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Definite Integration formulas

22 formulas and 24 common traps for MHT-CET Maths Definite Integration, grouped by subtopic.

Full notes with worked examples

Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts

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The Fundamental Theorem: Evaluate the Antiderivative at the Limits

Fundamental theorem and two properties

∫abf(x) dx=F(b)−F(a)∫baf=−∫abf∫abf=∫acf+∫cbf\int_a^b f(x)\,dx = F(b) - F(a) \qquad \int_b^a f = -\int_a^b f \qquad \int_a^b f = \int_a^c f + \int_c^b f
  • FFany antiderivative of ff

Standard Forms with Limits — Split the Numerator, Complete the Square, Partial Fractions

Linear numerator over a quadratic

∫px+qx2+a2 dx=p2log⁡(x2+a2)+qatan⁡−1xa∫dx(u2+k2)3/2=uk2u2+k2\int\frac{px + q}{x^2 + a^2}\,dx = \frac{p}{2}\log\left(x^2 + a^2\right) + \frac{q}{a}\tan^{-1}\frac{x}{a} \qquad \int\frac{dx}{(u^2 + k^2)^{3/2}} = \frac{u}{k^2\sqrt{u^2 + k^2}}

Substitution — Change the Limits, Never Substitute Back

Substitution with limits

∫abf(g(x)) g′(x) dx=∫g(a)g(b)f(t) dt∫011−x1+x dx=∫011−x1−x2 dx=π2−1\int_a^b f(g(x))\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(t)\,dt \qquad \int_0^1\sqrt{\frac{1 - x}{1 + x}}\,dx = \int_0^1\frac{1 - x}{\sqrt{1 - x^2}}\,dx = \frac{\pi}{2} - 1

By Parts with Limits — Inverse Trig Integrands and eˣ(f + f′)

By parts and the e^x(f + f′) shortcut

∫abu dv=[uv]ab−∫abv du∫ex[f(x)+f′(x)]dx=exf(x)∫01tan⁡−1x dx=π4−12log⁡2\int_a^b u\,dv = \big[uv\big]_a^b - \int_a^b v\,du \qquad \int e^x\left[f(x) + f'(x)\right]dx = e^x f(x) \qquad \int_0^1\tan^{-1}x\,dx = \frac{\pi}{4} - \frac12\log 2

Reduction: I_n + I_{n−2} for Powers of tan

tan-power reduction on [0, π/4]

In=∫0π/4tan⁡nθ dθ ⇒ In+In−2=1n−1I0=π4, I1=12log⁡2I_n = \int_0^{\pi/4}\tan^n\theta\,d\theta \ \Rightarrow\ I_n + I_{n-2} = \frac{1}{n - 1} \qquad I_0 = \frac{\pi}{4},\ I_1 = \frac12\log 2

Common traps

Dropping the lower limit's sign

[x3/3]−11=13−(−13)=23[x^3/3]_{-1}^{1} = \frac13 - \left(-\frac13\right) = \frac23, not 00. A negative value of F(a)F(a) is subtracted, which adds. This single slip is behind most wrong answers on otherwise easy definite integrals.

Forgetting the half in the log piece

∫3xx2+4 dx=32log⁡(x2+4)\int\dfrac{3x}{x^2 + 4}\,dx = \dfrac32\log(x^2 + 4), because the derivative of the denominator is 2x2x, not xx. The option built from 3log⁡(… )3\log(\dots) is always present.

Substituting back and using the old limits on the new variable

After t=tan⁡xt = \tan x, evaluating [tan⁡−1t][\tan^{-1}t] at x=π4x = \frac{\pi}{4} instead of at t=1t = 1 gives nonsense. Either convert the limits, or convert the antiderivative back — never mix the two.

The option that hides in the log

2−4log⁡322 - 4\log\frac32 is the same number as 2log⁡4e92\log\dfrac{4e}{9}: write 2=2log⁡e2 = 2\log e and combine. When your answer is not in the list, rewrite it before deciding it is wrong.

Not spotting f + f′

exx(1+xlog⁡x)\dfrac{e^x}{x}(1 + x\log x) integrated by parts from scratch is three lines of work that should be one: distribute to ex(log⁡x+1/x)e^x(\log x + 1/x) and read off exlog⁡xe^x\log x. Whenever exe^x multiplies a sum, test whether one term is the derivative of the other.

Answering 1/(n+1) instead of 1/(n−1)

tan⁡n−2sec⁡2\tan^{n-2}\sec^2 integrates to tan⁡n−1n−1\dfrac{\tan^{n-1}}{n - 1}. For I12+I10I_{12} + I_{10} the answer is 111\dfrac{1}{11}; 113\dfrac{1}{13} and 112\dfrac{1}{12} are the distractors.

Trigonometric Definite Integrals — tan x = t, Half-Angle Forms and Powers

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Divide by cos^n x and Put tan x = t

The tan substitution

t=tan⁡x,dt=sec⁡2x dx,sin⁡2x=2t1+t2,x: 0,π6,π4,π3 → t: 0,13,1,3t = \tan x,\quad dt = \sec^2x\,dx,\qquad \sin 2x = \frac{2t}{1 + t^2},\qquad x:\ 0,\tfrac{\pi}{6},\tfrac{\pi}{4},\tfrac{\pi}{3} \ \to\ t:\ 0,\tfrac{1}{\sqrt3},1,\sqrt3

Half-Angle Forms: 1 + cos x and the a + b cos x Standard Result

Half-angle results

∫dx1+cos⁡x=tan⁡x2∫0πdxa+bcos⁡x=πa2−b2 (a>∣b∣)tan⁡π8=2−1\int\frac{dx}{1 + \cos x} = \tan\frac{x}{2} \qquad \int_0^\pi\frac{dx}{a + b\cos x} = \frac{\pi}{\sqrt{a^2 - b^2}} \ (a > |b|) \qquad \tan\frac{\pi}{8} = \sqrt2 - 1

Spot the Derivative Pair: csc x cot x, sin x with 1 − cos x

Derivative pairs

d(csc⁡x)=−csc⁡xcot⁡x dxd(1−cos⁡x)=sin⁡x dxtan⁡−1A−tan⁡−1B=tan⁡−1A−B1+ABd(\csc x) = -\csc x\cot x\,dx \qquad d(1 - \cos x) = \sin x\,dx \qquad \tan^{-1}A - \tan^{-1}B = \tan^{-1}\frac{A - B}{1 + AB}

√tan x + √cot x: the sin x − cos x Substitution

The sin x − cos x substitution

t=sin⁡x−cos⁡x ⇒ dt=(cos⁡x+sin⁡x) dx,sin⁡xcos⁡x=1−t22t = \sin x - \cos x \ \Rightarrow\ dt = (\cos x + \sin x)\,dx,\quad \sin x\cos x = \frac{1 - t^2}{2}

Common traps

The paper's own typo: sen^{2/3}

The 14 May 2024 Shift 1 paper prints sen2/3xcsc⁡4/3x\text{sen}^{2/3}x\csc^{4/3}x; its key works with sec⁡2/3x\sec^{2/3}x and reaches 37/6−35/63^{7/6} - 3^{5/6}. With sin⁡2/3x\sin^{2/3}x the integrand has no elementary antiderivative — if a trig power integral looks impossible, suspect a misprint and try the sec version.

A positive integrand cannot give a negative answer

∫π/43π/4dx1+cos⁡x=2\int_{\pi/4}^{3\pi/4}\dfrac{dx}{1 + \cos x} = 2; the 2022 sitting's stored key once read −2-2 and the official key says 22. Sanity-check the sign of every definite integral against the sign of its integrand before choosing.

π/7 versus π/√7

πa2−b2\dfrac{\pi}{\sqrt{a^2 - b^2}} has the ROOT in the denominator. For 4+3cos⁡x4 + 3\cos x that is π/7\pi/\sqrt7; π/7\pi/7 is the distractor built for students who forget it.

Reading the negative of the integral off the option list

tan⁡−12−tan⁡−11\tan^{-1}2 - \tan^{-1}1 is positive, and equals tan⁡−113\tan^{-1}\frac13. The 2021 paper offered π4−tan⁡−12\frac{\pi}{4} - \tan^{-1}2 — the NEGATIVE — as a distractor, and the stored key once pointed at it. The integrand is positive on the interval, so any negative option is out before you compute.

Rationalising √tan + √cot term by term

Integrating tan⁡x\sqrt{\tan x} alone is a long substitution. The sum is far easier than either part — combine first, then substitute sin⁡x−cos⁡x\sin x - \cos x.

Odd and Even Integrands — Symmetric Limits

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The Odd/Even Test on Symmetric Limits

Symmetric-limit rule

∫−aaf(x) dx={0,f(−x)=−f(x)2∫0af(x) dx,f(−x)=f(x)\int_{-a}^{a} f(x)\,dx = \begin{cases} 0, & f(-x) = -f(x) \\[2pt] 2\displaystyle\int_0^a f(x)\,dx, & f(-x) = f(x) \end{cases}

Split a Mixed Integrand into Its Odd and Even Parts

Odd part vanishes

∫−aa[odd(x)+even(x)]dx=2∫0aeven(x) dx∫0π/2x2cos⁡x dx=π24−2\int_{-a}^{a}\big[\text{odd}(x) + \text{even}(x)\big]dx = 2\int_0^a \text{even}(x)\,dx \qquad \int_0^{\pi/2}x^2\cos x\,dx = \frac{\pi^2}{4} - 2

Symmetry Under x → 1/x on [1/2, 2]

The reciprocal symmetry

x=1t:∫1/aa1x g ⁣(x−1x)dx=−∫1/aa1t g ⁣(t−1t)dt  for odd g ⇒ I=0x = \tfrac1t:\quad \int_{1/a}^{a}\frac{1}{x}\,g\!\left(x - \frac1x\right)dx = -\int_{1/a}^{a}\frac{1}{t}\,g\!\left(t - \frac1t\right)dt \ \text{ for odd } g \ \Rightarrow\ I = 0

Even Does Not Mean Convergent: the csc⁴x Trap

Check the domain first

∫−π/4π/4csc⁡4x dx diverges;[−cot⁡x−cot⁡3x3]−π/4π/4=−83 is the exam’s formal key\int_{-\pi/4}^{\pi/4}\csc^4x\,dx \text{ diverges}; \quad \left[-\cot x - \tfrac{\cot^3x}{3}\right]_{-\pi/4}^{\pi/4} = -\tfrac83 \text{ is the exam's formal key}

Common traps

Testing the limits instead of the function

Symmetric limits are necessary, not sufficient. ∫−11x2 dx\int_{-1}^{1}x^2\,dx is 23\frac23, not 00: the integrand must be odd for the integral to vanish. Write f(−x)f(-x) explicitly every time.

Discarding the constant with the odd terms

t3+4+tcos⁡tt^3 + 4 + t\cos t over [−1,1][-1, 1]: the t3t^3 and tcos⁡tt\cos t vanish, but the 44 integrates to 88. Zero is the answer for the ODD part, not for the question.

Applying it without the 1/x

∫1/22sin⁡ ⁣(x−1x)dx\int_{1/2}^{2}\sin\!\left(x - \frac1x\right)dx is NOT zero: without the 1x\dfrac1x the substitution produces an extra 1t2\dfrac{1}{t^2} and the symmetry breaks. Check for the factor before claiming the cancellation.

Marking the mathematically honest option

There is no honest option — the integral diverges and the key is −83-\frac83. On this paper choose the key; on any paper, a negative value for a positive integrand should make you re-read the interval for a blow-up.

King's Property — f(a + b − x) and the f/(f + g) Family

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King's Property: ∫ f(x) = ∫ f(a + b − x)

King's property

∫abf(x) dx=∫abf(a+b−x) dx∫0π/4log⁡(1+tan⁡x) dx=π8log⁡2\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx \qquad \int_0^{\pi/4}\log(1 + \tan x)\,dx = \frac{\pi}{8}\log 2

The f/(f + g) Family: ∫ f(x)/(f(x) + f(a + b − x)) = (b − a)/2

The f/(f + g) result

∫abf(x)f(x)+f(a+b−x) dx=b−a2∫0π/2asin⁡x+bcos⁡xsin⁡x+cos⁡x dx=π4(a+b)\int_a^b\frac{f(x)}{f(x) + f(a + b - x)}\,dx = \frac{b - a}{2} \qquad \int_0^{\pi/2}\frac{a\sin x + b\cos x}{\sin x + \cos x}\,dx = \frac{\pi}{4}(a + b)

The x·f(sin x) Trick on [0, π]: Pull the x Out as π/2

Pulling x out

∫0πx f(sin⁡x) dx=π2∫0πf(sin⁡x) dx∫0πsin⁡x1+cos⁡2x dx=π2\int_0^\pi x\,f(\sin x)\,dx = \frac{\pi}{2}\int_0^\pi f(\sin x)\,dx \qquad \int_0^\pi\frac{\sin x}{1 + \cos^2x}\,dx = \frac{\pi}{2}

Functional Symmetry Given in the Stem: f(x) = f(1 − x), g(x) + g(a − x) = 4

Symmetry handed to you

f(a+b−x)=f(x) ⇒ ∫abxf(x) dx=a+b2∫abf(x) dxf(a + b - x) = f(x) \ \Rightarrow\ \int_a^b x f(x)\,dx = \frac{a + b}{2}\int_a^b f(x)\,dx

Integrands with 1/(1 + aˣ) over Symmetric Limits

The 1/(1 + aˣ) cancellation

11+ax+11+a−x=1f even: ∫−aaf(x)1+ex dx=∫0af(x) dx\frac{1}{1 + a^{x}} + \frac{1}{1 + a^{-x}} = 1 \qquad f \text{ even}:\ \int_{-a}^{a}\frac{f(x)}{1 + e^{x}}\,dx = \int_0^a f(x)\,dx

An Inverse-Trig Identity Before the Reflection

Unpack, then reflect

cot⁡−1(1−x+x2)=tan⁡−1x+tan⁡−1(1−x)∫01tan⁡−1(1−x+x2) dx=log⁡2\cot^{-1}(1 - x + x^2) = \tan^{-1}x + \tan^{-1}(1 - x) \qquad \int_0^1\tan^{-1}(1 - x + x^2)\,dx = \log 2

Common traps

Reflecting about the wrong point

On [π/3,2π/3][\pi/3, 2\pi/3] the reflection is x→π−xx \to \pi - x (endpoints add to π\pi), not x→π2−xx \to \frac{\pi}{2} - x. Always add the two endpoints first; that sum is the only thing the substitution uses.

Missing the disguised reflection in the denominator

log⁡(16x2−8x3+x4)=log⁡x2+log⁡(4−x)2\log(16x^2 - 8x^3 + x^4) = \log x^2 + \log(4 - x)^2: on [1,3][1, 3] the second term IS the reflection of the first (1+3=41 + 3 = 4). Factor the quartic before deciding the family does not apply.

Using the trick with cos x

cos⁡(π−x)=−cos⁡x\cos(\pi - x) = -\cos x, so ∫0πx g(cos⁡x) dx\int_0^\pi x\,g(\cos x)\,dx does NOT reduce this way unless gg is even. The trick needs the rest of the integrand to be unchanged by the reflection.

Treating R₂ as an integral of x f(x)

R2R_2 is the AREA under ff, i.e. ∫f\int f, with no xx. The relation R2=2R1R_2 = 2R_1 comes from 2R1=∫f2R_1 = \int f; reading it the other way round gives 12R1\frac12 R_1, which is offered.

Forgetting the halving

After adding, 2I=∫−aaf2I = \int_{-a}^{a}f, which for even ff is 2∫0af2\int_0^a f — so I=∫0afI = \int_0^a f, not 2∫0af2\int_0^a f. The option π2\frac{\pi}{2} beside the correct π4\frac{\pi}{4} is this slip.

Integrating tan⁻¹(1 − x + x²) directly

By parts on the quadratic argument produces a rational integral that takes minutes. The identity is the intended route and reduces the question to a known value of ∫01tan⁡−1x dx\int_0^1\tan^{-1}x\,dx.

Modulus and Greatest-Integer Integrands — Split the Interval

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Modulus: Split Where the Inside Changes Sign

Modulus splitting

∫ab∣u(x)∣ dx=∫ac±u dx+∫cb∓u dx  where u(c)=0, sign chosen to make each piece ≥0\int_a^b|u(x)|\,dx = \int_a^c\pm u\,dx + \int_c^b\mp u\,dx \ \text{ where } u(c) = 0,\ \text{sign chosen to make each piece } \ge 0

Greatest Integer: Piecewise Constant, So Sum the Pieces

Integrating a staircase

∫ab[x] g(x) dx=∑n  n∫max⁡(a,n)min⁡(b,n+1)g(x) dx∫02[x2] dx=5−2−3\int_a^b[x]\,g(x)\,dx = \sum_{n}\; n\int_{\max(a,n)}^{\min(b,n+1)} g(x)\,dx \qquad \int_0^2[x^2]\,dx = 5 - \sqrt2 - \sqrt3

Piecewise Constant by an Identity: tan⁻¹u + tan⁻¹(1/u)

The reciprocal arctangent identity

tan⁡−1u+tan⁡−11u={π2,u>0−π2,u<0\tan^{-1}u + \tan^{-1}\frac{1}{u} = \begin{cases} \dfrac{\pi}{2}, & u > 0 \\[4pt] -\dfrac{\pi}{2}, & u < 0 \end{cases}

Common traps

Integrating the bare expression

∫04(2x−5) dx=−4\int_0^4(2x - 5)\,dx = -4, but ∫04∣2x−5∣ dx=172\int_0^4|2x - 5|\,dx = \frac{17}{2}. A modulus integral is never negative and is never the signed integral of the inside; if you did not split, you answered a different question.

Using the endpoint's floor for the whole last piece

On [3,3.5][3, 3.5] the floor is 33 throughout, contributing 3×0.5=1.53 \times 0.5 = 1.5 — not 3.5×3.5 \times anything. And [x][x] on [0.2,1)[0.2, 1) is 00, so that piece contributes nothing however long it is.

The key that ignores the sign of x

Both sittings of this question mark 2π2\pi, which assumes tan⁡−1u+tan⁡−1(1/u)=π/2\tan^{-1}u + \tan^{-1}(1/u) = \pi/2 even for negative uu. Mathematically the answer on [−1,3][-1, 3] is π\pi; on the paper choose 2π2\pi, and know why.

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