By now you have every Extension 2 integration technique in your kit: $u$-substitution, trig substitution, partial fractions, integration by parts, the IBP reduction loop, and the $t = \tan(x/2)$ trick. This lesson stacks them, most HSC questions in MEX-C1 combine two or three techniques, and the definite-integral marks are won or lost on careful limit substitution. You will work three full-length problems start to finish.
+5 XP warm-up For each integrand below, name the first technique you would reach for.
For each integrand below, name the first technique you would reach for. Don't compute, just commit to a strategy. (a) $\int x^2 e^{x}\,dx$ (b) $\int \dfrac{1}{\sqrt{4-x^2}}\,dx$ (c) $\int \dfrac{2x+3}{x^2+x-2}\,dx$ (d) $\int_0^{\pi/2} \sin^5 x\,dx$.
+5 XP to read Two habits separate a clean HSC solution from a messy one: declare the technique before computing, and change...
Two habits separate a clean HSC solution from a messy one: declare the technique before computing, and change the limits the moment you substitute (never carry old limits onto new variables). Mixed problems usually need a chain, sub then parts, or sub then partial fractions, so each layer must be stated explicitly.
The declare-substitute-relimit-resolve habit: (1) declare $u = \ldots$ and write $du$; (2) substitute the integrand AND $dx$; (3) change limits $x = a, b \mapsto u = u(a), u(b)$; (4) finish with the new technique on the new variable.
$\int_a^b f(g(x))g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du$
An integrand like $\int \sin(\sqrt{x})\,dx$ resists IBP directly (you cannot easily integrate $\sin(\sqrt{x})$).
An integrand like $\int \sin(\sqrt{x})\,dx$ resists IBP directly (you cannot easily integrate $\sin(\sqrt{x})$). Substitution clears the inner function first; then IBP handles the polynomial-times-trig that remains.
Worked through the hook: $\int_0^{1} \dfrac{x^3}{\sqrt{1-x^2}}\,dx$. Trig substitution $x = \sin\theta$ gives $dx = \cos\theta\,d\theta$ and $\sqrt{1-x^2} = \cos\theta$. The integrand becomes $\sin^3\theta\,d\theta$. New limits: $x=0 \to \theta=0$, $x=1 \to \theta = \pi/2$. Then write $\sin^3\theta = \sin\theta(1-\cos^2\theta)$ and substitute again $w = \cos\theta$. Two layers, one clean answer.
Layering rule: substitution first when there is an inner function; IBP/partial fractions next · Change limits with each substitution (or back-substitute at the end, pick one) · Trig substitution: $\sqrt{a^2 - x^2} \Rightarrow x = a\sin\theta$; $\sqrt{a^2 + x^2} \Rightarrow x = a\tan\theta$; $\sqrt{x^2 - a^2} \Rightarrow x = a\sec\theta$
Pause, copy the layering rule (substitution first, then IBP or partial fractions), the three trig-substitution pairs ($\sqrt{a^2-x^2}\to a\sin\theta$ etc.), and the limit-changing instruction into your book.
Quick check: For $\int_0^{1} x^2 \sqrt{1-x^2}\,dx$, which first move is best?
We just saw the layering rule for substitution + IBP: apply substitution first when an inner function is present, then IBP;...
We just saw the layering rule for substitution + IBP: apply substitution first when an inner function is present, then IBP; change limits with each substitution or back-substitute at the end. That raises a question: what about substitution followed by partial fractions, what are the common forms that arise? This card answers it → $e^x/e^x$ multiplication converts an exponential denominator; complete the square for a quadratic; split the numerator for $\ln + \arctan$; $t = \tan(x/2)$ for rational trig.
Rational expressions with an exponential or trigonometric inner function become standard rationals after a clever substitution. Two recurring patterns:
Completing the square as a partial-fractions preprocessor. If the denominator is an irreducible quadratic like $x^2 + 2x + 5$, complete the square: $(x+1)^2 + 4$. Then $u = x + 1$ converts to $u^2 + 4$, which has standard $\arctan$ antiderivative.
Exponential-in-denominator: multiply by $e^x/e^x$, then $u = e^x$ · Irreducible quadratic denominator: complete the square, then $u = x + b/2$ · Linear over quadratic: split numerator into (derivative-of-denom part) + (constant part), $\ln + \arctan$ · $t = \tan(x/2)$ Weierstrass: $\sin x = \tfrac{2t}{1+t^2}$, $\cos x = \tfrac{1-t^2}{1+t^2}$, $dx = \tfrac{2}{1+t^2}dt$
Paste, copy the four substitution + partial-fraction patterns (exponential denominator → multiply $e^x/e^x$; irreducible quadratic → complete the square + shift; linear/quadratic → split numerator; rational trig → $t$-sub with three Weierstrass formulas) into your book.
Did you get this? True or false: to evaluate $\int_0^{\ln 2} \dfrac{e^{x}}{e^{2x} + 1}\,dx$, the substitution $u = e^{x}$ converts the integral to $\int_1^{2} \dfrac{du}{u^2 + 1}$.
Worked examples · 3 in a row, reveal as you go
Worked examples · 3 in a row, reveal as you go
Evaluate $\displaystyle\int_0^{1} \frac{x^3}{\sqrt{1 - x^2}}\,dx$. Leave the answer in exact form.
Evaluate $\displaystyle\int_0^{1} e^{\sqrt{x}}\,dx$. Leave the answer in exact form.
Evaluate $\displaystyle\int_0^{1} \frac{2x + 3}{x^2 + 2x + 5}\,dx$. Leave the answer in exact form.
Fill the gap: When the substitution $u = g(x)$ converts $\int_a^b f(g(x))g'(x)\,dx$ to a $u$-integral, the new lower limit is and the new upper limit is .
Misconceptions to fix · the 3 traps that cost marks
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: after the substitution $u = \cos\theta$ with $\theta \in [0, \pi/2]$, the limits become $u \in [1, 0]$ and the minus sign from $du = -\sin\theta\,d\theta$ can be absorbed by swapping the limits to $\int_0^1$.
Activities · practice with the ideas
Activities · practice with the ideas
Evaluate $\displaystyle\int_0^{\pi/2} \sin^3\theta \cos^2\theta\,d\theta$. (Hint: keep one $\sin$, convert the rest, then substitute $u = \cos\theta$.)
Evaluate $\displaystyle\int_0^{\ln 2} \frac{e^{x}}{e^{2x} + 4}\,dx$ in exact form. (Hint: substitute $u = e^{x}$.)
Evaluate $\displaystyle\int_1^{e} x\ln x\,dx$. (Hint: integration by parts.)
Evaluate $\displaystyle\int_0^{1} \frac{x}{x^2 + 4x + 5}\,dx$. (Hint: split numerator, then complete the square.)
Evaluate $\displaystyle\int_0^{1} \arctan x\,dx$. (Hint: IBP with $f = \arctan x$, $dg = dx$.)
Odd one out: Three of these integrals are best tackled by integration by parts as the FIRST move. Which one is NOT?
Earlier you committed to a first technique for $\int_0^{1} \dfrac{x^3}{\sqrt{1-x^2}}\,dx$ and asked what happens to the limits under substitution.
The two-layer plan, trig substitution $x = \sin\theta$ (limits $0 \to \pi/2$), then $w = \cos\theta$ (limits $1 \to 0$), produced a clean $\tfrac{2}{3}$. The key lesson is that mixed problems reward an up-front plan: declare each layer, change limits each time, and resist the urge to do algebra before stating the strategy.
Multiple choice, then short answer under exam conditions.
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Use the substitution $u = 1 + x^2$ to evaluate $\displaystyle\int_0^{1} \frac{x^3}{1 + x^2}\,dx$. (3 marks)
Q2. Evaluate $\displaystyle\int_0^{\pi/4} \sec^2 x \tan x\,dx$ in two ways: (a) substitution $u = \tan x$; (b) substitution $u = \sec x$. Confirm both give the same exact value. (3 marks)
Q3. Evaluate $\displaystyle\int_0^{1} x^2 e^{-x}\,dx$ using integration by parts twice. Leave the answer in exact form. (4 marks)
Activity answers:
1. $\sin^3\theta\cos^2\theta = \sin\theta(1-\cos^2\theta)\cos^2\theta$. Sub $u = \cos\theta$, $du = -\sin\theta\,d\theta$, limits $1 \to 0$. Integral $= \int_0^1 (1-u^2)u^2\,du = \int_0^1 (u^2 - u^4)\,du = \tfrac{1}{3} - \tfrac{1}{5} = \tfrac{2}{15}$.
2. $u = e^x$, $du = e^x\,dx$, limits $1 \to 2$. Integral $= \int_1^2 \dfrac{du}{u^2+4} = \tfrac{1}{2}[\arctan(u/2)]_1^2 = \tfrac{1}{2}[\arctan 1 - \arctan(1/2)] = \tfrac{\pi}{8} - \tfrac{1}{2}\arctan(1/2)$.
3. IBP: $f = \ln x$, $dg = x\,dx \Rightarrow df = dx/x$, $g = x^2/2$. $\int x\ln x\,dx = \tfrac{x^2}{2}\ln x - \int \tfrac{x}{2}\,dx = \tfrac{x^2}{2}\ln x - \tfrac{x^2}{4}$. Evaluate $1 \to e$: $\left(\tfrac{e^2}{2} - \tfrac{e^2}{4}\right) - \left(0 - \tfrac{1}{4}\right) = \tfrac{e^2}{4} + \tfrac{1}{4} = \tfrac{e^2 + 1}{4}$.
4. $x = \tfrac{1}{2}(2x + 4) - 2$. Split: $\int_0^1 \dfrac{x}{x^2+4x+5}\,dx = \tfrac{1}{2}[\ln(x^2+4x+5)]_0^1 - 2\int_0^1 \dfrac{dx}{(x+2)^2 + 1} = \tfrac{1}{2}\ln\!\left(\tfrac{10}{5}\right) - 2[\arctan(x+2)]_0^1 = \tfrac{1}{2}\ln 2 - 2(\arctan 3 - \arctan 2)$.
5. $f = \arctan x$, $dg = dx \Rightarrow df = dx/(1+x^2)$, $g = x$. $\int_0^1 \arctan x\,dx = [x\arctan x]_0^1 - \int_0^1 \dfrac{x}{1+x^2}\,dx = \tfrac{\pi}{4} - \tfrac{1}{2}\ln 2$.
Q1 (3 marks): $u = 1+x^2$, $du = 2x\,dx$, $x^2 = u-1$, limits $1 \to 2$ [1]. $\int_0^1 \dfrac{x^3}{1+x^2}\,dx = \tfrac{1}{2}\int_1^2 \dfrac{u-1}{u}\,du = \tfrac{1}{2}\int_1^2 (1 - 1/u)\,du$ [1]. $= \tfrac{1}{2}[u - \ln u]_1^2 = \tfrac{1}{2}(1 - \ln 2) = \tfrac{1 - \ln 2}{2}$ [1].
Q2 (3 marks): (a) $u = \tan x$, $du = \sec^2 x\,dx$, limits $0 \to 1$. $\int_0^1 u\,du = \tfrac{1}{2}$ [1]. (b) $u = \sec x$, $du = \sec x \tan x\,dx$, so $\sec^2 x \tan x\,dx = u\,du$, limits $1 \to \sqrt{2}$. $\int_1^{\sqrt 2} u\,du = \tfrac{1}{2}(2 - 1) = \tfrac{1}{2}$ [1]. Both equal $\tfrac{1}{2}$, consistent (the two antiderivatives differ by a constant) [1].
Q3 (4 marks): Let $I = \int_0^1 x^2 e^{-x}\,dx$. IBP with $f = x^2$, $dg = e^{-x}\,dx \Rightarrow df = 2x\,dx$, $g = -e^{-x}$. $I = [-x^2 e^{-x}]_0^1 + 2\int_0^1 x e^{-x}\,dx = -e^{-1} + 2J$ [1]. IBP on $J = \int_0^1 x e^{-x}\,dx$ with $f = x$, $dg = e^{-x}\,dx$. $J = [-xe^{-x}]_0^1 + \int_0^1 e^{-x}\,dx = -e^{-1} + (1 - e^{-1}) = 1 - 2e^{-1}$ [1]. So $I = -e^{-1} + 2(1 - 2e^{-1}) = 2 - 5e^{-1} = 2 - \tfrac{5}{e}$ [1]. Final exact form: $\boxed{2 - \tfrac{5}{e}}$ [1].
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