Algebra

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Analytic geometry

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• Related to Circle
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Derivatives and Limits

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Indefinite Integrals

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Definite integrals

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Series formulas

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Basic formulas : $$ \int^{\pi/2}_0 \sin^2x\,dx = \int^{\pi/2}_0 \cos^2x\,dx = \frac{\pi}{4} $$ $$\int^\infty_0 \frac{\sin(px)}{x} \,dx = \left\{ \begin{array}{l l l} \pi/2 & p > 0 \\ ~0 & p = 0 \\ -\pi/2 & p < 0 \\ \end{array} \right. $$ $$ \int^\infty_0 \frac{\sin^2px}{x^2} = \frac{\pi\,p}{2}$$ $$ \int^\infty_0 \frac{1 - \cos(px)}{x^2}dx = \frac{\pi\,p}{2}$$ $$ \int^\infty_0 \frac{\cos(px) - \cos(qx)}{x}dx = ln\frac{q}{p} $$ $$ \int^\infty_0 \frac{\cos(px) - \cos(qx)}{x^2}dx = \frac{\pi(q-p)}{2} $$ $$\int^{2\pi}_0 \frac{dx}{a + b\,\sin x} = \frac{2\pi}{\sqrt{a^2-b^2}}$$ $$\int^{2\pi}_0 \frac{dx}{a + b\,\cos(x)} = \frac{2\pi}{\sqrt{a^2-b^2}}$$ $$\int^\infty_0 \sin ax^2\,dx = \int^\infty_0 \cos(ax^2)\,dx = \frac{1}{2}\sqrt{\frac{\pi}{2a}} $$ $$\int^\infty_0 \frac{\sin x}{\sqrt{x}} dx = \int^\infty_0 \frac{\cos x}{\sqrt{x}} dx = \sqrt{\frac{\pi}{2}}$$ $$\int^\infty_0 \frac{\sin^3x}{x^3} dx = \frac{3\pi}{8}$$ $$\int^\infty_0 \frac{\sin^4x}{x^4} dx = \frac{\pi}{3}$$ $$\int^\infty_0 \frac{\tan x}{x} dx = \frac{\pi}{2}$$ $$\int^{\pi/2}_0 \frac{dx}{a + b\,\cos x} = \frac{\arccos(b/a)}{\sqrt{a^2-b^2}}$$ Advanced formulas: $$\int^\pi_0 \sin (mx) \cdot \sin (nx)\,dx = \left\{ \begin{array}{l l} 0 & \quad m , n \text{ integers and } m\ne n \\ \pi/2 & \quad m , n \text{ integers and } m = n \end{array} \right. $$ $$\int^\pi_0 \cos (mx) \cdot \cos (nx)\,dx = \left\{ \begin{array}{l l} 0 & \quad m , n \text{ integers and } m\ne n \\ \pi/2 & \quad m , n \text{ integers and } m = n \end{array} \right. $$ $$\int^\pi_0 \sin (mx) \cdot \cos (nx)\,dx = \left\{ \begin{array}{l l} 0 & \quad m , n \text{ integers and } m + n \text{ odd} \\ 2m/(m^2 - n^2) & \quad m , n \text{ integers and } m + n \text{ even} \end{array} \right. $$ $$ \int^{\pi/2}_0 \sin^{2m}x\,dx = \int^{\pi/2}_0 \cos^{2m}x\,dx = \frac{1\cdot3\cdot5\dots 2m-1}{2\cdot 4 \cdot 6 \dots 2m} \frac{\pi}{2} $$ $$ \int^{\pi/2}_0 \sin^{2m+1}x\,dx = \int^{\pi/2}_0 \cos^{2m+1}x\,dx = \frac{2\cdot 4 \cdot 6 \dots 2m}{1 \cdot 3 \cdot 5 \dots 2m + 1}$$ $$ \int^{\pi}_0 \sin^{2p-1}x\,\cos^{2q-1}x\,dx = \frac{\Gamma(p)\,\Gamma{q}}{2\,\Gamma(p+q)} $$ $$\int^\infty_0 \frac{\sin (px) \cdot \cos (qx)}{x} \,dx = \left\{ \begin{array}{l l l} ~0 & p > q >0 \\ \pi/2 & 0 < p < q \\ \pi/4 & p = q > 0 \\ \end{array} \right. $$ $$\int^\infty_0 \frac{\sin (px) \cdot \sin (qx)}{x^2} \,dx = \left\{ \begin{array}{l l l} \pi\,p/2 & 0 < p \le q \\ \pi\,q/2 & p \ge q > 0 \\ \end{array} \right. $$ $$ \int^\infty_0 \frac{\cos (mx)}{x^2 + a^2}dx = \frac{\pi}{2a}e^{-ma} $$ $$ \int^\infty_0 \frac{x\,\sin (mx)}{x^2 + a^2} dx = \frac{\pi}{2}e^{-ma} $$ $$\int^\infty_0 \frac{\sin (mx)}{x\left(x^2 + a^2\right)} dx = \frac{\pi}{2a^2}\left(1-e^{-ma}\right)$$ $$\int^{2\pi}_0 \frac{dx}{(a + b\,\sin x)^2} = \int^{2\pi}_0 \frac{dx}{(a + b\,\cos x)^2} = \frac{2\pi\,a}{(a^2 - b^2)^{3/2}}$$ $$\int^{2\pi}_0 \frac{dx}{1 - 2a\,\cos x + a^2} = \frac{2\pi}{1 - a^2}, ~~ 0 < a < 1$$ $$\int^{\pi}_0 \frac{x\,\sin x\,dx}{1 - 2a\,\cos x + a^2} = \left\{ \begin{array}{l l l} \frac{\pi}{a} ln(1+a) & |a| < 1 \\ \pi \, ln(1 + \frac{1}{a}) & |a| > 1 \\ \end{array} \right. $$ $$\int^{\pi}_0 \frac{\cos (mx)\,dx}{1 - 2a\,\cos x + a^2} = \frac{\pi a^m}{1 - a^2}, ~~ a^2 < 1$$ $$\int^\infty_0 \sin (ax^n)\,dx = \frac{1}{na^{1/n}} \Gamma(1/n)\,\sin \frac{\pi}{2n} , ~~ n > 1 $$ $$\int^\infty_0 \cos (ax^n)\,dx = \frac{1}{na^{1/n}} \Gamma(1/n)\,\cos \frac{\pi}{2n} , ~~ n > 1 $$ $$\int^\infty_0 \frac{\sin x}{x^p} dx = \frac{\pi}{2\,\Gamma(p)\, \sin (p\pi/2)}, ~~ 0 < p < 1$$ $$\int^\infty_0 \frac{\cos x}{x^p} dx = \frac{\pi}{2\,\Gamma(p)\, \cos (p\pi/2)}, ~~ 0 < p < 1$$ $$\int^\infty_0 \sin (ax^2)\,\cos (2bx) \, dx = \frac{1}{2} \sqrt{\frac{\pi}{2a}} \left(\cos\frac{b^2}{a} - \sin \frac{b^2}{a} \right)$$ $$\int^\infty_0 \cos (ax^2)\,\cos (2bx) \, dx = \frac{1}{2} \sqrt{\frac{\pi}{2a}} \left(\cos\frac{b^2}{a} + \sin \frac{b^2}{a} \right)$$ $$\int^\infty_0 \frac{dx}{1 + \tan^mx} dx = \frac{\pi}{4}$$