# P3 Integration - identities!Watch

This discussion is closed.
#1
OK, I need to work out:

∫ cos2xsinx .dx

I use the identity 2sinAcosB = sin(A + B) + sin(A - B) and I keep getting the answer -1/6.cos3x + 1/2cosx. But the answer given in the book is -2/3.cosÂ³x + cosx! Where did the power come from? Where have I gone wrong? Or are the two answers different forms of the same thing? Any help is appreciated.
0
13 years ago
#2
(Original post by mathsforsaken)
OK, I need to work out:

∫ cos2xsinx .dx

I use the identity 2sinAcosB = sin(A + B) + sin(A - B) and I keep getting the answer -1/6.cos3x + 1/2cosx. But the answer given in the book is -2/3.cosÂ³x + cosx! Where did the power come from? Where have I gone wrong? Or are the two answers different forms of the same thing? Any help is appreciated.
use cos (2A) = 2cos^2 x - 1
and substitution u = cos x
13 years ago
#3
the integral of sin(x)*(2cos^2(x)-1) gives the answer in the book. it expands the brackets and then integrates.
0
13 years ago
#4
(Original post by mathsforsaken)
OK, I need to work out:

∫ cos2xsinx .dx
∫cos2xsinx . dx

= ∫(2cos^2 x - 1)sinx . dx

let u = cos x
du/dx = -sin x

so ∫-(2u^2 - 1) . du
= -2/3u^3 + u + c

= -2/3cos^3 x + cos x + c

hope that helps
13 years ago
#5
cos2xsinx.dx cos2x=(2cos^2-1)
(2cos^2-1)sinx.dx u=cosx du/dx=-sinx
(2u^2-1).du

= -2/3u^3 + u + c
-2/3cos^3x + cosx
0
13 years ago
#6
(Original post by Freeway)
cos2xsinx.dx cos2x=(2cos^2-1)
(2cos^2-1)sinx.dx u=cosx du/dx=-sinx
(2u^2-1).du

= -2/3u^3 + u + c
-2/3cos^3x + cosx + c
:P..
13 years ago
#7
i always forget to do that!
0
#8
Thanks everyone , makes perfect sense now. I see it's all about using the identity cos2x = 2cosÂ²x-1 then. Silly me for not seeing that before!
0
13 years ago
#9
Your answer is correct, by the way. It's only a constant apart from the answer given in the book.
0
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