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Exercise 7.6 · Q11

Q.Integrate the following function: xcos⁡−1x1−x2\frac{x \cos^{-1}x}{\sqrt{1-x^2}}

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Concept understanding — Integration by Parts

Integration by Parts

The idea: reverse the product rule

Some integrands are a product of two very different functions — xexx e^x, xcos⁡xx\cos x, log⁡x\log x, xsin⁡−1xx\sin^{-1}x — where substitution gets you nowhere. Integration by parts is the tool for these. It comes straight from reversing the product rule for differentiation.

Starting from ddx(uv)=u v′+u′ v\dfrac{d}{dx}(uv)=u\,v' + u'\,v and integrating both sides gives the working formula:

∫u dvdx dx=uv−∫v dudx dx.\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx.

In words: integral of (first ×\times derivative-of-second) == first ×\times integral-of-second −- integral of (derivative-of-first ×\times integral-of-second).

Choosing uu: the ILATE rule

The whole game is picking which factor is uu (to differentiate) and which is dvdv (to integrate). Pick uu by ILATE — the first type that appears:

  • Inverse trig (sin⁡−1x\sin^{-1}x), Logarithmic (log⁡x\log x), Algebraic (x2x^2), Trigonometric (sin⁡x\sin x), Exponential (exe^x).

Whatever comes first in ILATE becomes uu; the rest is dvdv. This makes the new integral ∫v du\int v\,du simpler than the one you started with.

Worked idea

For ∫xex dx\int x e^x\,dx: algebraic before exponential, so u=xu=x, dv=exdxdv=e^x dx. Then du=dxdu=dx, v=exv=e^x:

∫xex dx=xex−∫ex dx=xex−ex+C=ex(x−1)+C.\int x e^x\,dx = x e^x - \int e^x\,dx = x e^x - e^x + C = e^x(x-1)+C. …

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