Integration by Parts (1 Viewer)

bored of sc · Apr 21, 2009

Having trouble with the following integrations. Explanations are neccesary please as I have worked solutions, but don't quite understand them (from TERRY LEE 7th Edition page 98, Ex 4.7).

Q 3 (f) Integrate e^axcos(bx) with respect to x.

Q 3 (h) Integrate e^xsin²x with respect to x.

My solution for (f)
e^ax(bsin[bx]-acos[bx]) / (a²+b²) +C

Terry Lee's solution for (f)
e^ax(bsin[bx]+acos[bx]) / (a²+b²) +C

My solution for (h)
e^x(5sin²x+2sinxcosx-2) +C

Terry Lee's solution for (h)
[e^x(sin²x-2sinxcosx+2)]/5 +C

I'm getting the numbers and symbols wrong. Hopefully I'm on the right track and it is simply errors giving the incorrect solution. But I doubt it.

If need be, I can post up my working.

Cheers y'all.

EvoRevolution · Apr 21, 2009

you answers for both f and h are not entirely different from terry lees answers, maybe u differentiate cosx or sinx which produced a different sign or maybe you just missed a negative or positive sign.

just check over your working

jet · Apr 21, 2009

$\int e^{ax} cos bx \, dx \\ \text{Let, } u = e^{ax}, du = ae^{ax} \, dx \\ dv = cosbx \, dx, v = \frac{1}{b}sinbx \\ \int u \, dv = uv - \int v \, du \\ \therefore \int e^{ax} cosbx \, dx =\frac{e^{ax}sinbx}{b} - \frac{a}{b}\int e^{ax}sinbx dx \\ \text{Now, let } u_2 = e^{ax}, du_2 = ae^{ax} \\ dv_2 = sinbx \, dx, v_2 = -\frac{1}{b}cosbx \\ \therefore \int e^{ax} cosbx \, dx =\frac{e^{ax}sinbx}{b} - \frac{a}{b} \left [u_2v_2 - \int v_2 \, du_2 \right ] \\ = \frac{e^{ax}sinbx}{b} - \frac{a}{b}\left[ -\frac{e^{ax}cosbx}{b} + \frac{a}{b} \int e^{ax}cosbx \, dx \right ] \\ = \frac{e^{ax}sinbx}{b} - \frac{a^2}{b^2}\int e^{ax}cosbx \, dx + \frac{a}{b^2}e^{ax}cosbx \\ \left (1 + \frac{a^2}{b^2} \right )\int e^{ax}cosbx \, dx = \frac{be^{ax}sinbx+ ae^{ax}cosbx}{b^2} \\ \left (\frac{a^2 + b^2}{b^2} \right ) \int e^{ax}cosbx \, dx = \frac{e^{ax} \left (bsinbx + acosbx \right )}{b^2} \\ \therefore \int e^{ax}cosbx \, dx = \frac{e^{ax} \left (bsinbx + acosbx \right )}{a^2 + b^2}$

$\int e^x sin^2 x \, dx \\ \text{Let } u = sin^2x, du = 2sinxcosx\, dx \\ dv = e^x \, dx, v = e^x \\ \int u \, dv = uv - \int v \, du + C \\ \therefore \int e^x sin^2 x \, dx = e^xsin^2 x - 2\int e^x sinxcosx \, dx + C \\ \text{Now, I will do another application of integration of parts, with } u_2 = sinxcosx, du_2 = cos^2x - sin^2x \, dx = 1 - 2sin^2x \, dx \\ \text{, and, } dv_2 = e^x \, dx, v = e^x \\ \therefore \int e^x sin^2 x \, dx = e^x sin^2 x - 2 \left [u_2v_2 - \int v_2 \, du_2 \right ] + C\\ = e^x sin^2x - 2 \left [ e^xsinxcosx - \int e^x \left (1 - 2sin^2x\right ) \, dx \right ] \ +C\\ \therefore \int e^x sin^2 x \, dx = e^x sin^2 x - 2e^xsinxcosx + 2\int e^x \, dx - 4\inte^xsin^2x \, dx + C \\ 5\int e^x sin^2 x \,dx = e^x \left (sin^2 x - 2sinxcosx + 2 \right ) + C\\ \int e^x sin^2 x \, dx = \frac{e^x \left (sin^2 x - 2sinxcosx + 2 \right )}{5} + D \qquad \text{(where } D = \frac{C}{5} \text{), though this does not really matter}$

It just seems like you have made some 'silly mistakes'. With practice they will go away.

EvoRevolution · Apr 21, 2009

for f) if u let u = cos(bx) v' = e^ax

it works fine, there is no point me putting up solution

bored of sc · Apr 21, 2009

jetblack2007 said:
$\int e^{ax} cos bx \, dx \\ \text{Let, } u = e^{ax}, du = ae^{ax} \, dx \\ dv = cosbx \, dx, v = \frac{1}{b}sinbx \\ \int u \, dv = uv - \int v \, du \\ \therefore \int e^{ax} cosbx \, dx =\frac{e^{ax}sinbx}{b} - \frac{a}{b}\int e^{ax}sinbx dx \\ \text{Now, let } u_2 = e^{ax}, du_2 = ae^{ax} \\ dv_2 = sinbx \, dx, v_2 = -\frac{1}{b}cosbx \\ \therefore \int e^{ax} cosbx \, dx =\frac{e^{ax}sinbx}{b} - \frac{a}{b} \left [u_2v_2 - \int v_2 \, du_2 \right ] \\ = \frac{e^{ax}sinbx}{b} - \frac{a}{b}\left[ -\frac{e^{ax}cosbx}{b} + \frac{a}{b} \int e^{ax}cosbx \, dx \right ] \\ = \frac{e^{ax}sinbx}{b} - \frac{a^2}{b^2}\int e^{ax}cosbx \, dx + \frac{a}{b^2}e^{ax}cosbx \\ \left (1 + \frac{a^2}{b^2} \right )\int e^{ax}cosbx \, dx = \frac{be^{ax}sinbx+ ae^{ax}cosbx}{b^2} \\ \left (\frac{a^2 + b^2}{b^2} \right ) \int e^{ax}cosbx \, dx = \frac{e^{ax} \left (bsinbx + acosbx \right )}{b^2} \\ \therefore \int e^{ax}cosbx \, dx = \frac{e^{ax} \left (bsinbx + acosbx \right )}{a^2 + b^2}$

$\int e^x sin^2 x \, dx \\ \text{Let } u = sin^2x, du = 2sinxcosx\, dx \\ dv = e^x \, dx, v = e^x \\ \int u \, dv = uv - \int v \, du + C \\ \therefore \int e^x sin^2 x \, dx = e^xsin^2 x - 2\int e^x sinxcosx \, dx + C \\ \text{Now, I will do another application of integration of parts, with } u_2 = sinxcosx, du_2 = cos^2x - sin^2x \, dx = 1 - 2sin^2x \, dx \\ \text{, and, } dv_2 = e^x \, dx, v = e^x \\ \therefore \int e^x sin^2 x \, dx = e^x sin^2 x - 2 \left [u_2v_2 - \int v_2 \, du_2 \right ] + C\\ = e^x sin^2x - 2 \left [ e^xsinxcosx - \int e^x \left (1 - 2sin^2x\right ) \, dx \right ] \ +C\\ \therefore \int e^x sin^2 x \, dx = e^x sin^2 x - 2e^xsinxcosx + 2\int e^x \, dx - 4\inte^xsin^2x \, dx + C \\ 5\int e^x sin^2 x \,dx = e^x \left (sin^2 x - 2sinxcosx + 2 \right ) + C\\ \int e^x sin^2 x \, dx = \frac{e^x \left (sin^2 x - 2sinxcosx + 2 \right )}{5} + D \qquad \text{(where } D = \frac{C}{5} \text{), though this does not really matter}$

It just seems like you have made some 'silly mistakes'. With practice they will go away.

Indeed.

Cheers for the clear working out. It makes perfect sense and I realise I have simply made silly errors (which are not so simple to erase).

Integration by Parts (1 Viewer)

bored of sc

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EvoRevolution

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jet

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EvoRevolution

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bored of sc

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