TheIBAxolotl
Integration by substitution
Substitution undoes the chain rule. Look for an inner function whose derivative also appears (up to a constant) in the integrand. Call the inner function \(u\), replace \(dx\) using \(\frac{du}{dx}\), and the integral becomes simpler.
Example. \(\int 2x(x^2 + 1)^5\,dx\). Let \(u = x^2 + 1\), so \(du = 2x\,dx\). Then
\[\int u^5\,du = \frac{u^6}{6} + C = \frac{(x^2 + 1)^6}{6} + C.\]
Every \(x\) must disappear after substituting. If some \(x\) is left over, either express it in terms of \(u\) (for example, if \(u = x + 1\) then \(x = u - 1\)) or choose a different \(u\).
Definite integrals: change the limits
With a definite integral, convert the limits to \(u\)-values and never go back to \(x\).
Worked example. Evaluate \(\int_0^1 x e^{x^2}\,dx\).
- Let \(u = x^2\), so \(du = 2x\,dx\) and \(x\,dx = \frac{1}{2}du\).
- Limits: \(x = 0 \Rightarrow u = 0\); \(x = 1 \Rightarrow u = 1\).
- \(\frac{1}{2}\int_0^1 e^u\,du = \frac{1}{2}\big[e^u\big]_0^1 = \frac{e - 1}{2}.\)
Integration by parts
Parts undoes the product rule. Rearranging \(\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}\) and integrating gives
\[\int u\,\frac{dv}{dx}\,dx = uv - \int v\,\frac{du}{dx}\,dx.\]
The aim is to make the new integral easier. Choose \(u\) to be the factor that gets simpler when differentiated. A common priority order is: logarithms, then algebraic powers, then trig and exponentials (which stay about as complicated either way).
Example. \(\int x e^x\,dx\): take \(u = x\), \(\frac{dv}{dx} = e^x\), so \(\frac{du}{dx} = 1\), \(v = e^x\):
\[\int x e^x\,dx = x e^x - \int e^x\,dx = x e^x - e^x + C.\]
Example. \(\int \ln x\,dx\): write it as \(\int 1 \cdot \ln x\,dx\) with \(u = \ln x\), \(\frac{dv}{dx} = 1\). Result: \(x\ln x - x + C\).
Example. \(\int x\cos x\,dx = x\sin x + \cos x + C\) (check by differentiating: \(\sin x + x\cos x - \sin x = x\cos x\)).
Repeated and cyclic parts
For \(\int x^2 e^x\,dx\), one round leaves \(\int 2x e^x\,dx\), which needs a second round. Result: \(e^x(x^2 - 2x + 2) + C\).
For \(I = \int e^x \sin x\,dx\), two rounds bring the original integral back: \(I = e^x\sin x - e^x\cos x - I\). Treat \(I\) as an unknown and solve: \(I = \frac{e^x(\sin x - \cos x)}{2} + C\). Keep the same type of choice for \(u\) in both rounds or the work will undo itself.
Common mistakes
- Leaving some \(x\) in a substituted integral, or forgetting to convert \(dx\).
- Changing variable but keeping the old \(x\)-limits.
- Choosing \(u = e^x\) in \(\int x e^x\,dx\), which makes the integral harder.
- Sign errors in the \(-\int v\,u'\,dx\) term, especially in repeated parts.
- Dropping \(+C\) at the end of an indefinite integral.
Exam tips
- If the question names the substitution, use exactly that one.
- Differentiate your answer to check it; for by-parts results this takes seconds.
- In cyclic questions, label the integral \(I\) from the start so the final rearrangement is clear.