Indefinite integrals

At a Glance

• Frequency: 7 sub-parts across 6 of 13 years (2013, 2014, 2015, 2016, 2020, 2021)
• Priority tier: T2
• Marks (count): 10 (4), 15 (1), 5 (1), 8 (1)
• Average solve time: ~6 min
• Difficulty mix: medium 4, easy 3
• Section: A | Dominant type: computation

Why This Chapter Matters

Integration trick questions appear in compulsory Q1 (Section A) and occasionally Q2/Q3. They are universally solvable — no new theorem is needed, just pattern recognition. Five archetypes account for every UPSC question on this atom. The hardest (King’s reflection, Beta/Gamma reduction) are worth 10 marks and take 5–8 minutes. Recognising the archetype at a glance is the single most valuable skill here: the wrong technique can consume 15 minutes without progress, while the correct one reaches the answer in 3.

Minimum Theory

The Fundamental Theorem (FTC) and its subtlety. If $F'(x)=f(x)$ on $(a,b]$ and $F$ extends continuously to $a$, then $\int_a^b f(x)\,dx = F(b)-F(a)$. The key subtlety: $F'(a)$ need not exist (the antiderivative can be non-differentiable at a single point), yet FTC still holds. The improper-FTC version via $\lim_{\varepsilon\to0}\int_{a+\varepsilon}^b$ is how such cases are handled rigorously.

King’s (reflection) property. For any integrable $f$ and $a\le b$: $\int_a^b f(x)\,dx = \int_a^b f(a+b-x)\,dx.$ Adding the two expressions and dividing by 2 gives a self-referential integral that collapses cleanly when $f(x)+f(a+b-x)$ simplifies to a constant. This is the most versatile trick in UPSC integration questions.

Beta and Gamma functions. $\Gamma(s)=\int_0^\infty t^{s-1}e^{-t}\,dt$ (valid for $s>0$), with $\Gamma(n)=(n-1)!$ for positive integers and $\Gamma(1/2)=\sqrt\pi$. The master reduction: $\int_0^1 x^m(-\log x)^n\,dx = \dfrac{\Gamma(n+1)}{(m+1)^{n+1}}$, obtained by the substitution $x=e^{-t}$. The Beta function $B(p,q)=\int_0^1 t^{p-1}(1-t)^{q-1}\,dt=\Gamma(p)\Gamma(q)/\Gamma(p+q)$, with the scaling $\int_a^b (x-a)^m(b-x)^n\,dx=(b-a)^{m+n+1}B(m+1,n+1)$.

Question Archetypes

recognize-antiderivative (1 question(s); 2013)

Recognise the integrand as an exact derivative and apply FTC

Recognition Cues

The integrand consists of two terms that look “mismatched” but together form the derivative of a product. Look for structures like $u'v + uv'$ (product rule) or $\frac{d}{dx}[uv]$ after grouping. A telltale sign: direct integration of each piece is intractable, but their sum is immediately an exact derivative.

Solution Template

1. Stare at the two additive terms and ask: could this be $(fg)'=f'g+fg'$ for some $f,g$?
2. Identify $f$ and $g$; verify by differentiating $fg$.
3. Extend $fg$ continuously to the endpoint if needed.
4. Apply FTC (possibly as a limit $\varepsilon\to0$).

Worked Example

2013 Paper 1, 2013-P1-Q1c (10 marks)

Evaluate $\displaystyle\int_0^1\!\left(2x\sin\frac{1}{x}-\cos\frac{1}{x}\right)dx$.

Step 1 — Spot the antiderivative. For $x>0$: $\frac{d}{dx}\!\left[x^2\sin\frac{1}{x}\right] = 2x\sin\frac{1}{x} + x^2\cos\frac{1}{x}\cdot\!\left(-\frac{1}{x^2}\right) = 2x\sin\frac{1}{x} - \cos\frac{1}{x}.$ The integrand is exactly $F'(x)$ with $F(x)=x^2\sin(1/x)$.

Step 2 — Handle $x=0$. Set $F(0)=0$; then $|F(x)|\le x^2\to 0$, so $F$ is continuous at $0$. The integrand is bounded (with an essential discontinuity at $0$, hence Riemann integrable).

Step 3 — Apply FTC via limit: $\int_0^1 = \lim_{\varepsilon\to0}\int_\varepsilon^1 = F(1)-F(\varepsilon) = \sin 1 - \varepsilon^2\sin(1/\varepsilon)\to\sin 1.$ $\boxed{\int_0^1\!\left(2x\sin\frac{1}{x}-\cos\frac{1}{x}\right)dx = \sin 1.}$

Common Traps

• Trying to integrate $2x\sin(1/x)$ and $\cos(1/x)$ separately — neither has an elementary antiderivative alone. They are engineered to give a clean primitive only in combination.
• The antiderivative $F$ is continuous at $0$ (bounded by $x^2$) but not differentiable there ($\cos(1/x)$ has no limit). FTC applies on $[\varepsilon,1]$; the limit as $\varepsilon\to0$ is justified by continuity of $F$, not differentiability.

symmetry-property (2 question(s); 2014, 2015)

Use a reflection/King property of definite integrals

Recognition Cues

The integral has limits $[a,b]$ symmetric about some midpoint, and $f(x)+f(a+b-x)$ simplifies to a constant. Canonical form: $\int_{\pi/6}^{\pi/3}$ (midpoint $\pi/4$, where $\sin\leftrightarrow\cos$) or $\int_0^1$ (midpoint $1/2$). Trig functions with complementary arguments, logarithms, and cube roots of trig ratios all signal this archetype.

Solution Template

1. Call the integral $I$.
2. Substitute $x \to a+b-x$ (limits stay $[a,b]$; $dx \to -dx$ with a sign flip from reversing limits). Call the result $I$ again (same value).
3. Add the two expressions for $I$; the integrand simplifies (often to a constant or constant times a simpler function).
4. Divide by 2.

Worked Example 1

2015 Paper 1, 2015-P1-Q1d (10 marks)

Evaluate $I = \displaystyle\int_{\pi/6}^{\pi/3}\frac{\sqrt[3]{\sin x}}{\sqrt[3]{\sin x}+\sqrt[3]{\cos x}}\,dx$.

The limits are symmetric about $\pi/4$ (since $\pi/6+\pi/3=\pi/2$). Substitute $x\to\pi/2-x$; then $\sin x \leftrightarrow \cos x$: $I = \int_{\pi/6}^{\pi/3}\frac{\sqrt[3]{\cos x}}{\sqrt[3]{\cos x}+\sqrt[3]{\sin x}}\,dx.$ Add: $2I = \int_{\pi/6}^{\pi/3}\frac{\sqrt[3]{\sin x}+\sqrt[3]{\cos x}}{\sqrt[3]{\sin x}+\sqrt[3]{\cos x}}\,dx = \int_{\pi/6}^{\pi/3}1\,dx = \frac{\pi}{3}-\frac{\pi}{6}=\frac{\pi}{6}.$ $\boxed{I = \frac{\pi}{12}.}$

Note: the cube-root is a red herring — any $g(\sin x)/[g(\sin x)+g(\cos x)]$ yields $I=(\text{interval length})/2$.

Worked Example 2

2014 Paper 1, 2014-P1-Q1d (10 marks)

Evaluate $\displaystyle\int_0^1\frac{\log(1+x)}{1+x^2}\,dx$.

Substitute $x=\tan\theta$: the integral becomes $I = \int_0^{\pi/4}\log(1+\tan\theta)\,d\theta$.

Apply King with $a+b-\theta = \pi/4-\theta$; use $\tan(\pi/4-\theta)=(1-\tan\theta)/(1+\tan\theta)$: $1+\tan(\pi/4-\theta) = \frac{2}{1+\tan\theta}, \quad \log(\cdot) = \log 2-\log(1+\tan\theta).$ Then $I = \int_0^{\pi/4}[\log 2-\log(1+\tan\theta)]\,d\theta = \frac{\pi}{4}\log 2 - I$, giving: $\boxed{I = \frac{\pi}{8}\log 2.}$

The key: the upper limit is $\pi/4$ precisely because $\tan(\pi/4)=1$ makes the symmetry work.

Common Traps

• The reflection trick requires the sum of limits to give the argument that creates the symmetry ($\pi/6+\pi/3=\pi/2$ gives $\sin\leftrightarrow\cos$; for $\int_0^{\pi/4}$ with $\tan$, the key is $\pi/4$ as the “fixed point” of $x\to\pi/4-x$). The trick fails for asymmetric limits.
• The integrand may look complicated, but after applying King, the whole function often cancels, leaving a simple integrand.

beta-gamma-reduction (2 question(s); 2016, 2021)

Reduce an integral to a Beta/Gamma function via substitution

Recognition Cues

The integrand contains $x^m(-\log x)^n$ on $[0,1]$, or $(x-a)^m(b-x)^n$ on $[a,b]$, or a product of two power functions on an interval. The substitution $x=e^{-t}$ converts the first type to a Gamma integral; linear rescaling converts the second to a Beta integral.

Solution Template

For $\int_0^1 x^m(-\log x)^n\,dx$: substitute $x=e^{-t}$, $-\log x = t$, $dx = -e^{-t}dt$, limits $[0,1]\to[\infty,0]$ (flip sign): $\int_0^\infty (e^{-t})^m t^n e^{-t}\,dt = \int_0^\infty t^n e^{-(m+1)t}\,dt = \frac{\Gamma(n+1)}{(m+1)^{n+1}}.$

For $\int_a^b (x-a)^m(b-x)^n\,dx$: substitute $x=a+(b-a)t$, $t\in[0,1]$: $= (b-a)^{m+n+1}\int_0^1 t^m(1-t)^n\,dt = (b-a)^{m+n+1}B(m+1,n+1).$

Worked Example 1

2016 Paper 1, 2016-P1-Q1c (10 marks)

Evaluate $I=\displaystyle\int_0^1\!\sqrt[3]{x\log\!\left(\frac{1}{x}\right)}\,dx$.

Rewrite: $I = \int_0^1 x^{1/3}(-\log x)^{1/3}\,dx$ (since $\log(1/x)=-\log x\ge0$ on $(0,1)$).

Substitute $x=e^{-t}$: $I = \int_0^\infty e^{-t/3}\cdot t^{1/3}\cdot e^{-t}\,dt = \int_0^\infty t^{1/3}e^{-4t/3}\,dt.$ Using $\int_0^\infty t^{s-1}e^{-pt}\,dt=\Gamma(s)/p^s$ with $s=4/3$ and $p=4/3$: $I = \frac{\Gamma(4/3)}{(4/3)^{4/3}} = \left(\frac{3}{4}\right)^{4/3}\Gamma\!\left(\frac{4}{3}\right) \approx 0.608.$ $\boxed{I = \left(\frac{3}{4}\right)^{4/3}\Gamma\!\!\left(\frac{4}{3}\right).}$

The most common error: forgetting the $(4/3)^{4/3}$ denominator (writing $\Gamma(4/3)$ alone is wrong because the formula is $\Gamma(s)/p^s$, not $\Gamma(s)$).

Worked Example 2

2021 Paper 1, 2021-P1-Q3a-iii (8 marks)

Express $\displaystyle\int_a^b(x-a)^m(b-x)^n\,dx$ in terms of the Beta function.

Substitute $x=a+(b-a)t$, $dx=(b-a)\,dt$. Then $x-a=(b-a)t$ and $b-x=(b-a)(1-t)$: $\int_a^b = (b-a)^m t^m\cdot(b-a)^n(1-t)^n\cdot(b-a)\,dt\bigg|_{t\in[0,1]} = (b-a)^{m+n+1}\int_0^1 t^m(1-t)^n\,dt.$ $\boxed{\int_a^b(x-a)^m(b-x)^n\,dx = (b-a)^{m+n+1}B(m+1,n+1).}$

Special case $a=0,b=1$: recovers the standard Beta integral $B(m+1,n+1)$ directly.

Common Traps

• Beta uses $p-1,q-1$ as exponents. So $t^m\to p-1=m$, i.e. $p=m+1$. Students who write $B(m,n)$ instead of $B(m+1,n+1)$ lose the answer.
• Count the $(b-a)$ powers. The factor $(b-a)^{m+n+1}$ accumulates as: $m$ from $(x-a)^m$, $n$ from $(b-x)^n$, and $1$ from $dx$.

integration-by-parts (1 question(s); 2020)

Evaluate a definite integral using integration by parts

Recognition Cues

The integrand contains an inverse-trig function (e.g., $\arctan$) or logarithm that resists direct integration. Integration by parts with $u=$ the transcendental and $dv=dx$ (so $v=x$) is the standard entry. The boundary term at the singular endpoint may be a $0\cdot(\text{finite})$ limit.

Solution Template

1. Set $u=$ (transcendental), $dv=dx$, compute $v=x$ and $du$.
2. $I = [xu]_a^b - \int_a^b x\,du$; evaluate boundary terms with limits.
3. Reduce $\int x\,du$ to a standard form (partial fractions, completing the square, $\arctan$ integral).

Worked Example

2020 Paper 1, 2020-P1-Q2a (15 marks)

Evaluate $\displaystyle\int_0^1\tan^{-1}\!\!\left(1-\frac{1}{x}\right)dx$.

IBP: $u=\arctan\!\bigl(1-\tfrac1x\bigr)$, $dv=dx$, $v=x$: $du = \frac{1}{1+(1-1/x)^2}\cdot\frac{1}{x^2}\,dx = \frac{dx}{2x^2-2x+1}.$ Boundary terms: at $x=1$: $\arctan(0)=0$; at $x\to0^+$: $\arctan(-\infty)=-\pi/2$, so product $\to 0$.

$I = 0 - \int_0^1\frac{x\,dx}{2x^2-2x+1}.$

Split $x=\frac{1}{4}(4x-2)+\frac{1}{2}$ to expose the logarithm-derivative and arctan parts:

$\int_0^1\frac{x\,dx}{2x^2-2x+1} = \frac{1}{4}\underbrace{\Big[\ln(2x^2-2x+1)\Big]_0^1}_{=0} + \frac{1}{2}\int_0^1\frac{dx}{2x^2-2x+1}.$

Complete the square: $2x^2-2x+1 = 2(x-\tfrac12)^2+\tfrac12$, so $\frac{1}{2x^2-2x+1}=\frac{2}{(2x-1)^2+1}$.

$\frac{1}{2}\int_0^1\frac{2\,dx}{(2x-1)^2+1}\stackrel{u=2x-1}{=}\frac{1}{2}\int_{-1}^1\frac{du}{u^2+1}=\frac{1}{2}\cdot\frac{\pi}{2}=\frac{\pi}{4}.$ $\boxed{I = -\frac{\pi}{4}.}$

The integral is negative because $1-1/x < 0$ for $x\in(0,1)$, so the arctan is negative throughout.

Common Traps

• Boundary term at $x=0$. It is $0\cdot(-\pi/2)=0$ — a $0\cdot\text{finite}$ limit, not indeterminate. Treating it as $0/0$ or $\infty\cdot0$ without checking that $x\to0$ and $\arctan\to-\pi/2$ (finite) causes unnecessary alarm.
• Sign of the final answer. The IBP produces $I=-J$, so $I=-\pi/4$, not $+\pi/4$.

integral-functional-equation (1 question(s); 2021)

Differentiate an integral identity to recover $f(x)$

Recognition Cues

The question gives an equation of the form $\int_0^x f(t)\,dt = (\text{expression in }x)$ and asks for $f$ at a point (usually $f(1)$). The trick is immediate: differentiate both sides w.r.t. $x$ using FTC.

Solution Template

1. Differentiate both sides w.r.t. $x$: LHS becomes $f(x)$ (FTC); RHS becomes the derivative of the expression.
2. Solve for $f(x)$ pointwise (often from $f(x)(1+h(x))=1$ or similar).
3. Evaluate at the requested $x$.

Worked Example

2021 Paper 1, 2021-P1-Q3a-ii (5 marks)

If $\displaystyle\int_0^x f(t)\,dt = x + \int_x^1 tf(t)\,dt$, find $f(1)$.

Differentiate both sides w.r.t. $x$: $f(x) = 1 - xf(x).$ (LHS: FTC gives $f(x)$. RHS: $d/dx[\int_x^1 tf(t)\,dt] = -xf(x)$ — FTC with $x$ as the lower limit gives a minus sign.)

Solve: $f(x)(1+x)=1 \;\Rightarrow\; f(x)=\frac{1}{1+x}.$ $\boxed{f(1) = \frac{1}{2}.}$

Common Traps

• Sign of the lower-limit FTC. $\frac{d}{dx}\int_x^1 g(t)\,dt = -g(x)$, not $+g(x)$. Getting a $+$ here gives $f(x)=1+xf(x)$, which has no real solution for positive $x$.
• The full identity may be inconsistent. The differentiated relation gives $f$ pointwise; checking whether the original integral identity holds globally with this $f$ may fail (as verified in the source solution). UPSC expects only $f(1)$, which the differentiated form uniquely determines.

Marks-Aware Writing

For a 10-mark Q1 integral: Identify the archetype in the first 30 seconds. State the technique (e.g., “King’s property with $x\to\pi/2-x$”), show the key algebraic step that makes both expressions add to something simple, then write the answer. All three parts earn marks; a bare answer earns at most 3.

For a 15-mark integral (Q2/Q3): Expect two or three distinct stages (e.g., IBP followed by partial fractions). Lay out each stage with a label and sub-result. Boundary terms at singular endpoints need a limit argument, not just substitution.

Naming the technique (“substituting $x=e^{-t}$,” “using the reflection $x\to a+b-x$,” “differentiating both sides”) signals to the examiner that you understand what you are doing — this is worth 1–2 marks beyond the computation.

Practice Set

• 2015-P1-Q4a (13 m) — — integrate a rational+trig combination; look for partial fractions
• 2022-P1-Q1d (10 m) — — symmetry property; identify the midpoint carefully
• 2018-P1-Q4b (12 m) — — Beta/Gamma reduction with a non-standard power