JEE Integration Cheatsheet

Cheatsheet Content

1. Concept of Integration Integration is the reverse process of differentiation. It finds the accumulated quantity or area under a curve. 2. Essential Formulas & Standard Integrals Integral Result $\int x^n dx$ $\frac{x^{n+1}}{n+1} + C \quad (n \ne -1)$ $\int \frac{1}{x} dx$ $\ln|x| + C$ $\int e^x dx$ $e^x + C$ $\int a^x dx$ $\frac{a^x}{\ln a} + C$ $\int \sin x dx$ $-\cos x + C$ $\int \cos x dx$ $\sin x + C$ $\int \sec^2 x dx$ $\tan x + C$ $\int \csc^2 x dx$ $-\cot x + C$ $\int \sec x \tan x dx$ $\sec x + C$ $\int \csc x \cot x dx$ $-\csc x + C$ $\int \tan x dx$ $\ln|\sec x| + C$ $\int \cot x dx$ $\ln|\sin x| + C$ $\int \sec x dx$ $\ln|\sec x + \tan x| + C$ $\int \csc x dx$ $\ln|\csc x - \cot x| + C$ $\int \frac{1}{\sqrt{a^2-x^2}} dx$ $\sin^{-1}\left(\frac{x}{a}\right) + C$ $\int \frac{1}{a^2+x^2} dx$ $\frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C$ $\int \frac{1}{x\sqrt{x^2-a^2}} dx$ $\frac{1}{a}\sec^{-1}\left(\frac{x}{a}\right) + C$ $\int \frac{1}{x^2-a^2} dx$ $\frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right| + C$ $\int \frac{1}{a^2-x^2} dx$ $\frac{1}{2a}\ln\left|\frac{a+x}{a-x}\right| + C$ $\int \frac{1}{\sqrt{x^2 \pm a^2}} dx$ $\ln|x+\sqrt{x^2 \pm a^2}| + C$ 3. Techniques of Integration Substitution Rule: If $g(x)$ is differentiable, then $\int f(g(x))g'(x)dx = \int f(u)du$ by letting $u=g(x)$. Example: $\int 2x \cos(x^2) dx$. Let $u=x^2$, $du=2x dx$. Integral becomes $\int \cos u du = \sin u + C = \sin(x^2) + C$. Integration by Parts Rule: $\int u dv = uv - \int v du$. Use LIATE (Log, Inverse Trig, Algebraic, Trig, Exp) for choosing $u$. Example: $\int x e^x dx$. Let $u=x, dv=e^x dx \implies du=dx, v=e^x$. Result: $xe^x - \int e^x dx = xe^x - e^x + C$. Partial Fractions Rule: Decompose rational functions $\frac{P(x)}{Q(x)}$ into simpler fractions if degree of $P(x)$ Example: $\int \frac{1}{(x+1)(x-2)} dx = \int \left(\frac{A}{x+1} + \frac{B}{x-2}\right) dx$. Find $A, B$. Trigonometric Integrals & Identities Rule: Use identities to simplify powers or products of trig functions. E.g., $\sin^2 x = \frac{1-\cos(2x)}{2}$. Example: $\int \sin^2 x dx = \int \frac{1-\cos(2x)}{2} dx = \frac{1}{2}x - \frac{1}{4}\sin(2x) + C$. Trigonometric Substitution Rule: For $\sqrt{a^2-x^2}$ use $x=a\sin\theta$; for $\sqrt{a^2+x^2}$ use $x=a\tan\theta$; for $\sqrt{x^2-a^2}$ use $x=a\sec\theta$. Example: $\int \frac{1}{\sqrt{1-x^2}} dx$. Let $x=\sin\theta, dx=\cos\theta d\theta$. Integral is $\int \frac{\cos\theta}{\sqrt{1-\sin^2\theta}} d\theta = \int d\theta = \theta + C = \sin^{-1}x + C$. Rationalizing Substitutions Rule: Weierstrass substitution $t=\tan(x/2)$ for rational functions of $\sin x, \cos x$. Then $\sin x = \frac{2t}{1+t^2}, \cos x = \frac{1-t^2}{1+t^2}, dx = \frac{2dt}{1+t^2}$. Example: $\int \frac{1}{1+\sin x} dx$. Use $t=\tan(x/2)$ to convert to rational function of $t$. Reduction Formulas Rule: Formulas to express an integral in terms of a similar integral with a lower power or index. Example: $\int \sin^n x dx = -\frac{\sin^{n-1}x \cos x}{n} + \frac{n-1}{n}\int \sin^{n-2}x dx$. Improper Integrals Rule: Integrals with infinite limits or discontinuous integrands. Evaluate using limits. Example: $\int_0^\infty e^{-x} dx = \lim_{b\to\infty} \int_0^b e^{-x} dx = \lim_{b\to\infty} [-e^{-x}]_0^b = \lim_{b\to\infty} (-e^{-b} + e^0) = 0+1=1$. 4. Key Shortcuts & Common Transforms $\int e^x (f(x) + f'(x)) dx = e^x f(x) + C$. $\int \frac{f'(x)}{f(x)} dx = \ln|f(x)| + C$. $\int \frac{1}{ax+b} dx = \frac{1}{a}\ln|ax+b| + C$. $\int \frac{\sin x}{\cos x} dx = -\ln|\cos x| + C$. $\int \frac{dx}{x^2+a^2} = \frac{1}{a}\tan^{-1}(\frac{x}{a}) + C$. 5. Worked Examples Mains Examples (Mains) $\int \frac{e^x(1+x)}{\cos^2(xe^x)} dx$ Insight: Recognize $xe^x$ as a function whose derivative is $e^x(1+x)$. Let $t = xe^x$. Then $dt = (e^x \cdot 1 + x \cdot e^x) dx = e^x(1+x) dx$. The integral becomes $\int \frac{dt}{\cos^2 t} = \int \sec^2 t dt = \tan t + C$. Substitute back: $\boxed{\tan(xe^x) + C}$. (Mains) $\int \frac{\sin^6 x + \cos^6 x}{\sin^2 x \cos^2 x} dx$ Insight: Simplify the numerator using $a^3+b^3=(a+b)(a^2-ab+b^2)$. Numerator: $(\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x) = 1 \cdot (( \sin^2 x + \cos^2 x)^2 - 3\sin^2 x \cos^2 x) = 1 - 3\sin^2 x \cos^2 x$. Integral becomes $\int \frac{1 - 3\sin^2 x \cos^2 x}{\sin^2 x \cos^2 x} dx$. Split: $\int \left(\frac{1}{\sin^2 x \cos^2 x} - 3\right) dx = \int \left(\frac{\sin^2 x + \cos^2 x}{\sin^2 x \cos^2 x} - 3\right) dx$. Simplify: $\int (\sec^2 x + \csc^2 x - 3) dx = \tan x - \cot x - 3x + C$. Result: $\boxed{\tan x - \cot x - 3x + C}$. (Mains) $\int \frac{dx}{\sqrt{9x^2-12x+8}}$ Insight: Complete the square in the denominator. $9x^2-12x+8 = (3x)^2 - 2(3x)(2) + 2^2 - 2^2 + 8 = (3x-2)^2 + 4$. Let $u = 3x-2$, $du = 3dx \implies dx = \frac{1}{3}du$. Integral is $\int \frac{1}{\sqrt{u^2+4}} \frac{1}{3}du = \frac{1}{3} \int \frac{1}{\sqrt{u^2+2^2}} du$. Using $\int \frac{1}{\sqrt{x^2+a^2}} dx = \ln|x+\sqrt{x^2+a^2}| + C$: $\frac{1}{3} \ln|u+\sqrt{u^2+4}| + C$. Substitute back: $\boxed{\frac{1}{3} \ln|3x-2+\sqrt{9x^2-12x+8}| + C}$. (Mains) $\int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$ Insight: Use King's Rule: $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$. Let $I = \int_0^{\pi/2} \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$. By King's Rule, $I = \int_0^{\pi/2} \frac{\sqrt{\sin(\pi/2-x)}}{\sqrt{\sin(\pi/2-x)} + \sqrt{\cos(\pi/2-x)}} dx = \int_0^{\pi/2} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$. Add the two expressions for $I$: $2I = \int_0^{\pi/2} \frac{\sqrt{\sin x} + \sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx = \int_0^{\pi/2} 1 dx = [x]_0^{\pi/2} = \frac{\pi}{2}$. So, $2I = \frac{\pi}{2} \implies I = \frac{\pi}{4}$. Result: $\boxed{\frac{\pi}{4}}$. (Mains) $\int \frac{dx}{x(x^n+1)}$ Insight: Multiply numerator and denominator by $x^{n-1}$ to facilitate substitution. $\int \frac{x^{n-1}}{x^n(x^n+1)} dx$. Let $t=x^n$, then $dt = nx^{n-1}dx \implies x^{n-1}dx = \frac{1}{n}dt$. Integral becomes $\frac{1}{n} \int \frac{dt}{t(t+1)} = \frac{1}{n} \int \left(\frac{1}{t} - \frac{1}{t+1}\right) dt$. Integrate: $\frac{1}{n} (\ln|t| - \ln|t+1|) + C = \frac{1}{n} \ln\left|\frac{t}{t+1}\right| + C$. Substitute back: $\boxed{\frac{1}{n} \ln\left|\frac{x^n}{x^n+1}\right| + C}$. Advanced Examples (Advanced) $\int \frac{x^2-1}{x^4+1} dx$ Insight: Divide numerator and denominator by $x^2$. Then use substitution. Divide by $x^2$: $\int \frac{1-1/x^2}{x^2+1/x^2} dx$. Denominator: $x^2+1/x^2 = (x+1/x)^2 - 2$ or $(x-1/x)^2 + 2$. Since numerator is $1-1/x^2$, let $t = x+1/x$. Then $dt = (1-1/x^2) dx$. The integral becomes $\int \frac{dt}{t^2-2} = \int \frac{dt}{t^2-(\sqrt{2})^2}$. Using $\int \frac{1}{x^2-a^2} dx = \frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right| + C$: $\frac{1}{2\sqrt{2}}\ln\left|\frac{t-\sqrt{2}}{t+\sqrt{2}}\right| + C$. Substitute back: $\boxed{\frac{1}{2\sqrt{2}}\ln\left|\frac{x+1/x-\sqrt{2}}{x+1/x+\sqrt{2}}\right| + C = \frac{1}{2\sqrt{2}}\ln\left|\frac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}\right| + C}$. (Advanced) If $I_n = \int_0^{\pi/4} \tan^n x dx$, prove $I_n + I_{n-2} = \frac{1}{n-1}$. Insight: Split $\tan^n x$ as $\tan^{n-2} x \tan^2 x$ and use identity $\tan^2 x = \sec^2 x - 1$. $I_n = \int_0^{\pi/4} \tan^{n-2} x \tan^2 x dx = \int_0^{\pi/4} \tan^{n-2} x (\sec^2 x - 1) dx$. $I_n = \int_0^{\pi/4} \tan^{n-2} x \sec^2 x dx - \int_0^{\pi/4} \tan^{n-2} x dx$. The second term is $I_{n-2}$. So $I_n + I_{n-2} = \int_0^{\pi/4} \tan^{n-2} x \sec^2 x dx$. For the remaining integral, let $u = \tan x$, $du = \sec^2 x dx$. Limits $x=0 \implies u=0$, $x=\pi/4 \implies u=1$. So $I_n + I_{n-2} = \int_0^1 u^{n-2} du = \left[\frac{u^{n-1}}{n-1}\right]_0^1 = \frac{1^{n-1}}{n-1} - \frac{0^{n-1}}{n-1} = \frac{1}{n-1}$. (for $n>1$) Result: $\boxed{I_n + I_{n-2} = \frac{1}{n-1}}$. (Advanced) $\int_0^1 \ln x dx$ Insight: This is an improper integral. Evaluate using limits and integration by parts. $\int_0^1 \ln x dx = \lim_{\epsilon \to 0^+} \int_\epsilon^1 \ln x dx$. Use integration by parts for $\int \ln x dx$: Let $u=\ln x, dv=dx \implies du=\frac{1}{x}dx, v=x$. $\int \ln x dx = x \ln x - \int x \cdot \frac{1}{x} dx = x \ln x - x$. Evaluate definite integral: $[x \ln x - x]_\epsilon^1 = (1 \ln 1 - 1) - (\epsilon \ln \epsilon - \epsilon) = (0-1) - (\epsilon \ln \epsilon - \epsilon)$. Now, $\lim_{\epsilon \to 0^+} (-1 - \epsilon \ln \epsilon + \epsilon)$. We know $\lim_{\epsilon \to 0^+} \epsilon \ln \epsilon = 0$. So, $-1 - 0 + 0 = -1$. Result: $\boxed{-1}$. (Advanced) $\int \frac{dx}{(x^2+a^2)^{3/2}}$ Insight: Use trigonometric substitution $x=a\tan\theta$. Let $x = a\tan\theta$, then $dx = a\sec^2\theta d\theta$. $x^2+a^2 = a^2\tan^2\theta + a^2 = a^2(\tan^2\theta+1) = a^2\sec^2\theta$. $(x^2+a^2)^{3/2} = (a^2\sec^2\theta)^{3/2} = a^3\sec^3\theta$. Integral becomes $\int \frac{a\sec^2\theta d\theta}{a^3\sec^3\theta} = \frac{1}{a^2} \int \frac{1}{\sec\theta} d\theta = \frac{1}{a^2} \int \cos\theta d\theta$. Integrate: $\frac{1}{a^2} \sin\theta + C$. From $x=a\tan\theta$, $\tan\theta = x/a$. Draw a right triangle: opposite $x$, adjacent $a$, hypotenuse $\sqrt{x^2+a^2}$. So $\sin\theta = \frac{x}{\sqrt{x^2+a^2}}$. Result: $\boxed{\frac{x}{a^2\sqrt{x^2+a^2}} + C}$. (Advanced) $\int_0^{\pi/2} \ln(\sin x) dx$ Insight: Use King's Rule and properties of logarithms. Let $I = \int_0^{\pi/2} \ln(\sin x) dx$. By King's Rule, $I = \int_0^{\pi/2} \ln(\sin(\pi/2-x)) dx = \int_0^{\pi/2} \ln(\cos x) dx$. Add the two expressions for $I$: $2I = \int_0^{\pi/2} (\ln(\sin x) + \ln(\cos x)) dx = \int_0^{\pi/2} \ln(\sin x \cos x) dx$. $2I = \int_0^{\pi/2} \ln\left(\frac{2\sin x \cos x}{2}\right) dx = \int_0^{\pi/2} \ln\left(\frac{\sin(2x)}{2}\right) dx$. $2I = \int_0^{\pi/2} (\ln(\sin(2x)) - \ln 2) dx = \int_0^{\pi/2} \ln(\sin(2x)) dx - \int_0^{\pi/2} \ln 2 dx$. For $\int_0^{\pi/2} \ln(\sin(2x)) dx$: Let $u=2x$, $du=2dx$. When $x=0, u=0$; when $x=\pi/2, u=\pi$. This becomes $\frac{1}{2} \int_0^\pi \ln(\sin u) du$. Use property $\int_0^{2a} f(x) dx = 2\int_0^a f(x) dx$ if $f(2a-x)=f(x)$. Here, $\sin(\pi-u)=\sin u$. So $\frac{1}{2} \cdot 2 \int_0^{\pi/2} \ln(\sin u) du = \int_0^{\pi/2} \ln(\sin x) dx = I$. Thus, $2I = I - \ln 2 \cdot \frac{\pi}{2}$. $I = -\frac{\pi}{2}\ln 2 = \frac{\pi}{2}\ln(1/2)$. Result: $\boxed{-\frac{\pi}{2}\ln 2}$. 6. Common Mistakes Forgetting the constant of integration '$C$' in indefinite integrals. Distributing integral over product/quotient: $\int f(x)g(x)dx \ne \int f(x)dx \int g(x)dx$. Incorrectly applying substitution limits in definite integrals. Ignoring absolute values for $\ln|x|$ or $\ln|f(x)|$. Algebraic errors, especially with signs, during simplification or partial fractions. Not handling improper integrals with limits correctly. 7. Quick Practice Set (Mains) $\int \frac{x^2 \tan^{-1}(x^3)}{1+x^6} dx$ Answer: $\boxed{\frac{1}{6}(\tan^{-1}(x^3))^2 + C}$ (Mains) $\int \frac{e^x}{e^x-1} dx$ Answer: $\boxed{\ln|e^x-1| + C}$ (Mains) $\int \frac{dx}{x\ln x \ln(\ln x)}$ Answer: $\boxed{\ln|\ln(\ln x)| + C}$ (Mains) $\int \sin^3 x \cos^2 x dx$ Answer: $\boxed{\frac{\cos^5 x}{5} - \frac{\cos^3 x}{3} + C}$ (Advanced) $\int \frac{x^2+1}{x^4+x^2+1} dx$ Answer: $\boxed{\frac{1}{\sqrt{3}}\tan^{-1}\left(\frac{x^2-1}{x\sqrt{3}}\right) + C}$ (Advanced) $\int \frac{dx}{x\sqrt{x^4-1}}$ Answer: $\boxed{\frac{1}{2}\sec^{-1}(x^2) + C}$ (Advanced) $\int_0^1 \frac{\ln(1+x)}{1+x^2} dx$ Answer: $\boxed{\frac{\pi}{8}\ln 2}$ (Advanced) $\int \sqrt{\tan x} dx$ Answer: $\boxed{\frac{1}{\sqrt{2}}\left[\tan^{-1}\left(\frac{\sqrt{\tan x}}{1-\sqrt{\tan x}}\right) - \tan^{-1}\left(\frac{\sqrt{\tan x}}{1+\sqrt{\tan x}}\right) + \ln\left|\frac{\tan x-\sqrt{2\tan x}+1}{\tan x+\sqrt{2\tan x}+1}\right|\right] + C}$ 8. One-page Cheatsheet Key Formulas & Substitution Hints $\int x^n dx = \frac{x^{n+1}}{n+1}$ $\int \frac{1}{x} dx = \ln|x|$ $\int e^x dx = e^x$ $\int \sin x dx = -\cos x$ $\int \cos x dx = \sin x$ $\int \sec^2 x dx = \tan x$ $\int \frac{1}{\sqrt{a^2-x^2}} dx = \sin^{-1}(\frac{x}{a})$ $\int \frac{1}{a^2+x^2} dx = \frac{1}{a}\tan^{-1}(\frac{x}{a})$ $\int \frac{1}{\sqrt{x^2 \pm a^2}} dx = \ln|x+\sqrt{x^2 \pm a^2}|$ $\int \frac{1}{x^2-a^2} dx = \frac{1}{2a}\ln\left|\frac{x-a}{x+a}\right|$ By Parts: $\int u dv = uv - \int v du$ (LIATE) $\int e^x(f(x)+f'(x))dx = e^x f(x)$ Substitution: $u=g(x)$, $du=g'(x)dx$ Trig Sub: $\sqrt{a^2-x^2} \implies x=a\sin\theta$; $\sqrt{a^2+x^2} \implies x=a\tan\theta$; $\sqrt{x^2-a^2} \implies x=a\sec\theta$ Weierstrass: $t=\tan(x/2)$ for rational $\sin x, \cos x$ King's Rule: $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$ Algebraic tricks: Complete square, add/subtract terms, divide by $x^n$. Common JEE types: $\int \frac{f'(x)}{f(x)} dx$, $\int \frac{x^2 \pm 1}{x^4+k x^2+1} dx$, symmetric integrals. Last-minute trick: Always check if the integrand is a derivative of a known function or can be simplified using basic identities.