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Twice the area of the purple triangle is the stereographic projection s = tan 1 / 2 ϕ = tanh 1 / 2 ψ. The blue point has coordinates (cosh ψ, sinh ψ). The red point has coordinates (cos ϕ, sin ϕ). The purple point has coordinates (0, s). The integral of the hyperbolic secant function defines the Gudermannian function:
These identities are useful whenever expressions involving trigonometric functions need to be simplified. An important application is the integration of non-trigonometric functions: a common technique involves first using the substitution rule with a trigonometric function, and then simplifying the resulting integral with a trigonometric identity.
For antiderivatives involving both exponential and trigonometric functions, see List of integrals of exponential functions. For a complete list of antiderivative functions, see Lists of integrals. For the special antiderivatives involving trigonometric functions, see Trigonometric integral. [1]
Integration is the basic operation in integral calculus.While differentiation has straightforward rules by which the derivative of a complicated function can be found by differentiating its simpler component functions, integration does not, so tables of known integrals are often useful.
The substitution is described in most integral calculus textbooks since the late 19th century, usually without any special name. [5] It is known in Russia as the universal trigonometric substitution , [ 6 ] and also known by variant names such as half-tangent substitution or half-angle substitution .
This geometric argument relies on definitions of arc length and area, which act as assumptions, so it is rather a condition imposed in construction of trigonometric functions than a provable property. [2] For the sine function, we can handle other values. If θ > π /2, then θ > 1. But sin θ ≤ 1 (because of the Pythagorean identity), so sin ...
Plot of Si(x) for 0 ≤ x ≤ 8π. Plot of the cosine integral function Ci(z) in the complex plane from −2 − 2i to 2 + 2i. The different sine integral definitions are = = .
This visualization also explains why integration by parts may help find the integral of an inverse function f −1 (x) when the integral of the function f(x) is known. Indeed, the functions x ( y ) and y ( x ) are inverses, and the integral ∫ x dy may be calculated as above from knowing the integral ∫ y dx .