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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 ...
The integral of the secant function was one of the "outstanding open problems of the mid-seventeenth century", solved in 1668 by James Gregory. [3] He applied his result to a problem concerning nautical tables. [1] In 1599, Edward Wright evaluated the integral by numerical methods – what today we would call Riemann sums. [4]
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.
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.
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 .
For a definite integral, one must figure out how the bounds of integration change. For example, as x {\displaystyle x} goes from 0 {\displaystyle 0} to a / 2 , {\displaystyle a/2,} then sin θ {\displaystyle \sin \theta } goes from 0 {\displaystyle 0} to 1 / 2 , {\displaystyle 1/2,} so θ {\displaystyle \theta } goes from 0 {\displaystyle 0 ...
Just as the definite integral of a positive function of one variable represents the area of the region between the graph of the function and the x-axis, the double integral of a positive function of two variables represents the volume of the region between the surface defined by the function (on the three-dimensional Cartesian plane where z = f(x, y)) and the plane which contains its domain. [1]
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