Mc29 - Integration by parts

Mc29 - Integration by parts

Choose factors that turn a difficult product into an easier integral. Derive the parts rule, repeat it for polynomial products, integrate logarithms and justify exact definite values.

Reversing the product rule

Why is xsin⁡xx\sin x almost, but not quite, an antiderivative of xcos⁡xx\cos x? Its derivative is xcos⁡x+sin⁡xx\cos x+\sin x. We need to cancel the extra sine term; integration by parts organises exactly that correction.

You already know

For differentiable functions, (uv)′=u′v+uv′(uv)'=u'v+uv'. Trigonometric differentiation and integration use radians.

Integrating and rearranging the product rule gives the useful exchange below.

Integration by parts

On an interval where the functions and integrals are defined,

∫udvdx dx=uv−∫vdudx dx.\int u\frac{\mathrm dv}{\mathrm dx}\,\mathrm dx =uv-\int v\frac{\mathrm du}{\mathrm dx}\,\mathrm dx.

In the formula sheet.

Choose uu to simplify under differentiation and v′v' to integrate easily. A poor choice is legal but may make the remaining integral harder. Use one arbitrary constant in the final indefinite result; any intermediate constant in vv cancels.

Worked example 1: cancelling the extra sine

Question. Find ∫xcos⁡x dx\int x\cos x\,\mathrm dx.

Plan. Take u=xu=x, u′=1u'=1, v′=cos⁡xv'=\cos x, v=sin⁡xv=\sin x.

I=xsin⁡x−∫sin⁡x dx=xsin⁡x+cos⁡x+C.I=x\sin x-\int\sin x\,\mathrm dx=x\sin x+\cos x+C.

Why differentiate xx? The polynomial factor disappears from the remaining integral.

Answer. xsin⁡x+cos⁡x+Cx\sin x+\cos x+C.

Check. Its derivative is sin⁡x+xcos⁡x−sin⁡x=xcos⁡x\sin x+x\cos x-\sin x=x\cos x.

Change only the angle multiplier from 11 to 22. Each integration must respect it.

Worked example 2: scaling the angle

Question. Find ∫xcos⁡2x dx\int x\cos2x\,\mathrm dx.

Plan. Keep u=xu=x, now using v=sin⁡2x/2v=\sin2x/2.

I=xsin⁡2x2−12∫sin⁡2x dx=xsin⁡2x2+cos⁡2x4+C.\begin{aligned} I&=\frac{x\sin2x}{2}-\frac12\int\sin2x\,\mathrm dx\\ &=\frac{x\sin2x}{2}+\frac{\cos2x}{4}+C. \end{aligned}

Why a quarter in the second term? Integrating the sine introduces another factor 1/21/2.

Answer. xsin⁡2x/2+cos⁡2x/4+Cx\sin2x/2+\cos2x/4+C.

Check. The two sine terms cancel under differentiation.

For exponential factors, the same decision gives ∫xe2x dx=e2x(x/2−1/4)+C\int x\mathrm e^{2x}\,\mathrm dx=\mathrm e^{2x}(x/2-1/4)+C. Taking u=e2xu=\mathrm e^{2x} instead would increase the polynomial degree in the remaining integral.

Watch out

Wrong: integrate the two factors separately and multiply. Differentiating that product produces an extra term. The parts formula subtracts the integral of that extra term.

Change the angle multiplier again and keep both scale factors.

Reducing a polynomial factor more than once

A second parts calculation can finish what the first starts. Keep its outer coefficient visible so a sign does not quietly change on the way back.

Worked example 3: a quadratic times sine

Question. Find ∫x2sin⁡x dx\int x^2\sin x\,\mathrm dx.

Plan. Differentiate the polynomial on each application.

I=−x2cos⁡x+2∫xcos⁡x dxfirst parts step∫xcos⁡x dx=xsin⁡x+cos⁡xsecond parts stepI=−x2cos⁡x+2xsin⁡x+2cos⁡x+C.\begin{aligned} I&=-x^2\cos x+2\int x\cos x\,\mathrm dx &&\text{first parts step}\\ \int x\cos x\,\mathrm dx&=x\sin x+\cos x &&\text{second parts step}\\ I&=-x^2\cos x+2x\sin x+2\cos x+C. \end{aligned}

Why an addition after the first step? The minus sign in parts multiplies the negative primitive of sine.

Answer. −x2cos⁡x+2xsin⁡x+2cos⁡x+C-x^2\cos x+2x\sin x+2\cos x+C.

Check. Differentiation cancels both the linear cosine and constant sine terms.

Change only the trigonometric factor from sine to cosine. Its primitive changes the signs you must track.

Worked example 4: a quadratic times cosine

Question. Find ∫x2cos⁡x dx\int x^2\cos x\,\mathrm dx.

Plan. Apply parts first to the quadratic product, then to the remaining linear product.

I=x2sin⁡x−2∫xsin⁡x dx∫xsin⁡x dx=−xcos⁡x+sin⁡xI=x2sin⁡x+2xcos⁡x−2sin⁡x+C.\begin{aligned} I&=x^2\sin x-2\int x\sin x\,\mathrm dx\\ \int x\sin x\,\mathrm dx&=-x\cos x+\sin x\\ I&=x^2\sin x+2x\cos x-2\sin x+C. \end{aligned}

Why bracket the second result? The outer −2-2 changes both of its signs.

Answer. x2sin⁡x+2xcos⁡x−2sin⁡x+Cx^2\sin x+2x\cos x-2\sin x+C.

Check. Its derivative simplifies to x2cos⁡xx^2\cos x.

For comparison, ∫x2ex dx=ex(x2−2x+2)+C\int x^2\mathrm e^x\,\mathrm dx=\mathrm e^x(x^2-2x+2)+C, obtained through x2ex−2∫xex dxx^2\mathrm e^x-2\int x\mathrm e^x\,\mathrm dx. These are finite repeated calculations; reduction formulae for indexed families are excluded. The resulting primitives are derived, not separately printed in the formula sheet.

Watch out

Wrong: let the second integral absorb the outer coefficient without recording it. Right: work on that integral separately, then substitute its whole result in brackets.

Keep the quadratic-times-cosine structure and change its angle multiplier.

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