it not be $\frac{1}{4} sin^4 y$ on the LHS?

Ah apologies! Thanks.

Mother*$&#ing indices.


Indeed it would, apologies.

$\frac{\mathrm{d}y}{\mathrm{d}x} = x \sec y \, \mathrm{cosec}^3 y[br][br]\Rightarrow \; \cos y \sin^3 y \; \mathrm{d}y = x \; \mathrm{d} x [br][br]\Rightarrow \; \int \cos y \sin^3 y \; \mathrm{d}y = \int x \; \mathrm{d} x$

The RHS should be very simple to integrate; but look at the LHS. And think of what differentiates to give it - this sort of simple 'recognition' of what differentiates to give what saves you a nasty Integration by Parts job:

$\Rightarrow \; \sin^4 y = \frac{x^2}{2} + c$

And then you sub in your numbers to get $c = \frac{-1}{4}$.

can you explain how this helps please ? Just im quite confused following through :/

So those are the corrected numbers, integrated properly. That now gives you a general solution, which is your answer, unless they've given you other values to substitute in...?

given that y=pi/4 when x=1 solve the differential equation

dy/dx = x sec y cosec^3y


....i've rearranged to get an integral but i'm struggling now

i have cos y sin^3y dy = x dx

help p

Would it not be $\frac{1}{4} sin^4 y$ on the LHS?

thank you !! lol cos i didn't understand this I kind of trawled around the book and now got it under my belt

i thought when you integrate cos y sin^3 y you get 1/4 sin y x


Does anyone know how to answer this question i don't see how to integrate get 1/4 sin^4 y

Like i get how sin^3 y integrates to give 1/4 sin^4 y but what happens to the cos y

Its 10 years old. Its better to start your own thread rather than resurrect old, zombie ones. But you can use substitution u = sin(y) or simply reverse chain rule to get
Int cos(y) sin^3(y) dy -> 1/4 sin^4(y) + c

I thought when you integrate cos y you get -siny