Indefinite Integration

Indefinite integration is fundamentally defined as the inverse operation of differentiation. If $f(x)$ is a given real-valued function and there exists a function $F(x)$ such that its derivative identically satisfies $F'(x) = f(x)$ for all $x$ within a specified domain, then $F(x)$ is called an anti-derivative or a primitive of $f(x)$. Since the derivative of any real constant $C$ vanishes ($\frac{d}{dx}[C] = 0$), the absolute generalized anti-derivative map is represented as an infinite family of curves: \[ \int f(x) \, dx = F(x) + C \] where $C \in \mathbb{R}$ is the arbitrary constant of integration.

A basic set of standard algebraic and transcendental integrals must be committed to memory. These form the baseline elements for complex multi-tier integral reductions: \[ \int x^n \, dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1), \quad \int \frac{1}{x} \, dx = \ln|x| + C \] \[ \int e^x \, dx = e^x + C, \quad \int a^x \, dx = \frac{a^x}{\ln a} + C \quad (a > 0, a \neq 1) \] \[ \int \sin x \, dx = -\cos x + C, \quad \int \cos x \, dx = \sin x + C \] \[ \int \sec^2 x \, dx = \tan x + C, \quad \int \csc^2 x \, dx = -\cot x + C \] \[ \int \sec x \tan x \, dx = \sec x + C, \quad \int \csc x \cot x \, dx = -\csc x + C \]

Advanced derivatives of trigonometric parent metrics yield the log-transcendental standard extensions: \[ \int \tan x \, dx = \ln|\sec x| + C = -\ln|\cos x| + C \] \[ \int \cot x \, dx = \ln|\sin x| + C \] \[ \int \sec x \, dx = \ln|\sec x + \tan x| + C = \ln\left|\tan\left(\frac{x}{2} + \frac{\pi}{4}\right)\right| + C \] \[ \int \csc x \, dx = \ln|\csc x - \cot x| + C = \ln\left|\tan\left(\frac{x}{2}\right)\right| + C \]

Standard algebraic expressions containing quadratic radical variables under denominators cannot be solved using simple polynomial integration. Instead, they require inverse trigonometric and log-radical primitive formulas. Memorizing these standard forms is essential for solving more advanced rational quadratics: \[ \int \frac{1}{\sqrt{a^2 - x^2}} \, dx = \sin^{-1}\left(\frac{x}{a}\right) + C \] \[ \int \frac{1}{a^2 + x^2} \, dx = \frac{1}{a} \tan^{-1}\left(\frac{x}{a}\right) + C \] \[ \int \frac{1}{x^2 - a^2} \, dx = \frac{1}{2a} \ln\left|\frac{x - a}{x + a}\right| + C \] \[ \int \frac{1}{\sqrt{x^2 + a^2}} \, dx = \ln\left|x + \sqrt{x^2 + a^2}\right| + C = \sinh^{-1}\left(\frac{x}{a}\right) + C \] \[ \int \frac{1}{\sqrt{x^2 - a^2}} \, dx = \ln\left|x + \sqrt{x^2 - a^2}\right| + C = \cosh^{-1}\left(\frac{x}{a}\right) + C \]

When evaluating these integrals in competitive exams, look out for the sign of the variable and constant terms. A mismatch can lead to a completely incorrect function type (e.g., mistaking an inverse sine form for a logarithmic radical form).

When an integrand cannot be integrated directly using standard forms, you can often transform it by changing the variable of integration. The Method of Substitution simplifies an integral by changing the variable $x$ to a new variable $u$ using a transformation function $x = g(u)$. By applying the chain rule to this transformation, the differential term becomes $dx = g'(u) \, du$. This maps the entire expression into a simpler form: \[ \int f(g(x)) \cdot g'(x) \, dx = \int f(u) \, du \] where $u = g(x)$ and $du = g'(x) \, dx$.

• Substitute $u = \ln x \implies du = \frac{1}{x} \, dx$
• Substitute $u = e^x \implies du = e^x \, dx$
• Substitute $u = \tan^{-1} x \implies du = \frac{1}{1+x^2} \, dx$
• Substitute $u = a x^2 + b x + c \implies du = (2ax + b) \, dx$

• For $\sqrt{a^2 - x^2}$, substitute $x = a\sin\theta \implies dx = a\cos\theta \, d\theta$. This simplifies the radical using the identity $\sqrt{a^2 - a^2\sin^2\theta} = a\cos\theta$.
• For $\sqrt{a^2 + x^2}$, substitute $x = a\tan\theta \implies dx = a\sec^2\theta \, d\theta$. This simplifies the radical using the identity $\sqrt{a^2 + a^2\tan^2\theta} = a\sec\theta$.
• For $\sqrt{x^2 - a^2}$, substitute $x = a\sec\theta \implies dx = a\sec\theta\tan\theta \, d\theta$. This simplifies the radical using the identity $\sqrt{a^2\sec^2\theta - a^2} = a\tan\theta$.

For rational fractions containing sine and cosine terms in the denominator, like $\int \frac{1}{a + b\cos x + c\sin x} \, dx$, use the Weierstrass substitution. This technique uses the half-angle tangent substitution $t = \tan\left(\frac{x}{2}\right)$ to transform trigonometric rational functions into standard algebraic fractions. The corresponding trigonometric terms change as follows: \[ \sin x = \frac{2t}{1+t^2}, \quad \cos x = \frac{1-t^2}{1+t^2}, \quad dx = \frac{2}{1+t^2} \, dt \]

Integration by Parts (IBP) is derived by reversing the product rule for differentiation. For two differentiable functions $u(x)$ and $v(x)$, the product rule states that $\frac{d}{dx}[uv] = u \frac{dv}{dx} + v \frac{du}{dx}$. Integrating both sides of this equation yields the standard Integration by Parts formula: \[ \int u \, dv = uv - \int v \, du \] Alternatively, written in terms of functions $f(x)$ and $g(x)$: \[ \int f(x) g(x) \, dx = f(x) \int g(x) \, dx - \int \left[ f'(x) \int g(x) \, dx \right] \, dx \]

• I: Inverse Trigonometric functions (e.g., $\sin^{-1} x, \tan^{-1} x$)
• L: Logarithmic functions (e.g., $\ln x, \log_a x$)
• A: Algebraic functions (e.g., $x^2, 3x, x^n$)
• T: Trigonometric functions (e.g., $\sin x, \cos x$)
• E: Exponential functions (e.g., $e^x, 2^x$)

A key shortcut form often tested in JEE Advanced involves the exponential product structure: \[ \int e^x [f(x) + f'(x)] \, dx = e^x f(x) + C \] This structure resolves instantly by distributing the integral and applying IBP to the $\int e^x f(x) \, dx$ component.

When integrands contain high integer powers of transcendental functions, like $\int \sin^n x \, dx$ or $\int \tan^n x \, dx$, evaluating them directly can be difficult. Instead, we use Reduction Formulae to express an integral containing a power $n$ in terms of a lower-power version of the same integral structure (such as $n-2$ or $n-1$). These formulas are derived by applying Integration by Parts or using standard trigonometric identities.

Let us analyze the derivation for $I_n = \int \sin^n x \, dx$. Split the integrand into two factors: \[ I_n = \int \sin^{n-1} x \cdot \sin x \, dx \] Apply Integration by Parts, choosing $u = \sin^{n-1} x$ and $dv = \sin x \, dx$: \[ I_n = -\sin^{n-1} x \cos x - \int \left( (n-1)\sin^{n-2} x \cos x \cdot (-\cos x) \right) \, dx \] \[ I_n = -\sin^{n-1} x \cos x + (n-1)\int \sin^{n-2} x \cos^2 x \, dx \] Substitute the identity $\cos^2 x = 1 - \sin^2 x$ into the remaining integral term: \[ I_n = -\sin^{n-1} x \cos x + (n-1)\int \sin^{n-2} x (1 - \sin^2 x) \, dx \] \[ I_n = -\sin^{n-1} x \cos x + (n-1)I_{n-2} - (n-1)I_n \] Collect all $I_n$ terms on the left side of the equation and solve for $I_n$ to find the final reduction formula: \[ I_n(1 + n - 1) = -\sin^{n-1} x \cos x + (n-1)I_{n-2} \implies I_n = -\frac{\sin^{n-1} x \cos x}{n} + \frac{n-1}{n} I_{n-2} \]

A rational function is defined as a ratio of two polynomials, $R(x) = \frac{P(x)}{Q(x)}$. If the degree of the numerator polynomial $P(x)$ is strictly less than the degree of the denominator polynomial $Q(x)$, $R(x)$ is a proper rational function. If the degree of $P(x)$ is greater than or equal to the degree of $Q(x)$, it is an improper rational function. Improper rational functions must first be transformed using polynomial long division: \[ \frac{P(x)}{Q(x)} = S(x) + \frac{T(x)}{Q(x)} \] where $S(x)$ is the polynomial quotient and $\frac{T(x)}{Q(x)}$ is a proper rational remainder fraction.

• Distinct Linear Factors: If $Q(x) = (x-a)(x-b)$, set up the decomposition as: \[ \frac{T(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b} \]
• Repeated Linear Factors: If $Q(x) = (x-a)^2(x-b)$, include a partial fraction term for every power of the repeated factor up to its multiplicity: \[ \frac{T(x)}{(x-a)^2(x-b)} = \frac{A}{x-a} + \frac{B}{(x-a)^2} + \frac{C}{x-b} \]
• Irreducible Quadratic Factors: If $Q(x) = (x-a)(x^2 + bx + c)$ where the quadratic term has no real roots ($b^2 - 4ac < 0$), use a linear numerator expression above the quadratic factor: \[ \frac{T(x)}{(x-a)(x^2 + bx + c)} = \frac{A}{x-a} + \frac{Bx + C}{x^2 + bx + c} \]

Two standard quadratic forms often appear in competitive exam problems: \[ \text{Type I: } \int \frac{px+q}{ax^2+bx+c} \, dx \quad \text{and} \quad \text{Type II: } \int \frac{px+q}{\sqrt{ax^2+bx+c}} \, dx \] To evaluate these forms, split the linear numerator expression into two parts: a scalar multiple of the derivative of the quadratic denominator, plus a remaining constant term: \[ px + q = \lambda \cdot \frac{d}{dx}(ax^2 + bx + c) + \mu = \lambda(2ax + b) + \mu \] Equate the linear coefficients to solve for the constants $\lambda$ and $\mu$: \[ \lambda = \frac{p}{2a}, \quad \mu = q - \frac{pb}{2a} \] This splits the original integral into two separate, simpler integrals that can be evaluated independently using standard forms: \[ \int \frac{px+q}{ax^2+bx+c} \, dx = \lambda \int \frac{2ax+b}{ax^2+bx+c} \, dx + \mu \int \frac{1}{ax^2+bx+c} \, dx \] The first integral can be solved directly using substitution, yielding a log term. The second integral can be resolved by completing the square in the denominator to match inverse trigonometric or log-radical standard forms.

Another core integration form involves rational trigonometric transformations: \[ \text{Type III: } \int \frac{a\cos x + b\sin x}{c\cos x + d\sin x} \, dx \] To solve this form, split the numerator expression into a linear combination of the denominator and the derivative of the denominator: \[ \text{Numerator} = \lambda(\text{Denominator}) + \mu(\text{Denominator}') \] \[ a\cos x + b\sin x = \lambda(c\cos x + d\sin x) + \mu(-c\sin x + d\cos x) \] Equate the sine and cosine coefficients to solve for the constants $\lambda$ and $\mu$. This splits the integral into two components: a constant integral term and a substitution integral term that yields a natural log.