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The integral of the secant function was historically one of the first integrals of its type ever evaluated, before most of the development of integral calculus. It is important because it is the vertical coordinate of the Mercator projection, used for marine navigation with constant compass bearing.
This is one of several integrals usually done in a first-year calculus course in which the most natural way to proceed involves integrating by parts and returning to the same integral one started with (another is the integral of the product of an exponential function with a sine or cosine function; yet another the integral of a power of the ...
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 following is a list of integrals (antiderivative functions) of trigonometric functions. 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 a definite integral, the bounds change once the substitution is performed and are determined using the equation = , with values in the range < <. Alternatively, apply the boundary terms directly to the formula for the antiderivative.
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.
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 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.