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Concept 01 / 10Integral Calculus
Indefinite Integrals: Antiderivatives and Standard Formulas
Integration as the reverse of differentiation, why the arbitrary constant appears, linearity, the table of standard integrals, checking an answer by differentiating, and the choice of constant from an initial condition.
Figure 1
A function F is an antiderivative (primitive) of f on an interval if F′(x)=f(x) there. If F is one antiderivative, then every antiderivative has the form F(x)+C, because two functions with the same derivative on an interval differ by a constant (a consequence of the mean value theorem). This family is the indefinite integral:
∫f(x)dx=F(x)+C.
The figure shows the family for f(x)=x: the curves 2x2+C are vertical translates of one another, and at any fixed x they all have the same slope. An initial condition such as F(0)=2 picks out one member of the family.
The phrase 'on an interval' matters. ∫x1dx=ln∣x∣+C, but because the domain x=0 is two separate intervals, the most general antiderivative can use different constants on x>0 and x<0. Exam answers use a single C by convention.
Linearity: ∫[af(x)+bg(x)]dx=a∫f+b∫g. There is no product rule or quotient rule for integrals in the direct sense; products are handled by substitution or by parts.
Standard integrals (each is a derivative read backwards; add +C to every one):
∫xndx=n+1xn+1 for n=−1; ∫x1dx=ln∣x∣.
∫exdx=ex; ∫axdx=lnaax.
∫sinxdx=−cosx; ∫cosxdx=sinx.
∫sec2xdx=tanx; ∫csc2xdx=−cotx.
∫secxtanxdx=secx; ∫cscxcotxdx=−cscx.
∫tanxdx=ln∣secx∣; ∫cotxdx=ln∣sinx∣.
∫secxdx=ln∣secx+tanx∣; ∫cscxdx=ln∣cscx−cotx∣=ln∣tan2x∣.
∫a2−x2dx=sin−1ax; ∫a2+x2dx=a1tan−1ax; ∫xx2−a2dx=a1sec−1ax.
Linear substitution rule: if ∫f(x)dx=F(x)+C, then ∫f(ax+b)dx=a1F(ax+b)+C. For example, ∫cos(3x+1)dx=31sin(3x+1)+C.
Preparing the integrand. Many integrals become standard after algebra or trigonometric identities: expand products, divide polynomials, split fractions, or use identities such as sin2x=21−cos2x, tan2x=sec2x−1, and product-to-sum formulas (2sinAcosB=sin(A+B)+sin(A−B)). For example, ∫1+x2x2dx=∫(1−1+x21)dx=x−tan−1x+C.
Checking. Every indefinite integral can be verified by differentiating the answer. Two correct answers can look different: ∫2sinxcosxdx can be written as sin2x+C, −cos2x+C or −21cos2x+C, since these differ only by constants.
Key Takeaways:
- ∫f=F+C where F′=f; all antiderivatives differ by constants.
- ∫xn=n+1xn+1 (n=−1); ∫x1=ln∣x∣.
- ∫tanx=ln∣secx∣; ∫secx=ln∣secx+tanx∣.
- ∫a2+x2dx=a1tan−1ax; ∫a2−x2dx=sin−1ax.
- ∫f(ax+b)dx=a1F(ax+b).
- Simplify with identities before integrating; check by differentiating.
- Pitfall: writing ∫x1=lnx, which is wrong for negative x.
Brain-Teaser 1: Evaluate ∫sin2xcos2xdx.
Solution: Write 1=sin2x+cos2x in the numerator: sin2xcos2xsin2x+cos2x=sec2x+csc2x. So the integral is tanx−cotx+C. The '1=sin2+cos2' trick splits a single awkward fraction into two standard ones. An equivalent form is −2cot2x+C.
Brain-Teaser 2: Find f if f′(x)=3x2−x2 and f(1)=4, for x>0.
Solution: f(x)=x3−2lnx+C. The condition f(1)=1−0+C=4 gives C=3. So f(x)=x3−2lnx+3. The initial value fixes the one free constant; for x<0 a separate condition would be needed, since the domain has two pieces.
Brain-Teaser 3: Evaluate ∫1+sin2xdx for 0<x<2π.
Solution: 1+sin2x=sin2x+cos2x+2sinxcosx=(sinx+cosx)2. On (0,2π), sinx+cosx>0, so the square root is sinx+cosx, and the integral is −cosx+sinx+C. On an interval where sinx+cosx<0, the answer would change sign, because u2=∣u∣. Recognising a perfect square under a root is common in JEE, and so is the need to track its sign.
