Definite Integration — MHT-CET Maths

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

Formulas (4)

• Divide by cos^n x and Put tan x = t · The tan substitution
  $$t = \tan x,\quad dt = \sec^2x\,dx,\qquad \sin 2x = \frac{2t}{1 + t^2},\qquad x:\ 0,\tfrac{\pi}{6},\tfrac{\pi}{4},\tfrac{\pi}{3} \ \to\ t:\ 0,\tfrac{1}{\sqrt3},1,\sqrt3$$
• Half-Angle Forms: 1 + cos x and the a + b cos x Standard Result · Half-angle results
  $$\int\frac{dx}{1 + \cos x} = \tan\frac{x}{2} \qquad \int_0^\pi\frac{dx}{a + b\cos x} = \frac{\pi}{\sqrt{a^2 - b^2}} \ (a > |b|) \qquad \tan\frac{\pi}{8} = \sqrt2 - 1$$
• Spot the Derivative Pair: csc x cot x, sin x with 1 − cos x · Derivative pairs
  $$d(\csc x) = -\csc x\cot x\,dx \qquad d(1 - \cos x) = \sin x\,dx \qquad \tan^{-1}A - \tan^{-1}B = \tan^{-1}\frac{A - B}{1 + AB}$$
• √tan x + √cot x: the sin x − cos x Substitution · The sin x − cos x substitution
  $$t = \sin x - \cos x \ \Rightarrow\ dt = (\cos x + \sin x)\,dx,\quad \sin x\cos x = \frac{1 - t^2}{2}$$

Watch out for (5)

• The paper's own typo: sen^{2/3}
  → Divide by cos^n x and Put tan x = t
• A positive integrand cannot give a negative answer
  → Half-Angle Forms: 1 + cos x and the a + b cos x Standard Result
• π/7 versus π/√7
  → Half-Angle Forms: 1 + cos x and the a + b cos x Standard Result
• Reading the negative of the integral off the option list
  → Spot the Derivative Pair: csc x cot x, sin x with 1 − cos x
• Rationalising √tan + √cot term by term
  → √tan x + √cot x: the sin x − cos x Substitution

Formulas (4)

• The Odd/Even Test on Symmetric Limits · Symmetric-limit rule
  $$\int_{-a}^{a} f(x)\,dx = \begin{cases} 0, & f(-x) = -f(x) \\[2pt] 2\displaystyle\int_0^a f(x)\,dx, & f(-x) = f(x) \end{cases}$$
• Split a Mixed Integrand into Its Odd and Even Parts · Odd part vanishes
  $$\int_{-a}^{a}\big[\text{odd}(x) + \text{even}(x)\big]dx = 2\int_0^a \text{even}(x)\,dx \qquad \int_0^{\pi/2}x^2\cos x\,dx = \frac{\pi^2}{4} - 2$$
• Symmetry Under x → 1/x on [1/2, 2] · The reciprocal symmetry
  $$x = \tfrac1t:\quad \int_{1/a}^{a}\frac{1}{x}\,g\!\left(x - \frac1x\right)dx = -\int_{1/a}^{a}\frac{1}{t}\,g\!\left(t - \frac1t\right)dt \ \text{ for odd } g \ \Rightarrow\ I = 0$$
• Even Does Not Mean Convergent: the csc⁴x Trap · Check the domain first
  $$\int_{-\pi/4}^{\pi/4}\csc^4x\,dx \text{ diverges}; \quad \left[-\cot x - \tfrac{\cot^3x}{3}\right]_{-\pi/4}^{\pi/4} = -\tfrac83 \text{ is the exam's formal key}$$

Watch out for (4)

• Testing the limits instead of the function
  → The Odd/Even Test on Symmetric Limits
• Discarding the constant with the odd terms
  → Split a Mixed Integrand into Its Odd and Even Parts
• Applying it without the 1/x
  → Symmetry Under x → 1/x on [1/2, 2]
• Marking the mathematically honest option
  → Even Does Not Mean Convergent: the csc⁴x Trap

Formulas (6)

• King's Property: ∫ f(x) = ∫ f(a + b − x) · King's property
  $$\int_a^b f(x)\,dx = \int_a^b f(a + b - x)\,dx \qquad \int_0^{\pi/4}\log(1 + \tan x)\,dx = \frac{\pi}{8}\log 2$$
• The f/(f + g) Family: ∫ f(x)/(f(x) + f(a + b − x)) = (b − a)/2 · The f/(f + g) result
  $$\int_a^b\frac{f(x)}{f(x) + f(a + b - x)}\,dx = \frac{b - a}{2} \qquad \int_0^{\pi/2}\frac{a\sin x + b\cos x}{\sin x + \cos x}\,dx = \frac{\pi}{4}(a + b)$$
• The x·f(sin x) Trick on [0, π]: Pull the x Out as π/2 · Pulling x out
  $$\int_0^\pi x\,f(\sin x)\,dx = \frac{\pi}{2}\int_0^\pi f(\sin x)\,dx \qquad \int_0^\pi\frac{\sin x}{1 + \cos^2x}\,dx = \frac{\pi}{2}$$
• Functional Symmetry Given in the Stem: f(x) = f(1 − x), g(x) + g(a − x) = 4 · Symmetry handed to you
  $$f(a + b - x) = f(x) \ \Rightarrow\ \int_a^b x f(x)\,dx = \frac{a + b}{2}\int_a^b f(x)\,dx$$
• Integrands with 1/(1 + aˣ) over Symmetric Limits · The 1/(1 + aˣ) cancellation
  $$\frac{1}{1 + a^{x}} + \frac{1}{1 + a^{-x}} = 1 \qquad f \text{ even}:\ \int_{-a}^{a}\frac{f(x)}{1 + e^{x}}\,dx = \int_0^a f(x)\,dx$$
• An Inverse-Trig Identity Before the Reflection · Unpack, then reflect
  $$\cot^{-1}(1 - x + x^2) = \tan^{-1}x + \tan^{-1}(1 - x) \qquad \int_0^1\tan^{-1}(1 - x + x^2)\,dx = \log 2$$

Watch out for (6)

• Reflecting about the wrong point
  → King's Property: ∫ f(x) = ∫ f(a + b − x)
• Missing the disguised reflection in the denominator
  → The f/(f + g) Family: ∫ f(x)/(f(x) + f(a + b − x)) = (b − a)/2
• Using the trick with cos x
  → The x·f(sin x) Trick on [0, π]: Pull the x Out as π/2
• Treating R₂ as an integral of x f(x)
  → Functional Symmetry Given in the Stem: f(x) = f(1 − x), g(x) + g(a − x) = 4
• Forgetting the halving
  → Integrands with 1/(1 + aˣ) over Symmetric Limits
• Integrating tan⁻¹(1 − x + x²) directly
  → An Inverse-Trig Identity Before the Reflection

Formulas (3)

• Modulus: Split Where the Inside Changes Sign · Modulus splitting
  $$\int_a^b|u(x)|\,dx = \int_a^c\pm u\,dx + \int_c^b\mp u\,dx \ \text{ where } u(c) = 0,\ \text{sign chosen to make each piece } \ge 0$$
• Greatest Integer: Piecewise Constant, So Sum the Pieces · Integrating a staircase
  $$\int_a^b[x]\,g(x)\,dx = \sum_{n}\; n\int_{\max(a,n)}^{\min(b,n+1)} g(x)\,dx \qquad \int_0^2[x^2]\,dx = 5 - \sqrt2 - \sqrt3$$
• Piecewise Constant by an Identity: tan⁻¹u + tan⁻¹(1/u) · The reciprocal arctangent identity
  $$\tan^{-1}u + \tan^{-1}\frac{1}{u} = \begin{cases} \dfrac{\pi}{2}, & u > 0 \\[4pt] -\dfrac{\pi}{2}, & u < 0 \end{cases}$$

Watch out for (3)

• Integrating the bare expression
  → Modulus: Split Where the Inside Changes Sign
• Using the endpoint's floor for the whole last piece
  → Greatest Integer: Piecewise Constant, So Sum the Pieces
• The key that ignores the sign of x
  → Piecewise Constant by an Identity: tan⁻¹u + tan⁻¹(1/u)