Foundations & Standard Forms

A function $F$ is an antiderivative of $f$ if $F'(x) = f(x)$. The indefinite integral $\int f(x)\,dx = F(x) + C$ denotes the whole family of antiderivatives, where $C$ is an arbitrary constant. Because $\dfrac{d}{dx}[F(x) + C] = f(x)$ for every constant $C$, the constant can never be recovered from $f$ alone — an extra condition (a boundary value) is needed to fix it.

The integrals you must recall instantly:

• $\int x^n\,dx = \dfrac{x^{n+1}}{n+1} + C$ for $n \neq -1$
• $\int \dfrac{1}{x}\,dx = \ln|x| + C$
• $\int e^x\,dx = e^x + C$, and $\int a^x\,dx = \dfrac{a^x}{\ln a} + C$
• $\int \sin x\,dx = -\cos x + C$, $\int \cos x\,dx = \sin x + C$
• $\int \sec^2 x\,dx = \tan x + C$, $\int \csc^2 x\,dx = -\cot x + C$
• $\int \sec x\tan x\,dx = \sec x + C$, $\int \tan x\,dx = \ln|\sec x| + C$
• $\int \dfrac{dx}{1+x^2} = \tan^{-1}x + C$, $\int \dfrac{dx}{\sqrt{1-x^2}} = \sin^{-1}x + C$

The second dozen — the rows the PYQs quote as "standard form" without proof:

• $\int \sec x\,dx = \ln|\sec x + \tan x| + C$, $\int \csc x\,dx = \ln|\csc x - \cot x| + C$, $\int \cot x\,dx = \ln|\sin x| + C$
• $\int \dfrac{dx}{x^2+a^2} = \dfrac1a\tan^{-1}\dfrac xa + C$, $\int \dfrac{dx}{\sqrt{a^2-x^2}} = \sin^{-1}\dfrac xa + C$, $\int \dfrac{dx}{x^2-a^2} = \dfrac{1}{2a}\ln\left|\dfrac{x-a}{x+a}\right| + C$
• $\int \dfrac{dx}{\sqrt{x^2+a^2}} = \ln\big|x+\sqrt{x^2+a^2}\big| + C$, $\int \dfrac{dx}{\sqrt{x^2-a^2}} = \ln\big|x+\sqrt{x^2-a^2}\big| + C$
• $\int \sqrt{x^2+a^2}\,dx = \dfrac x2\sqrt{x^2+a^2} + \dfrac{a^2}{2}\ln\big|x+\sqrt{x^2+a^2}\big| + C$; $\int \sqrt{x^2-a^2}\,dx = \dfrac x2\sqrt{x^2-a^2} - \dfrac{a^2}{2}\ln\big|x+\sqrt{x^2-a^2}\big| + C$; $\int \sqrt{a^2-x^2}\,dx = \dfrac x2\sqrt{a^2-x^2} + \dfrac{a^2}{2}\sin^{-1}\dfrac xa + C$
• And for every row, $\int f(ax+b)\,dx = \dfrac1a F(ax+b) + C$ — the $\tfrac1a$ is the most-forgotten factor in the chapter.

For constants $a, b$:

The collapsing identities the NDA tests most:

• $e^{\ln u} = u$ — the exponential and natural log undo each other, so $\int e^{\ln(\tan x)}\,dx = \int \tan x\,dx$.
• $\ln(u^k) = k\ln u$, so $e^{k\ln x} = x^k$ — a stacked log/exponent becomes a power.
• $\ln(ab)=\ln a+\ln b$ and $\ln\tfrac ab=\ln a-\ln b$ — so $\ln(10e)=1+\ln10$, and $\int\dfrac{dx}{x\ln(10e)}$ is $\dfrac{\ln|x|}{1+\ln10}$, a constant divisor, not a new function.
• A quotient like $\dfrac{P(x)}{x}$ splits into powers (linearity), and the $\dfrac{1}{x}$ term is exactly what forces a $\ln$ (or must vanish for a rational answer).

The base-$a$ rule:

Drive any $ax^2+bx+c$ into the form $(x-h)^2 + k^2$ by completing the square, then use:

The pattern:

The two standard results (each provable by parts — see Integration by Parts):

A function $g$ is periodic with period $p$ if $g(x+p) = g(x)$. Integrating a periodic function need NOT give a periodic result. Example: $\sin^2 x = \dfrac{1-\cos 2x}{2}$ is periodic, but $\int \sin^2 x\,dx = \dfrac{x}{2} - \dfrac{\sin 2x}{4} + C$ carries a $\dfrac{x}{2}$ term that grows without bound. So 'the integrand is periodic' is true, while 'the antiderivative is periodic' is false — two statements that look linked but are not. Two inverse relations underpin all of this: integrating a derivative returns the function (up to $+C$), and differentiating an integral returns the integrand.

Keep these collapses in reflex memory:

• Half-angle of $1\pm\cos x$: $1-\cos x = 2\sin^2\tfrac{x}{2}$, $1+\cos x = 2\cos^2\tfrac{x}{2}$; so $\dfrac{1-\cos x}{1+\cos x}=\tan^2\tfrac{x}{2}$, $\dfrac{1}{1-\cos x}=\tfrac12\csc^2\tfrac{x}{2}$, $\dfrac{1}{1+\cos x}=\tfrac12\sec^2\tfrac{x}{2}$.
• Power-reduction (double angle): $1-\cos 2x = 2\sin^2 x$, $1+\cos 2x = 2\cos^2 x$, $\sin 2x = 2\sin x\cos x$.
• Perfect square under a root: $1\pm\sin 2x=(\sin x\pm\cos x)^2$ and $1\pm\sin\theta=\left(\cos\tfrac\theta2\pm\sin\tfrac\theta2\right)^2$, so $\sqrt{1\pm\sin 2x}=|\sin x\pm\cos x|$ — keep the modulus; its sign depends on the interval.
• $\sec\pm\tan$: $\sec x+\tan x=\dfrac{1+\sin x}{\cos x}=\tan\!\left(\tfrac\pi4+\tfrac x2\right)$, $\sec x-\tan x=\tan\!\left(\tfrac\pi4-\tfrac x2\right)$.
• Harmonic form: $a\sin x+b\cos x=\sqrt{a^2+b^2}\,\sin(x+\alpha)$, so its extreme values are $\pm\sqrt{a^2+b^2}$.
• Weierstrass $t=\tan\tfrac{x}{2}$: $\sin x=\dfrac{2t}{1+t^2}$, $\cos x=\dfrac{1-t^2}{1+t^2}$, $dx=\dfrac{2\,dt}{1+t^2}$ — turns any rational function of $\sin x,\cos x$ into a rational function of $t$.
• The Pythagorean pair as named identities: $\sec^2x=1+\tan^2x$ and $\csc^2x=1+\cot^2x$ — they are what make $\int\sec^2x\,dx=\tan x$ and the divide-by-$\cos^2x$ move work, and $\int\tan^2x\,dx=\tan x-x$, $\int\cot^2x\,dx=-\cot x-x$ follow from them.
• Stripping an inverse: $\tan^{-1}(\tan\theta)=\theta$ only for $\theta\in(-\tfrac\pi2,\tfrac\pi2)$, so $\tan^{-1}(\sec x+\tan x)=\tan^{-1}\tan(\tfrac\pi4+\tfrac x2)=\tfrac\pi4+\tfrac x2$ on $-\tfrac{3\pi}{2}<x<\tfrac\pi2$, and its integral is $\tfrac{\pi x}{4}+\tfrac{x^2}{4}+C$.