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Concept 03 / 10Integral Calculus
Integration by Parts and the Standard Exponential Forms
Deriving integration by parts from the product rule, choosing u with the ILATE order, repeated and cyclic integration by parts, the integral of e^x (f(x) + f'(x)), integrals of e^(ax) sin bx and e^(ax) cos bx, and integrating inverse functions such as ln x and tan^-1 x.
From the product rule (uv)′=u′v+uv′, integrating both sides gives the formula for integration by parts:
∫udv=uv−∫vdu, or in function form ∫f(x)g(x)dx=f(x)∫gdx−∫[f′(x)∫gdx]dx.
It trades one integral for another. The goal is to choose u so that du is simpler than u, and dv so that v is easy to find.
Choosing u: the ILATE order. Take as u the function that comes first in: Inverse trigonometric, Logarithmic, Algebraic, Trigonometric, Exponential. Functions early in the list simplify when differentiated (the derivative of lnx is x1, the derivative of tan−1x is algebraic), while those at the end are easy to integrate. So ∫xexdx takes u=x: xex−∫exdx=(x−1)ex+C. And ∫x2lnxdx takes u=lnx: 3x3lnx−∫3x3⋅x1dx=3x3lnx−9x3+C. ILATE is a guideline, not a law; substitution sometimes comes first.
Integrals of a single inverse function. Take dv=dx (so v=x). ∫lnxdx=xlnx−x+C. ∫tan−1xdx=xtan−1x−21ln(1+x2)+C. ∫sin−1xdx=xsin−1x+1−x2+C.
Repeated parts. For ∫xnexdx or ∫xnsinxdx, apply parts n times, each time reducing the power of x. A tabular shortcut: list the successive derivatives of the polynomial and successive integrals of the other function, and combine diagonally with alternating signs.
Cyclic parts. For I=∫eaxsinbxdx, two applications of parts bring back I itself, and then you solve for it:
∫eaxsinbxdx=a2+b2eax(asinbx−bcosbx)+C,
∫eaxcosbxdx=a2+b2eax(acosbx+bsinbx)+C.
A compact form: a2+b2eaxsin(bx−ϕ) with tanϕ=ab. Similarly ∫sec3xdx=21(secxtanx+ln∣secx+tanx∣)+C comes from a cyclic application.
The ex(f+f′) form. Since dxd[exf(x)]=ex[f(x)+f′(x)],
∫ex[f(x)+f′(x)]dx=exf(x)+C.
Recognising this pattern saves a lot of work. The skill is splitting the integrand as a function plus its derivative. For example, ∫exx2x−1dx=∫ex(x1−x21)dx=xex+C, since (x1)′=−x21. The generalisation is ∫ekx[kf(x)+f′(x)]dx=ekxf(x)+C, and similarly ∫[f(x)+xf′(x)]dx=xf(x)+C.
Standard results by parts: ∫a2−x2dx=2xa2−x2+2a2sin−1ax+C; ∫x2±a2dx=2xx2±a2±2a2ln∣x+x2±a2∣+C.
Key Takeaways:
- ∫udv=uv−∫vdu.
- Choose u by ILATE: Inverse, Log, Algebraic, Trig, Exponential.
- ∫lnx=xlnx−x; ∫tan−1x=xtan−1x−21ln(1+x2).
- ∫ex(f+f′)=exf; ∫ekx(kf+f′)=ekxf.
- ∫eaxsinbx=a2+b2eax(asinbx−bcosbx).
- Cyclic cases: parts twice, then solve for the integral.
- Pitfall: adding the constant of integration to v at the intermediate step; it cancels anyway, so omit it until the end.
Brain-Teaser 1: Evaluate ∫ex1+cosx1+sinxdx.
Solution: Use half-angle identities: 1+cosx1+sinx=2cos22x1+2sin2xcos2x=21sec22x+tan2x. With f(x)=tan2x, f′(x)=21sec22x, so the integrand is ex(f+f′) and the integral is extan2x+C. The identity work is what reveals the hidden f+f′ structure.
Brain-Teaser 2: Evaluate ∫xsin−1xdx.
Solution: ILATE gives u=sin−1x, dv=xdx, v=2x2. So the integral is 2x2sin−1x−21∫1−x2x2dx. Write 1−x2x2=1−x21−1−x2. Using the standard results, ∫1−x2x2dx=sin−1x−2x1−x2−21sin−1x=21sin−1x−2x1−x2. The final answer is 2x2sin−1x−41sin−1x+4x1−x2+C=42x2−1sin−1x+4x1−x2+C.
Brain-Teaser 3: Evaluate ∫e2xsin3xdx using the cyclic method, and check with the formula.
Solution: Let I=∫e2xsin3xdx. Parts with u=sin3x: I=2e2xsin3x−23∫e2xcos3xdx. Parts again: ∫e2xcos3xdx=2e2xcos3x+23I. So I=2e2xsin3x−43e2xcos3x−49I, giving 413I=4e2x(2sin3x−3cos3x) and I=13e2x(2sin3x−3cos3x)+C. This matches the formula with a=2, b=3. Consistency in choosing u (the trigonometric factor both times) is essential; switching roles on the second step undoes the first.
