Now that the three substitutions are familiar, the harder skills follow: converting the limits on a definite integral, finishing a trig integral with a double-angle identity, and reshaping a non-standard quadratic radical into $\sqrt{a^2 - u^2}$ form by completing the square. These three techniques together unlock every Module 15 trig-substitution question.
+5 XP warm-up For the definite integral $\displaystyle\int_0^1 \sqrt{1 - x^2}\,dx$, choose $x = \sin\theta$.
For the definite integral $\displaystyle\int_0^1 \sqrt{1 - x^2}\,dx$, choose $x = \sin\theta$. Before checking what are the new $\theta$-limits? What standard area should this integral evaluate to? Sketch your reasoning.
+5 XP to read This lesson layers three skills onto the basic substitution: change the limits with the integrand (no...
This lesson layers three skills onto the basic substitution: change the limits with the integrand (no back-substitution required), use double-angle identities to integrate even powers of sine/cosine, and complete the square inside a radical to reach standard form $\sqrt{a^2 - u^2}$ or $a^2 + u^2$.
The limit-identity-square routine: (1) substitute $x$ and update the limits to $\theta$-values, (2) collapse the radical and integrate using double-angle identities if powers appear, (3) reshape via completing the square when the quadratic isn't already $a^2 \pm x^2$.
$\cos^2\theta = \tfrac{1+\cos 2\theta}{2}$ · $\sin^2\theta = \tfrac{1-\cos 2\theta}{2}$ · $ax^2+bx+c = a(x+\tfrac{b}{2a})^2 + (c - \tfrac{b^2}{4a})$
For a definite integral, the substitution rule is
For a definite integral, the substitution rule is
where $x = g(\theta)$. The new limits remove the need to convert the antiderivative back to $x$, you evaluate directly in $\theta$.
Worked through the hook: $\displaystyle\int_0^{\sqrt 3} \frac{dx}{1 + x^2}$. Let $x = \tan\theta$, $dx = \sec^2\theta\,d\theta$, $1+x^2 = \sec^2\theta$.
For a definite integral: substitute, then immediately convert each limit to $\theta$ · Evaluate directly in $\theta$, never convert back to $x$ when limits are changed · $\int_0^{\sqrt 3} dx/(1+x^2) = \pi/3$ (a standard value to memorise) · $\arctan 0 = 0$, $\arctan 1 = \pi/4$, $\arctan\sqrt 3 = \pi/3$, $\arctan(1/\sqrt 3) = \pi/6$
Pause, copy the definite-integral procedure (convert limits to $\theta$, evaluate in $\theta$), the standard values $\arctan\sqrt{3} = \pi/3$, $\arctan 1 = \pi/4$, $\arctan(1/\sqrt{3}) = \pi/6$, and $\int_0^{\sqrt 3} dx/(1+x^2) = \pi/3$ into your book.
Quick check: For $\displaystyle\int_0^{1/2} \frac{dx}{\sqrt{1-x^2}}$ using $x = \sin\theta$, what are the new $\theta$-limits?
We just saw that for a definite trig-substitution integral you convert each limit to immediately after substituting, then...
We just saw that for a definite trig-substitution integral you convert each limit to $\theta$ immediately after substituting, then evaluate entirely in $\theta$, never convert back to $x$. That raises a question: what if the quadratic in the radical has a linear $x$ term? This card answers it → complete the square first, shift $u = x+b/2$ (so $du = dx$), then apply the appropriate trig substitution to the standard form.
Not every problem arrives in standard form. When the integrand contains $\sqrt{\,\text{quadratic in } x\,}$ that is neither $\sqrt{a^2-x^2}$ nor $\sqrt{x^2-a^2}$, complete the square first.
Mini-example: $\sqrt{5 - 4x - x^2} = \sqrt{-(x^2 + 4x - 5)} = \sqrt{-((x+2)^2 - 9)} = \sqrt{9 - (x+2)^2}$. Let $u = x+2$: integral becomes one over $\sqrt{9 - u^2}$, solved by $u = 3\sin\theta$.
Complete-the-square formula: $x^2 + bx + c = (x + b/2)^2 + (c - b^2/4)$ · Always shift $u = x + b/2$, then $du = dx$ · For $-x^2 - bx + c$, factor out $-1$ first: $-(x^2 + bx) + c = c + b^2/4 - (x + b/2)^2$ · Check sign of constant after completing, determines sine vs tangent substitution
Pause, copy the completing-the-square formula $x^2+bx+c = (x+b/2)^2+(c-b^2/4)$, the shift $u = x+b/2$, the sign check (positive constant → trig sub, negative → factor), and the factor-out-negative rule for $-x^2$ terms into your book.
Did you get this? True or false: $\sqrt{x^2 + 6x + 13} = \sqrt{(x+3)^2 + 4}$.
Worked examples · 3 in a row, reveal as you go
Worked examples · 3 in a row, reveal as you go
Evaluate $\displaystyle\int_0^{1} \frac{dx}{\sqrt{4 - x^2}}$.
Evaluate $\displaystyle\int_0^{2} \sqrt{4 - x^2}\,dx$ using $x = 2\sin\theta$, and confirm the result equals the area of a quarter circle of radius 2.
Evaluate $\displaystyle\int \frac{dx}{\sqrt{5 - 4x - x^2}}$ by completing the square.
Fill the gap: Completing the square: $x^2 + 6x + 13 = (x + $ $)^2 + $ . The shift $u = x + 3$ converts $a^2 + u^2$ standard form with $a^2 = 4$.
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: $\displaystyle\int_0^{\pi/4} \cos^2\theta\,d\theta = \tfrac{\pi}{8} + \tfrac{1}{4}$.
Activities · practice with the ideas
Activities · practice with the ideas
Evaluate $\displaystyle\int_0^{1} \frac{dx}{1 + x^2}$ by changing the limits under $x = \tan\theta$.
Evaluate $\displaystyle\int_0^{\sqrt 2/2} \sqrt{1 - x^2}\,dx$ using $x = \sin\theta$ and a double-angle identity.
Rewrite $\sqrt{8 + 2x - x^2}$ in completed-square form, identify the resulting standard pattern, and state the substitution required.
Find $\displaystyle\int \frac{dx}{\sqrt{2x - x^2}}$ by completing the square.
Evaluate $\displaystyle\int \frac{dx}{x^2 + 4x + 13}$ by completing the square and applying the tangent substitution.
Odd one out: Three of these statements about definite-integral trig substitution are correct. Which one is NOT?
Earlier you set up $\displaystyle\int_0^{\sqrt 3} \frac{dx}{1+x^2}$ with $x = \tan\theta$ and predicted the new limits.
The limits go from $\theta = 0$ to $\theta = \pi/3$, the integrand collapses to the constant 1, and the integral evaluates to $\pi/3$ directly, no back-substitution needed. Combined with double-angle identities for $\sin^2\theta$ and $\cos^2\theta$, and the completing-the-square trick for non-standard quadratics, you can now handle the full range of MEX-C1 trig-substitution problems.
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. Evaluate $\displaystyle\int_0^{2} \frac{dx}{4 + x^2}$. (2 marks)
Q2. Evaluate $\displaystyle\int_0^{1} \sqrt{1 - x^2}\,dx$ using $x = \sin\theta$ and a double-angle identity. Confirm geometrically. (3 marks)
Q3. By completing the square, find $\displaystyle\int \frac{dx}{\sqrt{3 - 2x - x^2}}$. (3 marks)
Activity answers:
1. $x = \tan\theta$; limits $0 \to \pi/4$. $\int_0^{\pi/4} d\theta = \pi/4$.
2. $x = \sin\theta$; limits $0 \to \pi/4$. Integrand $\cos^2\theta\,d\theta = \tfrac{1+\cos 2\theta}{2}\,d\theta$. Evaluates to $\tfrac{\pi}{8} + \tfrac{1}{4}$.
3. $8 + 2x - x^2 = 9 - (x-1)^2$. Pattern $\sqrt{a^2 - u^2}$ with $a = 3$, $u = x-1$. Substitution $u = 3\sin\theta$.
4. $2x - x^2 = 1 - (x-1)^2$. $\int du/\sqrt{1-u^2} = \arcsin u + C = \arcsin(x-1) + C$.
5. $x^2 + 4x + 13 = (x+2)^2 + 9$. $\int du/(u^2 + 9) = \tfrac{1}{3}\arctan(u/3) + C = \tfrac{1}{3}\arctan\!\frac{x+2}{3} + C$.
Q1 (2 marks): $x = 2\tan\theta$, $dx = 2\sec^2\theta\,d\theta$, $4 + x^2 = 4\sec^2\theta$. Limits: $0 \to 0$ and $2 \to \pi/4$ [1]. Integral $= \int_0^{\pi/4} \tfrac{1}{2}\,d\theta = \tfrac{\pi}{8}$ [1].
Q2 (3 marks): $x = \sin\theta$, $\sqrt{1-x^2} = \cos\theta$, limits $0 \to \pi/2$. Integrand $= \cos^2\theta\,d\theta$ [1]. Use $\cos^2\theta = \tfrac{1}{2}(1+\cos 2\theta)$: $\int_0^{\pi/2}\tfrac{1}{2}(1+\cos 2\theta)\,d\theta = \left[\tfrac{\theta}{2}+\tfrac{\sin 2\theta}{4}\right]_0^{\pi/2} = \tfrac{\pi}{4}$ [1]. Matches area of quarter unit circle $= \tfrac{1}{4}\pi(1)^2 = \tfrac{\pi}{4}$ [1].
Q3 (3 marks): $3 - 2x - x^2 = -(x^2+2x-3) = -((x+1)^2 - 4) = 4 - (x+1)^2$ [1]. Let $u = x+1$; integral becomes $\int du/\sqrt{4-u^2}$ [1]. Standard result $= \arcsin(u/2) + C = \arcsin((x+1)/2) + C$ [1].
Sit the module quiz, then close the lesson off.
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
Start the module quiz →Tick when you've finished the practice and review.
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