Difference between revisions of "Table of indefinite integrals inversecircularfunctions" - Rhea

Latest revision as of 16:05, 26 February 2015

Indefinite Integrals with $\frac{1}{\sin x}$

\int \arcsin \frac {x }{ a}dx = x \arcsin \frac {x}{a} + \sqrt { a^2-x^2 } +C

\int x \arcsin \frac {x }{ a}dx = \left ( \frac{x^2}{2}-\frac {a^2}{4} \right) \arcsin \frac {x}{a} + \frac {x \sqrt { a^2-x^2 }}{4} +C

\int x^2 \arcsin \frac {x }{ a}dx = \frac{x^3}{3}\arcsin \frac {x}{a} + \frac {\left( x^2+2a^2 \right) \sqrt { a^2-x^2 }}{9} +C

\int \frac {\arcsin (x/a)}{x}dx = \frac {x}{a} + \frac {(a/x)^3}{2 \cdot 3 \cdot 3}+ \frac{1 \cdot 3(x/a)^5}{2 \cdot 4 \cdot 5 \cdot 5} + \frac {1 \cdot 3 \cdot 5 (x/a)^7}{2 \cdot 4 \cdot 6 \cdot 7 \cdot 7} + \cdot \cdot \cdot +C

\int \frac {\arcsin (x/a)}{x^2}dx = -\frac{\arcsin (x/a)}{x} - \frac {1}{a} \ln { \left( \frac {a +\sqrt{a^2-x^2}}{x} \right)} +C

\int \left( \arcsin \frac{x}{a} \right)^2 dx = x \left( \arcsin \frac{x}{a} \right)^2-2x + 2 \sqrt{a^2-x^2} \arcsin \frac{x}{a} +C

\int \arccos \frac {x }{ a}dx = x \arccos \frac {x}{a} - \sqrt { a^2-x^2 } +C

\int x \arccos \frac {x }{ a}dx = \frac{x^2}{2}-\frac {a^2}{4} \arccos \frac {x}{a} - \frac {x \sqrt { a^2-x^2 }}{4} +C

\int x^2 \arccos \frac {x }{ a}dx = \frac{x^3}{3}\arccos \frac {x}{a} - \frac {\left( x^2+2a^2 \right) \sqrt { a^2-x^2 }}{9} +C

\int \frac {\arccos (x/a)}{x}dx = \frac {\pi}{2}\ln{x} - \int \frac {\arcsin (x/a)}{x} dx +C

\int \frac {\arccos (x/a)}{x^2}dx = -\frac{\arccos (x/a)}{x} + \frac {1}{a} \ln { \left( \frac {a +\sqrt{a^2-x^2}}{x} \right)} +C

\int \left( \arccos \frac{x}{a} \right)^2 dx = x \left( \arccos \frac{x}{a} \right)^2-2x - 2 \sqrt{a^2-x^2} \arccos \frac{x}{a} +C

\int \arctan \frac {x }{ a}dx = x \arctan \frac {x}{a} - \frac {a}{2} \ln {x^2+a^2} +C

\int x \arctan \frac {x }{ a}dx = \frac{1}{2}(x^2+a^2)\arctan \frac {x}{a} - \frac {ax}{2} +C

\int x^2 \arctan \frac {x }{ a}dx = \frac {x^3}{3}\arctan \frac{x}{a} - \frac {ax^2}{6} +\frac{a^3}{6} \ln \left (x^2+a^2) \right) +C

\int \frac {\arctan (x/a)}{x}dx = \frac {x}{a} - \frac{(x/a)^3}{3^2} + \frac {(x/a)^5}{5^2} - \frac {(x/a)^7)}{7^2} + \cdot \cdot \cdot +C

\int \frac {\arctan (x/a)}{x^2}dx = -\frac{1}{x}\arctan \frac{x}{a} - \frac {a}{2} \ln { \left( \frac {\sqrt{a^2+x^2}}{x^2} \right)} +C

\int \arccot \frac {x }{ a}dx = x \arccot \frac {x}{a} + \frac {a}{2} \ln {x^2+a^2} +C

\int x \arccot \frac {x }{ a}dx = \frac{1}{2}(x^2+a^2)\arccot \frac {x}{a} + \frac {ax}{2} +C

\int x^2 \arccot \frac {x }{ a}dx = \frac {x^3}{3}\arccot \frac{x}{a} + \frac {ax^2}{6} - \frac{a^3}{6} \ln \left (x^2+a^2) \right) +C

\int \frac {\arccot (x/a)}{x}dx = \frac {\pi}{2}\ln{x} - \int \frac{\arctan(x/a)}{x}dx +C

\int \frac {\arccot (x/a)}{x^2}dx = -\frac{\arccot(x/a)}{x} + \frac {1}{2a} \ln { \left( \frac {a^2+x^2}{x^2} \right)} +C

\int \arccos \frac {a}{x}dx= \begin{cases} x \arccos \frac{a}{x} - a \ln \left( x + \sqrt {x^2-a^2} \right) \qquad 0<\arccos \frac{a}{x}< \frac{\pi}{2} +C\\ x \arccos \frac{a}{x} + a \ln \left( x + \sqrt {x^2-a^2} \right) \qquad \frac {\pi}{2}<\arccos \frac{a}{x}< \pi +C\\ \end{cases}

\int x \arccos \frac {a}{x}dx= \begin{cases} \frac {x^2}{2} \arccos \frac{a}{x} - \frac {a \sqrt {x^2-a^2}}{2} \qquad 0<\arccos \frac{a}{x}< \frac{\pi}{2} +C\\ \frac {x^2}{2} \arccos \frac{a}{x} + \frac {a \sqrt {x^2-a^2}}{2} \qquad \frac {\pi}{2}<\arccos \frac{a}{x}< \pi +C\\ \end{cases}

\int x^2 \arccos \frac {a}{x}dx= \begin{cases} \frac {x^3}{3} \arccos \frac{a}{x} - \frac {ax \sqrt {x^2-a^2}}{6} - \frac {a^3}{6} \ln {\left(x+\sqrt {x^2-a^2}\right) }\qquad 0<\arccos \frac{a}{x}< \frac{\pi}{2} +C\\ \frac {x^3}{3} \arccos \frac{a}{x} + \frac {ax \sqrt {x^2-a^2}}{6} - \frac {a^3}{6} \ln {\left(x+\sqrt {x^2-a^2}\right) }\qquad \frac {\pi}{2}<\arccos \frac{a}{x}< \pi +C\\ \end{cases}

\int \frac {\arccos (x/a)}{x}dx = \frac {\pi}{2} + \frac {a}{x}+ \frac {(a/x)^3}{2 \cdot 3 \cdot 3}+ \frac{1 \cdot 3(x/a)^5}{2 \cdot 4 \cdot 5 \cdot 5} + \frac {1 \cdot 3 \cdot 5 (x/a)^7}{2 \cdot 4 \cdot 6 \cdot 7 \cdot 7} + \cdot \cdot \cdot +C

\int \frac {\arccos (x/a)}{x^2} dx= \begin{cases} -\frac{\arccos(a/x)}{x}+\frac{\sqrt{x^2-a^2}}{ax} \qquad 0<\arccos \frac{a}{x}< \frac{\pi}{2} +C\\ -\frac{\arccos(a/x)}{x}-\frac{\sqrt{x^2-a^2}}{ax} \qquad \frac{\pi}{2}< \arccos \frac{a}{x}<{\pi} +C\\ \end{cases}

\int \arcsin \frac {a}{x}dx= \begin{cases} x \arcsin \frac{a}{x} + a \ln \left( x + \sqrt {x^2-a^2} \right) \qquad 0<\arcsin \frac{a}{x}< \frac{\pi}{2} +C\\ x \arcsin \frac{a}{x} + a \ln \left( x - \sqrt {x^2-a^2} \right) \qquad -\frac {\pi}{2}<\arcsin \frac{a}{x}< 0 +C \\ \end{cases}

\int x \arcsin \frac {a}{x}dx= \begin{cases} \frac {x^2}{2} \arcsin \frac{a}{x} + \frac {a \sqrt {x^2-a^2}}{2} \qquad 0<\arccos \frac{a}{x}< \frac{\pi}{2} \\ \frac {x^2}{2} \arcsin \frac{a}{x} + \frac {a \sqrt {x^2-a^2}}{2} \qquad -\frac {\pi}{2}<\arccos \frac{a}{x}< 0 \\ \end{cases}

\int x^2 \arcsin \frac {a}{x}dx= \begin{cases} \frac {x^3}{3} \arcsin \frac{a}{x} + \frac {a \sqrt {x^2-a^2}}{2} + \frac {a^3}{6} \ln {\left(x+\sqrt {x^2-a^2}\right) }\qquad 0<\arcsin \frac{a}{x}< \frac{\pi}{2} \\ \frac {x^3}{3} \arcsin \frac{a}{x} - \frac {a \sqrt {x^2-a^2}}{2} - \frac {a^3}{6} \ln {\left(x+\sqrt {x^2-a^2}\right) }\qquad -\frac {\pi}{2}<\arccos \frac{a}{x}< 0 \\ \end{cases}

\int \frac {\arcsin (x/a)}{x}dx = - \left( \frac {a}{x}+ \frac {(a/x)^3}{2 \cdot 3 \cdot 3}+ \frac{1 \cdot 3(x/a)^5}{2 \cdot 4 \cdot 5 \cdot 5} + \frac {1 \cdot 3 \cdot 5 (x/a)^7}{2 \cdot 4 \cdot 6 \cdot 7 \cdot 7} + \cdot \cdot \cdot \right)+C

\int \frac {\arcsin (x/a)}{x^2} dx= \begin{cases} -\frac{\arcsin(a/x)}{x}-\frac{\sqrt{x^2-a^2}}{ax} \qquad 0<\arcsin \frac{a}{x}< \frac{\pi}{2} \\ -\frac{\arcsin(a/x)}{x}+\frac{\sqrt{x^2-a^2}}{ax} \qquad -\frac{\pi}{2}< \arccos \frac{a}{x}<0 \\ \end{cases}

\int x^m \arcsin \frac {x}{a}dx=\frac{x^{m+1}}{m+1}\arcsin \frac {x}{a} - \frac {1}{m+1} \int \frac{x^{m+1}}{\sqrt{a^2-x^2}}dx +C

\int x^m \arccos \frac {x}{a}dx=\frac{x^{m+1}}{m+1}\arccos \frac {x}{a} + \frac {1}{m+1} \int \frac{x^{m+1}}{\sqrt{a^2-x^2}}dx +C

\int x^m \arctan \frac {x}{a}dx=\frac{x^{m+1}}{m+1}\arcsin \frac {x}{a} - \frac {a}{m+1} \int \frac{x^{m+1}}{a^2+x^2}dx +C

\int x^m \arccot \frac {x}{a}dx=\frac{x^{m+1}}{m+1}\arcsin \frac {x}{a} + \frac {a}{m+1} \int \frac{x^{m+1}}{a^2+x^2}dx +C

\int x^m\arccos \frac {a}{x}dx= \begin{cases} \frac {x^{m+1} \arccos (a/x)}{m+1} - \frac {a}{m+1} \int \frac {x^mdx}{\sqrt{x^2-a^2}} \qquad 0<\arccos \frac{a}{x}< \frac{\pi}{2} \\ \frac {x^{m+1} \arccos (a/x)}{m+1} + \frac {a}{m+1} \int \frac { x^mdx}{\sqrt {x^2-a^2} }\qquad \frac {\pi}{2}<\arccos \frac{a}{x}< \pi \\ \end{cases}

\int x^m \arcsin \frac {a}{x}dx= \begin{cases} \frac {x^{m+1} \arcsin (a/x)}{m+1} + \frac {a}{m+1} \int \frac {x^m dx}{\sqrt{x^2-a^2}} \qquad 0<\arcsin \frac{a}{x}< \frac{\pi}{2} \\ \frac {x^{m+1} \arcsin (a/x)}{m+1} - \frac {a}{m+1} \int \frac { x^m dx}{\sqrt {x^2-a^2}} \qquad -\frac {\pi}{2}<\arcsin \frac{a}{x}< 0 \\ \end{cases}