integration by parts

Explain integration by parts with a worked example

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The answer

Integration by parts is the reverse of the product rule for differentiation. It helps integrate products of two functions. The formula is:

udv=uvvdu\int u\,dv = uv - \int v\,du

Here, choose uu (the part to differentiate) and dvdv (the part to integrate). The goal is to make the new integral vdu\int v\,du simpler than the original. Use the ILATE rule to pick uu: Inverse trig, Logarithmic, Algebraic, Trigonometric, Exponential. Pick uu from the first category that appears in your product.

Steps: 1. Identify uu and dvdv. 2. Differentiate uu to get dudu. 3. Integrate dvdv to get vv. 4. Plug into the formula and simplify.

Worked example: Integrate xcosxdx\int x\cos x\,dx.

Here, product is xx (algebraic) and cosx\cos x (trigonometric). By ILATE, Algebraic comes before Trigonometric, so choose u=xu = x, dv=cosxdxdv = \cos x\,dx.

Differentiate: du=dxdu = dx.

Integrate: v=sinxv = \sin x (since ddxsinx=cosx\frac{d}{dx}\sin x = \cos x).

Now apply formula:

xcosxdx=xsinxsinxdx\int x\cos x\,dx = x\sin x - \int \sin x\,dx

Integrate sinx\sin x: sinxdx=cosx\int \sin x\,dx = -\cos x.

So the final answer is:

xcosxdx=xsinx+cosx+C\int x\cos x\,dx = x\sin x + \cos x + C

(Don't forget the constant CC!)

Key tip: If the product has a polynomial and an exponential, choose uu as the polynomial, because its derivative becomes simpler (degree reduces). For example, xexdx\int x e^x\,dx: pick u=xu=x, dv=exdxdv=e^x\,dx, then v=exv=e^x, du=dxdu=dx, giving xexex+Cx e^x - e^x + C.

Common mistake: Choosing uu and dvdv wrongly can make the integral harder. Always check that vdu\int v\,du is simpler. Practice with lnxdx\int \ln x\,dx (treat as lnx1\ln x \cdot 1, pick u=lnxu=\ln x, dv=1dxdv=1\,dx) to get xlnxx+Cx\ln x - x + C.

Remember, integration by parts is your friend for products, and ILATE is your guide. Happy integrating!

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