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For the special antiderivatives involving trigonometric functions, see Trigonometric integral. [ 1 ] Generally, if the function sin x {\displaystyle \sin x} is any trigonometric function, and cos x {\displaystyle \cos x} is its derivative,
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.
3.1 Integrals of hyperbolic tangent, cotangent, secant, cosecant functions 3.2 Integrals involving hyperbolic sine and cosine functions 3.3 Integrals involving hyperbolic and trigonometric functions
Sine integral in the complex plane, plotted with a variant of domain coloring. Cosine integral in the complex plane. Note the branch cut along the negative real axis. In mathematics, trigonometric integrals are a family of nonelementary integrals involving trigonometric functions.
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 .
At this point we can either integrate directly, or we can first change the integrand to 2 cos 6x − 4 cos 4x + 2 cos 2x and continue from there. Either method gives Either method gives ∫ sin 2 x cos 4 x d x = − 1 24 sin 6 x + 1 8 sin 4 x − 1 8 sin 2 x + C . {\displaystyle \int \sin ^{2}x\cos 4x\,dx=-{\frac {1}{24 ...
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.