EAPCET Mathematics · Intermediate second year · Paper IIB

Integration

Integration covers indefinite integrals: standard forms, substitution, integration by parts and partial fractions.

Subject
Mathematics
Taught in
Intermediate second year · Paper IIB
Mathematics in the paper
80 of 160 questions

Key formulas and results

  • ∫xn dx=xn+1n+1+C\displaystyle\int x^{n}\,dx = \dfrac{x^{n+1}}{n + 1} + C, n≠−1n \ne -1
  • ∫1x dx=log⁡∣x∣+C\displaystyle\int \dfrac{1}{x}\,dx = \log|x| + C
  • ∫f′(x)f(x) dx=log⁡∣f(x)∣+C\displaystyle\int \dfrac{f'(x)}{f(x)}\,dx = \log|f(x)| + C
  • ∫ex[f(x)+f′(x)] dx=exf(x)+C\displaystyle\int e^{x}[f(x) + f'(x)]\,dx = e^{x} f(x) + C
  • ∫dxa2+x2=1atan⁡−1xa+C\displaystyle\int \dfrac{dx}{a^{2} + x^{2}} = \dfrac{1}{a}\tan^{-1}\dfrac{x}{a} + C; ∫dxa2−x2=sin⁡−1xa+C\displaystyle\int \dfrac{dx}{\sqrt{a^{2} - x^{2}}} = \sin^{-1}\dfrac{x}{a} + C
  • By parts: ∫uv dx=u∫v dx−∫[u′∫v dx]dx\displaystyle\int uv\,dx = u\int v\,dx - \int\left[u'\int v\,dx\right]dx
  • ∫dxx2−a2=12alog⁡∣x−ax+a∣+C\displaystyle\int \dfrac{dx}{x^{2} - a^{2}} = \dfrac{1}{2a}\log\left|\dfrac{x - a}{x + a}\right| + C and ∫dxa2−x2=12alog⁡∣a+xa−x∣+C\displaystyle\int \dfrac{dx}{a^{2} - x^{2}} = \dfrac{1}{2a}\log\left|\dfrac{a + x}{a - x}\right| + C
  • ∫dxx2+a2=log⁡∣x+x2+a2∣+C\displaystyle\int \dfrac{dx}{\sqrt{x^{2} + a^{2}}} = \log\left|x + \sqrt{x^{2} + a^{2}}\right| + C and ∫dxx2−a2=log⁡∣x+x2−a2∣+C\displaystyle\int \dfrac{dx}{\sqrt{x^{2} - a^{2}}} = \log\left|x + \sqrt{x^{2} - a^{2}}\right| + C
  • ∫a2−x2 dx=x2a2−x2+a22sin⁡−1xa+C\displaystyle\int \sqrt{a^{2} - x^{2}}\,dx = \dfrac{x}{2}\sqrt{a^{2} - x^{2}} + \dfrac{a^{2}}{2}\sin^{-1}\dfrac{x}{a} + C
  • ∫x2±a2 dx=x2x2±a2±a22log⁡∣x+x2±a2∣+C\displaystyle\int \sqrt{x^{2} \pm a^{2}}\,dx = \dfrac{x}{2}\sqrt{x^{2} \pm a^{2}} \pm \dfrac{a^{2}}{2}\log\left|x + \sqrt{x^{2} \pm a^{2}}\right| + C
  • ∫eaxsin⁡bx dx=eax(asin⁡bx−bcos⁡bx)a2+b2+C\displaystyle\int e^{ax}\sin bx\,dx = \dfrac{e^{ax}(a\sin bx - b\cos bx)}{a^{2} + b^{2}} + C and ∫eaxcos⁡bx dx=eax(acos⁡bx+bsin⁡bx)a2+b2+C\displaystyle\int e^{ax}\cos bx\,dx = \dfrac{e^{ax}(a\cos bx + b\sin bx)}{a^{2} + b^{2}} + C
  • ∫sec⁡x dx=log⁡∣sec⁡x+tan⁡x∣+C\displaystyle\int \sec x\,dx = \log|\sec x + \tan x| + C and ∫cosec⁡x dx=log⁡∣cosec⁡x−cot⁡x∣+C\displaystyle\int \operatorname{cosec} x\,dx = \log|\operatorname{cosec} x - \cot x| + C
  • Reduction: ∫tan⁡nx dx=tan⁡n−1xn−1−∫tan⁡n−2x dx\displaystyle\int \tan^{n}x\,dx = \dfrac{\tan^{n-1}x}{n - 1} - \int \tan^{n-2}x\,dx
  • Reduction: ∫sin⁡nx dx=−sin⁡n−1xcos⁡xn+n−1n∫sin⁡n−2x dx\displaystyle\int \sin^{n}x\,dx = -\dfrac{\sin^{n-1}x\cos x}{n} + \dfrac{n - 1}{n}\int \sin^{n-2}x\,dx

What this chapter covers

  • Standard integrals
  • Integration by substitution
  • Integration by parts
  • Partial fractions
  • Integrals of special forms

How it is usually asked

  • Integrate using a substitution
  • Integrate by parts
  • Spot the eˣ[f(x) + f′(x)] pattern
  • Use partial fractions

Understanding Integration

Integration reverses differentiation: ∫ f(x) dx is a function whose derivative is f(x), plus an arbitrary constant C. The first tool is a table of standard integrals read backwards from the derivative table, including xⁿ, 1/x, eˣ, the trigonometric functions and forms like 1/(a² + x²) and 1/√(a² − x²). Any answer can be checked by differentiating it.

Substitution handles integrands that contain a function and its derivative. If you set t = g(x), then g′(x) dx becomes dt, and the integral becomes a standard one in t. This explains the useful patterns ∫ f′(x)/f(x) dx = log|f(x)| + C and ∫ [f(x)]ⁿ f′(x) dx = [f(x)]ⁿ⁺¹/(n + 1) + C, which can be applied directly once recognised.

Integration by parts, ∫ u v dx = u ∫v dx − ∫ [u′ ∫v dx] dx, handles products such as x eˣ or x sin x; choose u by the ILATE order. A special case is ∫ eˣ [f(x) + f′(x)] dx = eˣ f(x) + C. Rational functions are split by partial fractions first. EAPCET questions test which method fits and these shortcut patterns.

Worked example

EAPCET-style question

∫ex 1+xlog⁡xx dx\displaystyle\int e^{x}\,\dfrac{1 + x\log x}{x}\,dx equals

  1. (1)exlog⁡x+Ce^{x}\log x + C
  2. (2)exx+C\dfrac{e^{x}}{x} + C
  3. (3)ex(log⁡x+1x)+Ce^{x}\left(\log x + \dfrac{1}{x}\right) + C
  4. (4)xexlog⁡x+Cxe^{x}\log x + C

Solution

  1. Split the integrand: ex 1+xlog⁡xx=ex[log⁡x+1x]e^{x}\,\dfrac{1 + x\log x}{x} = e^{x}\left[\log x + \dfrac{1}{x}\right].
  2. With f(x)=log⁡xf(x) = \log x, f′(x)=1xf'(x) = \dfrac{1}{x}, so it has the form ex[f(x)+f′(x)]e^{x}[f(x) + f'(x)].
  3. Hence the integral is exf(x)+C=exlog⁡x+Ce^{x} f(x) + C = e^{x}\log x + C.
  4. Check: ddx(exlog⁡x)=exlog⁡x+exx\dfrac{d}{dx}(e^{x}\log x) = e^{x}\log x + \dfrac{e^{x}}{x}, which matches the integrand.

Answer: exlog⁡x+Ce^{x}\log x + C

Shortcuts and results to remember

  • Pick u in integration by parts with ILATE: Inverse trig, Logarithmic, Algebraic, Trigonometric, Exponential.
  • For ∫dxa+bsin⁡2x\displaystyle\int \dfrac{dx}{a + b\sin^{2}x} or ∫dxacos⁡2x+b\displaystyle\int \dfrac{dx}{a\cos^{2}x + b}, divide top and bottom by cos⁡2x\cos^{2}x and put tan⁡x=t\tan x = t; for ∫dxa+bsin⁡x\displaystyle\int \dfrac{dx}{a + b\sin x} or (a+bcos⁡x)(a + b\cos x), use t=tan⁡x2t = \tan\dfrac{x}{2}.
  • For ∫px+qquadratic\displaystyle\int \dfrac{px + q}{\text{quadratic}}, write px+q=A×ddx(quadratic)+Bpx + q = A \times \dfrac{d}{dx}(\text{quadratic}) + B and split.
  • For ∫x2±1x4+kx2+1 dx\displaystyle\int \dfrac{x^{2} \pm 1}{x^{4} + kx^{2} + 1}\,dx, divide by x2x^{2} and put t=x∓1xt = x \mp \dfrac{1}{x}.
  • In partial fractions an irreducible quadratic factor gets a (Bx+C)(Bx + C) numerator, and a repeated factor (x−a)3(x - a)^{3} needs Ax−a+B(x−a)2+C(x−a)3\dfrac{A}{x - a} + \dfrac{B}{(x - a)^{2}} + \dfrac{C}{(x - a)^{3}}.

More practice questions

Practice question 1

If In=∫tan⁡nx dxI_{n} = \displaystyle\int \tan^{n}x\,dx (n>1)(n > 1) and I4+I6=atan⁡5x+bx5+CI_{4} + I_{6} = a\tan^{5}x + bx^{5} + C, then (a,b)(a, b) is

  1. (1)(15,0)\left(\tfrac{1}{5}, 0\right)
  2. (2)(15,−1)\left(\tfrac{1}{5}, -1\right)
  3. (3)(−15,0)\left(-\tfrac{1}{5}, 0\right)
  4. (4)(16,0)\left(\tfrac{1}{6}, 0\right)

Solution

  1. From the reduction formula, In+In−2=tan⁡n−1xn−1I_{n} + I_{n-2} = \dfrac{\tan^{n-1}x}{n - 1}
  2. Put n=6n = 6: I6+I4=tan⁡5x5I_{6} + I_{4} = \dfrac{\tan^{5}x}{5}
  3. Compare with atan⁡5x+bx5a\tan^{5}x + bx^{5}
  4. a=15a = \tfrac{1}{5} and b=0b = 0

Answer: (15,0)\left(\tfrac{1}{5}, 0\right)

Practice question 2

If ∫dxcos⁡3x2sin⁡2x=(tan⁡x)A+C(tan⁡x)B+K\displaystyle\int \dfrac{dx}{\cos^{3}x\sqrt{2\sin 2x}} = (\tan x)^{A} + C(\tan x)^{B} + K, then A+B+CA + B + C equals

  1. (1)165\dfrac{16}{5}
  2. (2)215\dfrac{21}{5}
  3. (3)710\dfrac{7}{10}
  4. (4)3

Solution

  1. 2sin⁡2x=4sin⁡xcos⁡x=2cos⁡xtan⁡x\sqrt{2\sin 2x} = \sqrt{4\sin x\cos x} = 2\cos x\sqrt{\tan x}
  2. The integrand becomes sec⁡4x2tan⁡x=(1+tan⁡2x)sec⁡2x2tan⁡x\dfrac{\sec^{4}x}{2\sqrt{\tan x}} = \dfrac{(1 + \tan^{2}x)\sec^{2}x}{2\sqrt{\tan x}}
  3. Put t=tan⁡xt = \tan x: ∫1+t22t dt=∫(12t−1/2+12t3/2)dt\displaystyle\int \dfrac{1 + t^{2}}{2\sqrt{t}}\,dt = \int \left(\tfrac{1}{2}t^{-1/2} + \tfrac{1}{2}t^{3/2}\right)dt
  4. =t1/2+15t5/2+K= t^{1/2} + \tfrac{1}{5}t^{5/2} + K
  5. A=12A = \tfrac{1}{2}, B=52B = \tfrac{5}{2}, C=15C = \tfrac{1}{5}, so A+B+C=165A + B + C = \tfrac{16}{5}

Answer: 165\dfrac{16}{5}

Common mistakes to avoid

  • Forgetting to divide by the coefficient of x in a linear inside function; ∫ cos 3x dx = (sin 3x)/3 + C, not sin 3x + C.
  • Leaving out the 1/a in ∫ dx/(a² + x²); the result is (1/a) tan⁻¹(x/a) + C.
  • In the eˣ [f + f′] pattern, writing eˣ [f + f′] as the answer; the answer is only eˣ f(x) + C.
  • Choosing u badly in integration by parts, such as u = eˣ in ∫ x eˣ dx, which makes the integral harder; follow ILATE and take u = x.

How to prepare it

Look first for a function and its derivative in the integrand — f′(x)/f(x) and eˣ[f + f′] patterns are common.

Once the formulas feel familiar, test them the way the exam does: practise Integration questions chapter-wise, then attempt a full previous year paper to see whether it holds up under a 3-hour timer.

Related Mathematics chapters

See every chapter in the EAPCET Mathematics syllabus.

Practise Integration chapter-wise

Chapter-wise quizzes show exactly which topics in this chapter you still get wrong — before a full paper does.

Frequently asked questions

Is Integration part of the EAPCET Mathematics syllabus?
Yes. Integration is taught in Intermediate second year · Paper IIB Mathematics, and the EAPCET Engineering syllabus is based on the Intermediate first- and second-year syllabus. Check the official notification each year for any changes.
How should I prepare Integration for EAPCET?
Look first for a function and its derivative in the integrand — f′(x)/f(x) and eˣ[f + f′] patterns are common.
How many Mathematics questions are there in EAPCET?
80 of the 160 questions in the Engineering paper are Mathematics, spread across 37 chapters.
What mistakes do students make in Integration?
Forgetting to divide by the coefficient of x in a linear inside function; ∫ cos 3x dx = (sin 3x)/3 + C, not sin 3x + C. Leaving out the 1/a in ∫ dx/(a² + x²); the result is (1/a) tan⁻¹(x/a) + C. In the eˣ [f + f′] pattern, writing eˣ [f + f′] as the answer; the answer is only eˣ f(x) + C. Choosing u badly in integration by parts, such as u = eˣ in ∫ x eˣ dx, which makes the integral harder; follow ILATE and take u = x.