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MHT-CET Maths · Definite Integration

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

Definite integrals of trigonometric expressions reduce to four moves — divide by a power of cos x and put tan x = t, use a half-angle identity, spot a derivative pair, or rewrite sin x ± cos x — with the limits converted alongside.

All Definite Integration notesMHT-CET Maths strategy

Why this matters

11 PYQs at 73% HARD — the most expensive page in the chapter, and the one where the difficulty is technique rather than recognition. The same integral with tan⁵x and cot⁵x was set in two sittings a fortnight apart, and the sec-and-cosec fractional-power integral in two more; the limits are always π/6, π/4 or π/3, so tan x = t lands on 1/√3, 1 or √3. Two printed keys on this page are wrong or garbled in the paper itself; the bank carries the corrected form and the honest note.

Concept 1 of 4

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

Intuition

Any rational expression in sin⁡x\sin x and cos⁡x\cos x that is homogeneous — every term of the same total degree — becomes a rational function of tan⁡x\tan x after dividing top and bottom by a power of cos⁡x\cos x, and then sec⁡2x dx=dt\sec^2x\,dx = dt is the missing piece.

Definition

  • Pattern: cos⁡2xsin⁡2x(cos⁡3x+sin⁡3x)2\dfrac{\cos^2x\sin^2x}{(\cos^3x + \sin^3x)^2} — divide by cos⁡6x\cos^6x to get tan⁡2xsec⁡2x(1+tan⁡3x)2\dfrac{\tan^2x\sec^2x}{(1 + \tan^3x)^2}; with u=1+tan⁡3xu = 1 + \tan^3x, du=3tan⁡2xsec⁡2x dxdu = 3\tan^2x\sec^2x\,dx.
  • Limits convert: x=0→t=0x = 0 \to t = 0, x=π6→13x = \frac{\pi}{6} \to \frac{1}{\sqrt3}, x=π4→1x = \frac{\pi}{4} \to 1, x=π3→3x = \frac{\pi}{3} \to \sqrt3.
  • sec⁡2x(1+tan⁡x)(2+tan⁡x)\dfrac{\sec^2x}{(1 + \tan x)(2 + \tan x)} is already in the form: t=tan⁡xt = \tan x gives ∫01dt(1+t)(2+t)=log⁡43\int_0^1\dfrac{dt}{(1 + t)(2 + t)} = \log\dfrac43.
  • Fractional powers: sec⁡2/3xcsc⁡4/3x=sec⁡2xtan⁡4/3x\sec^{2/3}x\csc^{4/3}x = \dfrac{\sec^2x}{\tan^{4/3}x} (multiply and divide by sec⁡4/3x\sec^{4/3}x); then ∫t−4/3dt=−3t−1/3\int t^{-4/3}dt = -3t^{-1/3}.
  • 1sin⁡2x(tan⁡5x+cot⁡5x)\dfrac{1}{\sin 2x(\tan^5x + \cot^5x)}: write sin⁡2x=2tan⁡x1+tan⁡2x\sin 2x = \dfrac{2\tan x}{1 + \tan^2x}, giving tan⁡4xsec⁡2x2(tan⁡10x+1)\dfrac{\tan^4x\sec^2x}{2(\tan^{10}x + 1)}; with u=tan⁡5xu = \tan^5x, du=5tan⁡4xsec⁡2x dxdu = 5\tan^4x\sec^2x\,dx, the integral is 110[tan⁡−1u]\dfrac{1}{10}\left[\tan^{-1}u\right].

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

Worked example

Evaluate ∫0π/3sec⁡2x(1+tan⁡x)2 dx\int_0^{\pi/3}\dfrac{\sec^2x}{(1 + \tan x)^2}\,dx.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Definite IntegrationMODERATE
∫0π/4sec⁡2x(1+tan⁡x)(2+tan⁡x) dx=\int_0^{\pi/4}\frac{\sec^2 x}{(1+\tan x)(2+\tan x)}\,dx =

[Q147 · 15th May Shift 2 · 2023]

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.
Drill 5 more on divide by cos^n x and put tan x = t

Concept 2 of 4

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

Intuition

1+cos⁡x1 + \cos x is 2cos⁡2x22\cos^2\dfrac{x}{2}, so 11+cos⁡x\dfrac{1}{1 + \cos x} is 12sec⁡2x2\dfrac12\sec^2\dfrac{x}{2} — integrable on sight. For a+bcos⁡xa + b\cos x the half-angle substitution t=tan⁡x2t = \tan\dfrac{x}{2} does the same job and produces one memorable number.

Definition

  • 1+cos⁡x=2cos⁡2x21 + \cos x = 2\cos^2\dfrac{x}{2}, 1−cos⁡x=2sin⁡2x21 - \cos x = 2\sin^2\dfrac{x}{2}, 1+sin⁡x=(sin⁡x2+cos⁡x2)21 + \sin x = \left(\sin\dfrac{x}{2} + \cos\dfrac{x}{2}\right)^2.
  • ∫dx1+cos⁡x=tan⁡x2\int\dfrac{dx}{1 + \cos x} = \tan\dfrac{x}{2}; over [π4,3π4]\left[\dfrac{\pi}{4}, \dfrac{3\pi}{4}\right] this is tan⁡3π8−tan⁡π8=(2+1)−(2−1)=2\tan\dfrac{3\pi}{8} - \tan\dfrac{\pi}{8} = (\sqrt2 + 1) - (\sqrt2 - 1) = 2.
  • Weierstrass t=tan⁡x2t = \tan\dfrac{x}{2}: dx=2 dt1+t2dx = \dfrac{2\,dt}{1 + t^2}, cos⁡x=1−t21+t2\cos x = \dfrac{1 - t^2}{1 + t^2}; limits 0→π0 \to \pi become 0→∞0 \to \infty.
  • Standard result (a>∣b∣a > |b|): ∫0πdxa+bcos⁡x=πa2−b2\int_0^\pi\dfrac{dx}{a + b\cos x} = \dfrac{\pi}{\sqrt{a^2 - b^2}}. So ∫0πdx4+3cos⁡x=π7\int_0^\pi\dfrac{dx}{4 + 3\cos x} = \dfrac{\pi}{\sqrt7}.
  • tan⁡π8=2−1\tan\dfrac{\pi}{8} = \sqrt2 - 1 and tan⁡3π8=2+1\tan\dfrac{3\pi}{8} = \sqrt2 + 1 are worth knowing cold.

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

Worked example

Evaluate ∫0π/2dx1+cos⁡x\int_0^{\pi/2}\dfrac{dx}{1 + \cos x}.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Definite IntegrationHARD
∫0πdx4+3cos⁡x=\int_0^{\pi} \frac{dx}{4 + 3\cos x} =

[Q101 · 11th May Shift 1 · 2024]

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.
Drill 1 more on half-angle forms: 1 + cos x and the a + b cos x standard result

Concept 3 of 4

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

Intuition

csc⁡xcot⁡x dx\csc x\cot x\,dx is −d(csc⁡x)-d(\csc x), and sin⁡x dx\sin x\,dx is −d(cos⁡x)=d(1−cos⁡x)-d(\cos x) = d(1 - \cos x). When the rest of the integrand is a function of that same quantity, the substitution is already written for you.

Definition

  • csc⁡xcot⁡x1+csc⁡2x\dfrac{\csc x\cot x}{1 + \csc^2x}: with t=csc⁡xt = \csc x, dt=−csc⁡xcot⁡x dxdt = -\csc x\cot x\,dx, limits π6→2\frac{\pi}{6} \to 2, π2→1\frac{\pi}{2} \to 1; the integral is ∫12dt1+t2=tan⁡−12−tan⁡−11=tan⁡−113\int_1^2\dfrac{dt}{1 + t^2} = \tan^{-1}2 - \tan^{-1}1 = \tan^{-1}\dfrac13.
  • 1+cos⁡x(1−cos⁡x)5/2\dfrac{\sqrt{1 + \cos x}}{(1 - \cos x)^{5/2}}: multiply by 1−cos⁡x1−cos⁡x\dfrac{\sqrt{1 - \cos x}}{\sqrt{1 - \cos x}} to get sin⁡x(1−cos⁡x)3\dfrac{\sin x}{(1 - \cos x)^3}, then t=1−cos⁡xt = 1 - \cos x.
  • Derivative pairs to keep ready: (sin⁡x,cos⁡x)(\sin x, \cos x), (tan⁡x,sec⁡2x)(\tan x, \sec^2x), (sec⁡x,sec⁡xtan⁡x)(\sec x, \sec x\tan x), (csc⁡x,csc⁡xcot⁡x)(\csc x, \csc x\cot x), (cot⁡x,csc⁡2x)(\cot x, \csc^2x).
  • The difference of arctangents is simplified with tan⁡−1A−tan⁡−1B=tan⁡−1A−B1+AB\tan^{-1}A - \tan^{-1}B = \tan^{-1}\dfrac{A - B}{1 + AB} — the options are written in the collapsed form.

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}

Worked example

Evaluate ∫0π/2sin⁡x(1+cos⁡x)2 dx\int_0^{\pi/2}\dfrac{\sin x}{(1 + \cos x)^2}\,dx.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Definite IntegrationHARD
∫π/3π/21+cos⁡x(1−cos⁡x)5/2 dx=\int_{\pi/3}^{\pi/2} \frac{\sqrt{1+\cos x}}{(1-\cos x)^{5/2}}\,dx =

[Q121 · 15th May Shift 2 · 2023]

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.
Drill 1 more on spot the derivative pair: csc x cot x, sin x with 1 − cos x

Concept 4 of 4

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

Intuition

tan⁡x+cot⁡x=sin⁡x+cos⁡xsin⁡xcos⁡x\sqrt{\tan x} + \sqrt{\cot x} = \dfrac{\sin x + \cos x}{\sqrt{\sin x\cos x}}, and the numerator is the derivative of sin⁡x−cos⁡x\sin x - \cos x while sin⁡xcos⁡x\sin x\cos x is 1−(sin⁡x−cos⁡x)22\dfrac{1 - (\sin x - \cos x)^2}{2}. One substitution turns the whole thing into sin⁡−1\sin^{-1}.

Definition

  • Identities: (sin⁡x−cos⁡x)2=1−2sin⁡xcos⁡x(\sin x - \cos x)^2 = 1 - 2\sin x\cos x and (sin⁡x+cos⁡x)2=1+2sin⁡xcos⁡x(\sin x + \cos x)^2 = 1 + 2\sin x\cos x.
  • If the numerator is sin⁡x+cos⁡x\sin x + \cos x, substitute t=sin⁡x−cos⁡xt = \sin x - \cos x (its derivative); if the numerator is sin⁡x−cos⁡x\sin x - \cos x, substitute t=sin⁡x+cos⁡xt = \sin x + \cos x.
  • ∫0π/4(tan⁡x+cot⁡x)dx=2∫−10dt1−t2=2⋅π2=π2\int_0^{\pi/4}\left(\sqrt{\tan x} + \sqrt{\cot x}\right)dx = \sqrt2\int_{-1}^{0}\dfrac{dt}{\sqrt{1 - t^2}} = \sqrt2\cdot\dfrac{\pi}{2} = \dfrac{\pi}{\sqrt2}.
  • Limits: at x=0x = 0, t=−1t = -1; at x=π4x = \frac{\pi}{4}, t=0t = 0; at x=π2x = \frac{\pi}{2}, t=1t = 1.

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}

Worked example

Evaluate ∫0π/2sin⁡x+cos⁡x1+2sin⁡xcos⁡x dx\int_0^{\pi/2}\dfrac{\sin x + \cos x}{\sqrt{1 + 2\sin x\cos x}}\,dx.
Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 4Definite IntegrationMODERATE
∫0π4(tan⁡x+cot⁡x)dx=\int_{0}^{\frac{\pi}{4}} (\sqrt{\tan x}+\sqrt{\cot x})dx =

[Q119 · 22 April Shift II · 2025]

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.

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 (4)

  • 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}

Watch out for (5)

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