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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]
If the function f does not have any continuous antiderivative which takes the value zero at the zeros of f (this is the case for the sine and the cosine functions), then sgn(f(x)) ∫ f(x) dx is an antiderivative of f on every interval on which f is not zero, but may be discontinuous at the points where f(x) = 0.
Si(x) (blue) and Ci(x) (green) shown on the same plot. 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.
Leonhard Euler used it to evaluate the integral / (+ ) in his 1768 integral calculus textbook, [3] and Adrien-Marie Legendre described the general method in 1817. [ 4 ] The substitution is described in most integral calculus textbooks since the late 19th century, usually without any special name. [ 5 ]
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 ...
The following is a list of indefinite integrals (antiderivatives) of expressions involving the inverse hyperbolic functions. For a complete list of integral formulas, see lists of integrals. In all formulas the constant a is assumed to be nonzero, and C denotes the constant of integration.
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 ...
In this case, 1 / 3 + 1 / 5 + … + 1 / 111 < 2, but 1 / 3 + 1 / 5 + … + 1 / 113 > 2. The exact answer can be calculated using the general formula provided in the next section, and a representation of it is shown below. Fully expanded, this value turns into a fraction that involves two 2736 ...