MHT-CET Maths · Definite Integration

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

A definite integral is an antiderivative evaluated between two limits — every Indefinite Integration technique carries over, with one new discipline: when you substitute, move the limits with you.

All Definite Integration notes MHT-CET Maths strategy

Why this matters

45 PYQs at 42% HARD, the largest page in the chapter, and every one of them is a technique from the Indefinite Integration chapter with limits attached: splitting a numerator against a quadratic, completing a square, partial fractions, a root substitution, by parts on an inverse trig function. What is new is bookkeeping — changing the limits with the substitution and evaluating cleanly — and the reduction formula for powers of tan, which appears here and nowhere else. The page is worth working slowly once, because the three property pages that follow assume you can finish an integral once the property has reduced it.

Concept 1 of 5: The Fundamental Theorem: Evaluate the Antiderivative at the Limits

A definite integral is a number, not a family of functions: find any antiderivative

F

, and the integral from

a

b

F(b) - F(a)

. The constant of integration cancels, which is why it never appears in a definite answer.

Definition

$\int_a^b f(x)\,dx = \big[F(x)\big]_a^b = F(b) - F(a)$ , where $F' = f$ .
Reversing the limits changes the sign: $\int_b^a = -\int_a^b$ . Splitting at any point $c$ : $\int_a^b = \int_a^c + \int_c^b$ — this is what makes piecewise integrands possible.
The answer is a number (or an expression in the given constants); a $+c$ in a definite answer is always wrong.
Use the same standard formulae as Indefinite Integration: $\int x^n = \frac{x^{n+1}}{n+1}$ , $\int\frac{dx}{x^2 + a^2} = \frac1a\tan^{-1}\frac{x}{a}$ , $\int\frac{f'}{f} = \log|f|$ , and so on.
Keep every sign: $F(b) - F(a)$ with a negative $F(a)$ is where marks are lost.

Fundamental theorem and two properties

\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

$F$ any antiderivative of $f$

Worked example

Evaluate

\int_1^2 (3x^2 + 1)\,dx

Practise this concept4 quick reps

Dropping the lower limit's sign

[x^3/3]_{-1}^{1} = \frac13 - \left(-\frac13\right) = \frac23

, not

0

. A negative value of

F(a)

is subtracted, which adds. This single slip is behind most wrong answers on otherwise easy definite integrals.

Concept 2 of 5: Standard Forms with Limits — Split the Numerator, Complete the Square, Partial Fractions

\dfrac{3x + 1}{x^2 + 4}

is two integrals in one: the

3x

part is

\frac{f'}{f}

and gives a log, the

1

part is the arctan standard form. Recognising the split is the whole question; the limits are then plugged into each piece.

Definition

Linear over quadratic: write $\dfrac{px + q}{x^2 + a^2} = \dfrac{p}{2}\cdot\dfrac{2x}{x^2 + a^2} + \dfrac{q}{x^2 + a^2}$ ; the first gives $\frac{p}{2}\log(x^2 + a^2)$ , the second $\frac{q}{a}\tan^{-1}\frac{x}{a}$ .
Complete the square for $x^2 - 2x + 4 = (x - 1)^2 + 3$ ; with a $3/2$ power below use $\int\dfrac{dx}{(u^2 + k^2)^{3/2}} = \dfrac{u}{k^2\sqrt{u^2 + k^2}}$ .
Partial fractions for $\dfrac{x}{(x + 2)(x + 3)} = \dfrac{-2}{x + 2} + \dfrac{3}{x + 3}$ ; the answer is a combination of logs that the options write as a single $\log\dfrac{p}{q}$ , so combine: $-2\log\frac43 + 3\log\frac54 = \log\dfrac{(5/4)^3}{(4/3)^2}$ .
Evaluate each piece at both limits before simplifying logs — $\log 8 - \log 4 = \log 2$ is cleaner than carrying $\log(x^2 + 4)$ around.

Linear numerator over a quadratic

\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}}

Worked example

Evaluate

\int_0^1 \dfrac{2x + 3}{x^2 + 1}\,dx

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 22 April Shift I · Q129Moderate

Example 2 · Definite Integration · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts

\int_{0}^{2} \frac{3x+ 1}{x^{2}+ 4}dx=

Forgetting the half in the log piece

\int\dfrac{3x}{x^2 + 4}\,dx = \dfrac32\log(x^2 + 4)

, because the derivative of the denominator is

2x

, not

x

. The option built from

3\log(\dots)

is always present.

Drill 2 more on standard forms with limits — split the numerator, complete the square, partial fractions

Concept 3 of 5: Substitution — Change the Limits, Never Substitute Back

When you put

x + 1 = t^2

, the integral becomes an integral in

t

, and its limits must become

t

-limits:

x = 0

t = 1

x = 3

t = 2

. Then you never return to

x

at all, which is faster and removes a whole class of errors.

Definition

Procedure: choose $t$ ; write $dx$ in terms of $dt$ ; convert both limits; integrate in $t$ ; evaluate. Do not convert the antiderivative back to $x$ .
Root substitutions: $\sqrt{x + 1} = t$ turns $\dfrac{dx}{(x + 2)\sqrt{x + 1}}$ into $\dfrac{2\,dt}{t^2 + 1}$ ; $2 + \sqrt{x} = t$ turns $\dfrac{dx}{2 + \sqrt x}$ into $\dfrac{2(t - 2)\,dt}{t}$ .
Trigonometric substitution: $x = \sin\theta$ for $\sqrt{1 - x^2}$ (then $\dfrac{x^2}{(1 - x^2)^{3/2}}\,dx = \tan^2\theta\,d\theta$ ); $x = \cos\theta$ for $\sqrt{\dfrac{1 - x}{1 + x}}$ — or rationalise it to $\dfrac{1 - x}{\sqrt{1 - x^2}}$ and integrate directly.
Manufactured substitutions: $\dfrac{(x - x^3)^{1/3}}{x^4} = \dfrac{\left(\frac{1}{x^2} - 1\right)^{1/3}}{x^3}$ , so $t = \dfrac{1}{x^2} - 1$ with $dt = -\dfrac{2}{x^3}dx$ . Look for the derivative of the bracket sitting outside it.
A negative $dt$ or reversed limits after substitution is normal — carry the sign, then flip the limits.

Substitution with limits

\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

Worked example

Evaluate

\int_0^3 \dfrac{dx}{\sqrt{x + 1}\,(x + 5)}

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 20 April Shift II · Q121Moderate

Example 3 · Definite Integration · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts

\int_{0}^{3} \frac{dx}{(x+ 2)\sqrt{x+ 1}}=

Substituting back and using the old limits on the new variable

After

t = \tan x

, evaluating

[\tan^{-1}t]

x = \frac{\pi}{4}

instead of at

t = 1

gives nonsense. Either convert the limits, or convert the antiderivative back — never mix the two.

The option that hides in the log

2 - 4\log\frac32

is the same number as

2\log\dfrac{4e}{9}

: write

2 = 2\log e

and combine. When your answer is not in the list, rewrite it before deciding it is wrong.

Drill 6 more on substitution — change the limits, never substitute back

Concept 4 of 5: By Parts with Limits — Inverse Trig Integrands and eˣ(f + f′)

\int_0^1\tan^{-1}x\,dx

has no standard formula, but treating it as

1\cdot\tan^{-1}x

and differentiating the inverse function turns it into an ordinary rational integral. The boundary term

[x\tan^{-1}x]_0^1

is evaluated on the spot.

Definition

By parts with limits: $\int_a^b u\,dv = \big[uv\big]_a^b - \int_a^b v\,du$ . Evaluate the bracket immediately; only the remaining integral needs work.
Inverse trig alone: take $u = \tan^{-1}x$ (or $\cos^{-1}x$ ), $dv = dx$ . Then $\int_0^1\tan^{-1}x\,dx = \dfrac{\pi}{4} - \int_0^1\dfrac{x}{1 + x^2}dx = \dfrac{\pi}{4} - \dfrac12\log 2$ ; $\int_0^1\cos^{-1}x\,dx = 0 + \int_0^1\dfrac{x}{\sqrt{1 - x^2}}dx = 1$ .
$e^x(f + f')$ : $\int e^x[f(x) + f'(x)]\,dx = e^x f(x)$ . Recognise it before integrating: $\dfrac{e^x}{x}(1 + x\log x) = e^x\left(\log x + \dfrac1x\right)$ , so the answer is $[e^x\log x]_1^e = e^e$ .
Polynomial times exponential ( $x^2e^x$ ): by parts twice, or the tabular method; $\int_0^1 x^2e^x\,dx = [x^2e^x - 2xe^x + 2e^x]_0^1 = e - 2$ .
A stem that defines $f$ by $f' = f$ , $f(0) = 1$ means $f = e^x$ ; read the definition, then integrate.

By parts and the e^x(f + f′) shortcut

\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

Worked example

Evaluate

\int_0^1 x\,e^{x}\,dx

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2025 · 26 April Shift I · Q117Moderate

Example 4 · Definite Integration · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts

\int_{0}^{1} \tan^{- 1}x\text{ }dx=

Not spotting f + f′

\dfrac{e^x}{x}(1 + x\log x)

integrated by parts from scratch is three lines of work that should be one: distribute to

e^x(\log x + 1/x)

and read off

e^x\log x

. Whenever

e^x

multiplies a sum, test whether one term is the derivative of the other.

Drill 3 more on by parts with limits — inverse trig integrands and eˣ(f + f′)

Concept 5 of 5: Reduction: I_n + I_{n−2} for Powers of tan

\tan^{12}\theta

on its own is hopeless, but

\tan^{12}\theta + \tan^{10}\theta = \tan^{10}\theta\sec^2\theta

, and that is a one-line substitution. The question asks for the SUM precisely because the sum is easy and the parts are not.

Definition

With $I_n = \int_0^{\pi/4}\tan^n\theta\,d\theta$ : $I_n + I_{n-2} = \int_0^{\pi/4}\tan^{n-2}\theta\,\sec^2\theta\,d\theta = \left[\dfrac{\tan^{n-1}\theta}{n - 1}\right]_0^{\pi/4} = \dfrac{1}{n - 1}$ .
So $I_{12} + I_{10} = \dfrac{1}{11}$ , $I_8 + I_6 = \dfrac17$ : the answer is $\dfrac{1}{(\text{larger index}) - 1}$ .
The same trick with $\cot^n$ on $[\pi/4, \pi/2]$ , and with $\sec^n$ via $\sec^2 = 1 + \tan^2$ .
If a single $I_n$ is asked, apply the relation repeatedly down to $I_0 = \dfrac{\pi}{4}$ or $I_1 = \dfrac12\log 2$ .

tan-power reduction on [0, π/4]

I_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

Worked example

I_n = \int_0^{\pi/4}\tan^n\theta\,d\theta

, find

I_8 + I_6

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 2 · Q142Moderate

Example 5 · Definite Integration · Evaluating Definite Integrals — Standard Forms, Algebraic Substitution and By Parts

I_n = \int_0^{\pi/4} \tan^n\theta\,d\theta

, then

I_{12} + I_{10}

Answering 1/(n+1) instead of 1/(n−1)

\tan^{n-2}\sec^2

integrates to

\dfrac{\tan^{n-1}}{n - 1}

. For

I_{12} + I_{10}

the answer is

\dfrac{1}{11}

;

\dfrac{1}{13}

and

\dfrac{1}{12}

are the distractors.

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (5)

The Fundamental Theorem: Evaluate the Antiderivative at the Limits
Fundamental theorem and two properties
$\int_a^b f(x)\,dx = F(b) - F(a) \qquad \int_b^a f = -\int_a^b f \qquad \int_a^b f = \int_a^c f + \int_c^b f$
Standard Forms with Limits — Split the Numerator, Complete the Square, Partial Fractions
Linear numerator over a quadratic
$\int\frac{px + q}{x^2 + a^2}\,dx = \frac{p}{2}\log\left(x^2 + a^2\right) + \frac{q}{a}\tan^{-1}\frac{x}{a} \qquad \int\frac{dx}{(u^2 + k^2)^{3/2}} = \frac{u}{k^2\sqrt{u^2 + k^2}}$
Substitution — Change the Limits, Never Substitute Back
Substitution with limits
$\int_a^b f(g(x))\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(t)\,dt \qquad \int_0^1\sqrt{\frac{1 - x}{1 + x}}\,dx = \int_0^1\frac{1 - x}{\sqrt{1 - x^2}}\,dx = \frac{\pi}{2} - 1$
By Parts with Limits — Inverse Trig Integrands and eˣ(f + f′)
By parts and the e^x(f + f′) shortcut
$\int_a^b u\,dv = \big[uv\big]_a^b - \int_a^b v\,du \qquad \int e^x\left[f(x) + f'(x)\right]dx = e^x f(x) \qquad \int_0^1\tan^{-1}x\,dx = \frac{\pi}{4} - \frac12\log 2$
Reduction: I_n + I_{n−2} for Powers of tan
tan-power reduction on [0, π/4]
$I_n = \int_0^{\pi/4}\tan^n\theta\,d\theta \ \Rightarrow\ I_n + I_{n-2} = \frac{1}{n - 1} \qquad I_0 = \frac{\pi}{4},\ I_1 = \frac12\log 2$

Watch out for (6)

Dropping the lower limit's sign
→ The Fundamental Theorem: Evaluate the Antiderivative at the Limits
Forgetting the half in the log piece
→ Standard Forms with Limits — Split the Numerator, Complete the Square, Partial Fractions
Substituting back and using the old limits on the new variable
→ Substitution — Change the Limits, Never Substitute Back
The option that hides in the log
→ Substitution — Change the Limits, Never Substitute Back
Not spotting f + f′
→ By Parts with Limits — Inverse Trig Integrands and eˣ(f + f′)
Answering 1/(n+1) instead of 1/(n−1)
→ Reduction: I_n + I_{n−2} for Powers of tan

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