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Module 15 · L15 of 16 ~45 min ⚡ +95 XP available

Mixed Integration Problems II

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.

Today's hook — You are asked to evaluate $\int_0^{1} \dfrac{x^3}{\sqrt{1-x^2}}\,dx$. Which technique first — substitution or by parts? And once you substitute, what happens to the limits? Write down your plan before looking at Worked Example 1.

0/5QUESTS

Recall — your gut answer first

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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) $\int_0^{\pi/2} \sin^5 x\,dx$.

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The two moves for a definite mixed integral

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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.

$\int_a^b f(g(x))g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du$

Change the limits, or back-substitute

Either change limits to $u$ AND drop the $+C$ for a clean definite answer, OR keep old limits, integrate in $u$, then back-substitute to $x$ before evaluating. Never mix.

Layer techniques in order

Substitution often simplifies the integrand into a form where IBP or partial fractions becomes obvious. State each layer: "Let $u = \ldots$; then I'll use parts on the result."

Exact form only

HSC MEX-C1 wants exact answers — surds, $\pi$, $\ln$, $\arctan$ etc. Never write a decimal unless asked. Simplify $\ln a - \ln b$ to $\ln(a/b)$ when reasonable.

What you'll master

Know

Key facts

Limits in a definite integral must transform with every substitution
A linear-over-quadratic fraction usually needs a $u = $ denominator split followed by completing the square
IBP often turns up after a substitution simplifies the integrand
$\int_0^a f(x)\,dx + \int_0^a f(a-x)\,dx = 2\int_0^a f(x)\,dx$ when $f$ is symmetric about $a/2$

Understand

Concepts

Why mixed problems demand a written plan before any algebra
Why changing limits removes the need to back-substitute and reduces error
How to recognise when a substitution converts an unfamiliar form into a standard one

Can do

Skills

Combine substitution with by parts in a single problem
Combine substitution with partial fractions, including completing the square
Handle definite integrals with careful limit transformation, leaving exact-form answers

Key terms

Mixed integralA definite or indefinite integral that requires two or more techniques in sequence (e.g. substitution followed by by parts).

Limit transformationWhen substituting $u = g(x)$, the limits change from $x = a, b$ to $u = g(a), g(b)$. Required for clean definite integration in the new variable.

Back-substitutionAlternative to changing limits: integrate in $u$, rewrite the antiderivative in $x$, then evaluate at the original $a, b$.

Completing the squareRewriting $x^2 + bx + c$ as $(x + b/2)^2 + (c - b^2/4)$ to expose a $u^2 + k^2$ form for $\arctan$ or $u^2 - k^2$ for partial fractions.

Symmetric definite integralIdentities such as $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$ that let you replace the integrand with its reflection, sometimes solving the integral by addition.

Exact formAn answer written with surds, $\pi$, $e$, $\ln$, $\arctan$ etc. rather than a decimal approximation. Required by HSC MEX-C1.

MEX-C1NESA outcome (Further Integration): applies advanced integration techniques — substitution, by parts, partial fractions, and combinations thereof.

Layering: substitution + by parts

core concept

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.

Spot the inner mess. Identify a composition $f(g(x))$ where $g(x)$ is a root, exponential or trig.
Substitute to clear it. Let $u = g(x)$, compute $dx$ in terms of $du$.
Look at what's left. Usually a product of polynomial and elementary function — apply IBP.

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.

Strategic point. If after the first substitution the integral looks worse, the substitution was wrong — go back. The first substitution should always reduce complexity, not shuffle it.

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?

Layering: substitution + partial fractions

core concept

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:

Exponential rationals. $\int \dfrac{1}{e^{x} + e^{-x}}\,dx$ — multiply top and bottom by $e^{x}$, then substitute $u = e^{x}$ to get a standard $\dfrac{du}{u^2 + 1}$.
$t = \tan(x/2)$ Weierstrass. Converts $\sin x$, $\cos x$ into rational functions of $t$, which partial fractions can finish.

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.

$$\int \frac{dx}{x^2 + bx + c} \;\longrightarrow\; \int \frac{du}{u^2 + k^2} = \frac{1}{k}\arctan\!\left(\frac{u}{k}\right) + C$$

Common mistake. When the numerator is not constant but linear, split it: write the numerator as $\tfrac{1}{2}(\text{derivative of denom}) + \text{remainder}$. The first part gives $\ln$; the remainder gives $\arctan$.

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

PROBLEM 1 · TRIG SUB + INNER U-SUB

Evaluate $\displaystyle\int_0^{1} \frac{x^3}{\sqrt{1 - x^2}}\,dx$. Leave the answer in exact form.

Let $x = \sin\theta$, so $dx = \cos\theta\,d\theta$ and $\sqrt{1 - x^2} = \cos\theta$ on $\theta \in [0, \pi/2]$. New limits: $x=0 \Rightarrow \theta = 0$; $x=1 \Rightarrow \theta = \pi/2$. Integral becomes $\int_0^{\pi/2} \dfrac{\sin^3\theta}{\cos\theta}\cdot \cos\theta\,d\theta = \int_0^{\pi/2} \sin^3\theta\,d\theta$.

The $\sqrt{1-x^2}$ form is the trig-substitution flag. Always declare the substitution AND change limits in the same line.

PROBLEM 2 · SUBSTITUTION + IBP

Evaluate $\displaystyle\int_0^{1} e^{\sqrt{x}}\,dx$. Leave the answer in exact form.

Let $u = \sqrt{x}$, so $x = u^2$ and $dx = 2u\,du$. Limits: $x = 0 \Rightarrow u = 0$; $x = 1 \Rightarrow u = 1$. Integral becomes $\int_0^1 e^{u}\cdot 2u\,du = 2\int_0^1 u\,e^{u}\,du$.

The composition $e^{\sqrt{x}}$ has an inner $\sqrt{x}$ — clear it first with substitution. The new integrand $u e^u$ is the textbook IBP setup.

PROBLEM 3 · COMPLETE THE SQUARE + LINEAR-OVER-QUADRATIC

Evaluate $\displaystyle\int_0^{1} \frac{2x + 3}{x^2 + 2x + 5}\,dx$. Leave the answer in exact form.

Split the numerator. The derivative of the denominator is $2x + 2$. Write $2x + 3 = (2x + 2) + 1$. Then $\dfrac{2x+3}{x^2+2x+5} = \dfrac{2x+2}{x^2+2x+5} + \dfrac{1}{x^2+2x+5}$, so the integral splits into two.

Linear-over-quadratic always splits this way: derivative-of-denom part gives $\ln$, the remainder gives $\arctan$ after completing the square.

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

Trap 01

Carrying old limits onto a new variable

After $u = \sin x$, the limits $x = 0, \pi/2$ become $u = 0, 1$ — not $u = 0, \pi/2$. Forgetting to change limits is the single most common mark-loser in mixed problems. Either change them, or keep the indefinite integral and back-substitute at the end.

Trap 02

Choosing IBP before clearing the inner function

Trying IBP on $\int e^{\sqrt{x}}\,dx$ directly forces $dv = e^{\sqrt{x}}\,dx$ — which you cannot integrate. Substitute first to expose $u e^u$, then IBP. The order matters.

Trap 03

Splitting a linear numerator wrongly

For $\dfrac{2x + 3}{x^2 + 2x + 5}$, the numerator splits as (derivative of denom) + (remainder) = $(2x + 2) + 1$, not as $2x$ + $3$. The first part gives $\ln$ via direct substitution; the second part needs completing the square for $\arctan$.

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 $\int_0^1$.

Work mode · how are you completing this lesson?

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?

Revisit your thinking

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.

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Multiple choice

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Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

Short answer

ApplyBand 33 marks

Q1. Use the substitution $u = 1 + x^2$ to evaluate $\displaystyle\int_0^{1} \frac{x^3}{1 + x^2}\,dx$. (3 marks)

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ApplyBand 43 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)

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Analyse4 marks4 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)

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Comprehensive answers (click to reveal)

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].

Boss battle · The Mixed Integrator II

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Five timed questions combining substitution, by parts, partial fractions and limit handling. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

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Science Jump · platform challenge

Climb platforms by answering quick integration questions. Lighter alternative to the boss.

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Module overview · Extension 2