Principle: If the integrand contains a composite function $f(g(x)) \cdot g’(x)$, substitute $u = g(x)$:
$$\int f(g(x))\,g’(x)\,dx = \int f(u)\,du$$
Example 1: $\displaystyle\int \frac{2x}{x^2+3}\,dx$.
Let $u = x^2+3$, $du = 2x\,dx$:
$$\int \frac{du}{u} = \ln|u| + c = \ln(x^2+3) + c$$
Example 2: $\displaystyle\int \sin^3 x\cos x\,dx$.
Let $u = \sin x$, $du = \cos x\,dx$:
$$\int u^3\,du = \frac{u^4}{4} + c = \frac{\sin^4 x}{4} + c$$
Example 3 (definite): $\displaystyle\int_0^1 xe^{x^2}\,dx$.
Let $u = x^2$, $du = 2x\,dx$. When $x=0$: $u=0$; when $x=1$: $u=1$.
$$\frac{1}{2}\int_0^1 e^u\,du = \frac{1}{2}[e^u]_0^1 = \frac{e-1}{2}$$
$$\int u\,dv = uv - \int v\,du$$
LIATE rule for choosing $u$ (differentiate) and $dv$ (integrate):
Logarithmic $>$ Inverse trig $>$ Algebraic $>$ Trigonometric $>$ Exponential.
Example 4: $\displaystyle\int x\cos x\,dx$.
$u = x$, $dv = \cos x\,dx$; $du = dx$, $v = \sin x$:
$$x\sin x - \int \sin x\,dx = x\sin x + \cos x + c$$
Example 5: $\displaystyle\int \ln x\,dx$.
$u = \ln x$, $dv = dx$; $du = dx/x$, $v = x$:
$$x\ln x - \int x \cdot \frac{1}{x}\,dx = x\ln x - x + c$$
Example 6 (twice): $\displaystyle\int e^x \sin x\,dx$.
Let $I = \displaystyle\int e^x \sin x\,dx$.
Parts with $u = \sin x$, $dv = e^x\,dx$: $I = e^x\sin x - \displaystyle\int e^x\cos x\,dx$.
Apply parts again to $\displaystyle\int e^x\cos x\,dx$ (with $u=\cos x$, $dv = e^x\,dx$):
$$I = e^x\sin x - e^x\cos x - I$$
$$2I = e^x(\sin x - \cos x) \Rightarrow I = \frac{e^x(\sin x - \cos x)}{2} + c$$
$$\int \frac{1}{\sqrt{a^2 - x^2}}\,dx = \arcsin\frac{x}{a} + c, \quad |x| < a$$
$$\int \frac{1}{a^2 + x^2}\,dx = \frac{1}{a}\arctan\frac{x}{a} + c$$
Example 7: $\displaystyle\int \frac{1}{\sqrt{9-x^2}}\,dx = \arcsin\frac{x}{3} + c$.
Example 8: $\displaystyle\int_0^{\sqrt{3}} \frac{1}{1+x^2}\,dx = [\arctan x]_0^{\sqrt{3}} = \frac{\pi}{3} - 0 = \frac{\pi}{3}$.
Example 9: $\displaystyle\int \frac{1}{x^2 + 4x + 13}\,dx$.
Complete the square: $x^2+4x+13 = (x+2)^2 + 9$.
$$\int \frac{1}{(x+2)^2+9}\,dx = \frac{1}{3}\arctan\frac{x+2}{3} + c$$
KEY TAKEAWAY: Substitution reverses the chain rule; integration by parts reverses the product rule. Every standard integral technique has a corresponding differentiation rule.
EXAM TIP: For integration by parts that loops (Example 6), label the original integral $I$ and solve the resulting equation algebraically.
COMMON MISTAKE: In substitution, forgetting to change the limits for definite integrals, or failing to include the factor $du/dx$ when changing variables.
VCAA FOCUS: Integration questions on Paper 2 frequently combine techniques. A rational function may require partial fractions and then inverse-trig integration after completing the square.
