5.16 Integration by substitution and by partsIB Maths: Analysis and Approaches HL: Revision notes

Substitution reverses the chain rule. To integrate with $u = g(x)$:

1. Write $\dfrac{du}{dx}$ and replace $dx$ (e.g. $dx = du$ when $u = x + 3$).
2. Replace every $x$ in the integrand; you may need $x$ in terms of $u$ ($x = u - 3$).
3. For a definite integral, change the limits to $u$-values.
4. Integrate, and for an indefinite integral write the answer back in terms of $x$.

$\int_{1}^{6}x\sqrt{x + 3}\,dx = \int_{4}^{9}(u - 3)u^{1/2}\,du = \left[\tfrac{2}{5}u^{5/2} - 2u^{3/2}\right]_{4}^{9} = \frac{232}{5}.$ In IB exams, a substitution will be given if the integral is not of the recognisable form $\displaystyle\int kg'(x)f(g(x))\,dx$.

From the product rule: $\int u\frac{dv}{dx}\,dx = uv - \int v\frac{du}{dx}\,dx.$ Choose $u$ to be the factor that becomes simpler when differentiated (often a power of $x$), and $\dfrac{dv}{dx}$ something you can integrate.

$\int x\sin x\,dx = -x\cos x + \int\cos x\,dx = -x\cos x + \sin x + C.$ $\int x\,e^{3x}\,dx = \frac{x}{3}e^{3x} - \frac{1}{9}e^{3x} + C.$

Some single functions can only be integrated by parts, writing them as $1 \times$ the function and taking $\dfrac{dv}{dx} = 1$: $\int\ln x\,dx = x\ln x - \int x\cdot\frac{1}{x}\,dx = x\ln x - x + C.$ $\int\arcsin x\,dx = x\arcsin x - \int\frac{x}{\sqrt{1 - x^{2}}}\,dx = x\arcsin x + \sqrt{1 - x^{2}} + C.$ The last integral is done by inspection (reverse chain rule).

If the new integral still needs parts, apply it again with the same type of choice.

Powers reduce: $\displaystyle\int x^{2}e^{x}\,dx = x^{2}e^{x} - 2\int xe^{x}\,dx = e^{x}(x^{2} - 2x + 2) + C$.

Cyclic integrals: for $J = \displaystyle\int e^{-t}\sin t\,dt$, two applications bring back $J$ itself: $J = -e^{-t}\sin t - e^{-t}\cos t - J \Rightarrow J = -\frac{1}{2}e^{-t}(\sin t + \cos t) + C.$ Collect the $J$ terms and solve — don't keep integrating forever.