# Transcendental Integrals Reference

This is a reference of some transcendental integrals. The integrals are categorized based on the type of functions.

## Trigonometric Integrals

$$\displaystyle \int \sqrt{1+\cos x}\text{ }dx = 2\sqrt{1-\cos x} = 2\sqrt{2}\sin(x/2)$$

$$\displaystyle \int \sqrt{1-\cos x}\text{ }dx = -2\sqrt{1+\cos x} = 2\sqrt{2}\cos(x/2)$$

$$\displaystyle \int \sqrt{1+\sin x}\text{ }dx = 2\sin(x/2)-2\cos(x/2)$$

$$\displaystyle \int \sqrt{1-\sin x}\text{ }dx = -2\cos(x/2)-2\sin(x/2)$$

$$\displaystyle \int \dfrac{1}{\sqrt{1+\sin x}}\text{ }dx = \sqrt{2}\tanh^{-1}\sqrt{\dfrac{1-\sin x}{2}}$$

$$\displaystyle \int \dfrac{1}{\sqrt{1-\sin x}}\text{ }dx = \sqrt{2}\tanh^{-1}\sqrt{\dfrac{1+\sin x}{2}}$$

$$\displaystyle \int \dfrac{1}{\sqrt{1+\cos x}}\text{ }dx = \sqrt{2}\tanh^{-1}(\sin(x/2))$$

$$\displaystyle \int \dfrac{1}{\sqrt{1-\cos x}}\text{ }dx = \sqrt{2}\tanh^{-1}(\cos(x/2))$$

$$\displaystyle \int \dfrac{1}{1+\sin x}\text{ }dx = \tan x - \sec x$$

$$\displaystyle \int \dfrac{1}{1+\cos x}\text{ }dx = -\cot x + \csc x$$

$$\displaystyle \int \dfrac{1}{a+b\sin x}\text{ } dx = \dfrac{2}{\sqrt{a^2-b^2}}\arctan\left[ \dfrac{a\cdot\tan(x/2)+b}{\sqrt{a^2-b^2}} \right]$$

$$\displaystyle \int \dfrac{1}{a+b\cos x}\text{ } dx = \dfrac{2}{\sqrt{a^2-b^2}}\arctan\left[ \sqrt{\dfrac{a-b}{a+b}}\tan(x/2) \right]$$

$$\displaystyle \int \dfrac{1}{a+b\tan x}\text{ } dx = \dfrac{ax+b\ln \left|a\cos x+b\sin x \right|}{a^2+b^2}$$

$$\displaystyle \int \dfrac{1}{a+b\sec x}\text{ } dx = \dfrac{u}{a}+\dfrac{2b}{\sqrt{a^2-b^2}}\tanh^{-1}\left[ \dfrac{(b-a)\tan(x/2)}{\sqrt{a^2-b^2}} \right]$$

$$\displaystyle \int \dfrac{1}{a\sin x+b\cos x}\text{ } dx = \dfrac{1}{\sqrt{a^2+b^2}}\ln\left|\csc\left(x+\tan^{-1}\dfrac{b}{a}\right) - \cot\left(x+\tan^{-1}\dfrac{b}{a}\right)\right|$$

$$\displaystyle \int \dfrac{1}{a^2\cos^{2} x+b^2\sin^{2} x}\text{ } dx = \dfrac{1}{ab}\tan^{-1}\left(\dfrac{b}{a}\tan x\right)$$

## Exponential and Mixed Functions

$$\displaystyle \int \dfrac{1}{a+be^x}\text{ }dx = \dfrac{x}{a}-\dfrac{1}{a}\ln(a+be^x)$$

$$\displaystyle \int \dfrac{1}{a^2e^x + b^2e^{-x}}\text{ }dx = \dfrac{1}{ab}\tan^{-1}\left(\dfrac{a}{b}e^x\right)$$

$$\displaystyle \int \sqrt{1+e^{2x}}\text{ }dx = \sqrt{1+e^{2x}} - \tanh^{-1}\sqrt{1+e^{2x}} = \sqrt{1+e^{2x}} - \sinh^{-1}(e^{-x})$$

$$\displaystyle \int \dfrac{\ln x}{x}dx = \dfrac{1}{2}(\ln x)^2$$

$$\displaystyle \int \dfrac{1}{x\ln x}dx = \ln(\ln x)$$

$$\displaystyle \int \ln\left[a^2+(bx)^2\right]\text{ }dx = x\ln\left[a^2+(bx)^2\right]+\dfrac{2b}{a}\arctan\left(\dfrac{b}{a}x\right)-2x$$

$$\displaystyle \int \sin\left[a\ln(bx)\right]\text{ }dx = \dfrac{x\sin\left[a\ln(bu)\right]-ax\cos\left[a\ln(bx)\right]}{1+a^2}$$

$$\displaystyle \int e^{ax}\sin(bx)\text{ }dx = \dfrac{e^{ax}}{a^2+b^2}\left[a\sin(bx)-b\cos(bx)\right]$$

$$\displaystyle \int e^{ax}\cos(bx)\text{ }dx = \dfrac{e^{ax}}{a^2+b^2}\left[a\cos(bx)-b\sin(bx)\right]$$

$$\displaystyle \int (\cos x)\ln(\cos x) \text{ }dx = (\sin x)\ln(\cos x)+\ln(\sec x + \tan x)- \sin x$$

$$\displaystyle \int (\sin x)\ln(\sin x) \text{ }dx = -(\cos x)\ln(\sin x)-\ln(\csc x + \cot x) + \cos x$$

$$\displaystyle \int x^n\cdot\ln x \text{ }dx = \dfrac{x^{n+1}}{n+1}\ln x - \dfrac{x^{n+1}}{(n+1)^2}$$, (n ≠ 1)

## Transcendental Functions with Algebraic Derivative

$$\displaystyle \int \dfrac{\arcsin x}{x^2}dx = \dfrac{\arcsin x}{x}-\ln\left|\dfrac{1+\sqrt{1-x^2}}{x}\right|$$

$$\displaystyle \int \dfrac{\arccos x}{x^2}dx = -\dfrac{\arccos x}{x}+\ln\left|\dfrac{1+\sqrt{1-x^2}}{x}\right|$$

$$\displaystyle \int \dfrac{\arctan x}{x^2}dx = -\dfrac{\arctan x}{x}+\ln\left|\dfrac{x}{\sqrt{1+x^2}}\right|$$

$$\displaystyle \int \arctan(x^2)\text{ } dx = x\arctan(x^2)+\dfrac{\sqrt{2}}{2}\arctan(\sqrt{2}x+1)+$$$$\dfrac{\sqrt{2}}{2}\arctan(\sqrt{2}x-1)$$

## Natural Logarithm

$$\displaystyle \int \ln(1+x^n)\text{ }dx = x\ln(1+x^n) - nx + \displaystyle \int \dfrac{1}{1+x^n}dx$$

$$\displaystyle \int \ln(1+x^2)\text{ }dx = x\ln(1+x^2) - 2x + \arctan x$$

See Algebraic Integrals Reference for evaluation of n = 3, 4, and 6.