Integration of cosec inverse x can be calculated using integration by parts ,integration by substitution .Here is the formula for it \[ \int \csc^{-1}x \, dx = x \csc^{-1}x + \ln |x + \sqrt { x^2+ 1}| + C \] where (C) is the constant of integration. Proof of integration of cosec inverse x We […]

Integration of 1/a^2+ x^2, 1/a^2- x^2, 1/x^2 -a^2 can be obtained using trigonometric substitution, integration by partial fractions. The formula given are $\int \frac {1}{x^2 + a^2} dx = \frac {1}{a} \tan ^{-1} (\frac {x}{a}) + C$ $\int \frac {1}{x^2 – a^2} dx = \frac {1}{2a} ln |\frac {x-a}{x+a}| + C$ $\int \frac {1}{a^2 –

Integration of rational functions, which are quotients of polynomials, is a fundamental concept in calculus. The general form of a rational function is: $$ \frac{P(x)}{Q(x)} $$ where ( P(x) ) and ( Q(x) ) are polynomials. The strategy for integrating such functions typically involves a few steps: (1) Polynomial Division: If the degree of (

Integration of root tan x is little complex. This can be solve using integration by substitution method . The formula is given by $\int \sqrt {tan x} \; dx = \frac {1}{\sqrt 2} \tan^{-1} (\frac {tan x -1}{\sqrt {2tan x}}) + \frac {1}{2\sqrt 2} \ln \frac {(tan x + 1 -\sqrt {2tan x})}{(tan x +

The integration of $sin^7x$, can be found using integration substitution and trigonometry identities . The integral of $\sin^7 x $ with respect to (x) is: \[\int \sin^7 x \, dx =\frac{\cos^7(x)}{7} – \frac{3\cos^5(x)}{5} + \cos^3(x) – \cos(x) + C\] Here, (C) represents the constant of integration, which is added because the process of integration determines

The integral of the sec square function, $\sec ^2x$, can be found using basic calculus principles. The integral of ($\sec ^2x$) with respect to (x) is: \[\int \sec^2 x \, dx = \tan x + C\] Here, (C) represents the constant of integration, which is added because the process of integration determines the antiderivative up