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Integation FP1

Heya

hehe... last minute maths homework as always :redface: ...

Could anyone please help me integrate:

(e^2x)sinx

??:confused:

... I'm going round in circles! :redface:
Reply 1
Avatar for sahra-t
sahra-t
What syllabus are you doing FP1 on? That's not on mine (OCR)...

I tried doing integration by parts but it all ended up looking a bit too messy to be right =S
Starsplash
Heya

hehe... last minute maths homework as always :redface: ...

Could anyone please help me integrate:

(e^2x)sinx

??:confused:

... I'm going round in circles! :redface:


You call this last minute? Pfft.

Anyway, go by parts the same way twice see what happens, if you integrate or differentiate sinx twice you will end up with -sinx, and e2x always remains except with different coeeficients.

You should get integral = blah + k(Integral)
Hmm...

y = int((e^2x)sinx)dx

y = int(uv')dx = uv - int(u'v)dx

let u = sinx, v'=e^2x
so u'=cosx, v = (1/2)e^2x

so y = (1/2)(e^2x)(sinx) - (1/2)int((e^2x)(cosx))dx

now to find the second integral...

let u=cos x, v' = e^2x
so u'=-sinx, v = (1/2)e^2x

so y = (1/2)(e^2x)(sinx) - (1/4)(e^2x)(cosx) - (1/4)int((e^2x)(sinx))dx

but int(e^2x)(sinx)dx = y

so y = (1/4)(e^2x)(2sinx - cosx) - (y/4)
5y/4 = (1/4)(e^2x)(2sinx - cosx)
y = (1/5)(e^2x)(2sinx - cosx)
y = -(1/5)(e^2x)(cosx - 2sinx)

This is the answer The Integrator (http://integrals.wolfram.com/) gives too

PS I forgot the constants, remember to stick those in!
Reply 4
Avatar for Starsplash
Starsplash
OP
Thank you everyone!!!! :hugs:

Sahra-t - it's edexcel, but it's really c4 I think, just came up in an fp1 question and i've only got that text book with me! :rolleyes:

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