The Student Room Group

cos^(3)x substitution

How does the integral of
(8sin^2 x)(cos^3 x) dx
become the integral of
(8sin^2 x)(1-sin^2 x) dsinx?

I understand that cos^2 x= 1-sin^2 x but thats about it?
Original post by wkatie
How does the integral of
(8sin^2 x)(cos^3 x) dx
become the integral of
(8sin^2 x)(1-sin^2 x) dsinx?

I understand that cos^2 x= 1-sin^2 x but thats about it?


It can't be true, because using cos^2 x= 1-sin^2 x, if you multiply each term by cosx then you'd get cos^3x on the left hand side, which you subistiture for what's in the RHS, so the sinx at the end of your last line should be a cosx

Can you post the full question?

Edit: I've no idea about anything - see below. Thennotation of dsinx has caught me out.
(edited 6 years ago)
∫ 8sin2xcos3x dx


∫ 8sin2xcosxcos2x dx

∫ 8sin2xcosx{1 - sin2x} dx

let u = sinx

du/dx = cosx

cosx dx = du = d(sinx)...

∫ 8sin2x{1 - sin2x}cosx dx

∫ 8sin2x{1 - sin2x}d(sinx)
Reply 3
Original post by Kevin De Bruyne
It can't be true, because using cos^2 x= 1-sin^2 x, if you multiply each term by cosx then you'd get cos^3x on the left hand side, which you subistiture for what's in the RHS, so the sinx at the end of your last line should be a cosx

Can you post the full question?


ImageUploadedByStudent Room1493405259.486351.jpg


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Reply 4
Original post by the bear
∫ 8sin2xcos3x dx


∫ 8sin2xcosxcos2x dx

∫ 8sin2xcosx{1 - sin2x} dx

let u = sinx

du/dx = cosx

cosx dx = du = d(sinx)...

∫ 8sin2x{1 - sin2x}cosx dx

∫ 8sin2x{1 - sin2x}d(sinx)


Oh thank you v much😊 but how do you even integrate that now?


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Original post by wkatie
Oh thank you v much😊 but how do you even integrate that now?


Posted from TSR Mobile


∫ 8sin2x{1 - sin2x}d(sinx)

∫ 8u2{1 - u2}du
Reply 6
Original post by the bear
∫ 8sin2x{1 - sin2x}d(sinx)

∫ 8u2{1 - u2}du


Ah yeah cheers


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