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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,
Ptolemy's theorem states that the sum of the products of the lengths of opposite sides is equal to the product of the lengths of the diagonals. When those side-lengths are expressed in terms of the sin and cos values shown in the figure above, this yields the angle sum trigonometric identity for sine: sin(α + β) = sin α cos β + cos α sin β.
The integral can also be solved by manipulating the integrand and substituting twice. Using the definition sec θ = 1 / cos θ and the identity cos 2 θ + sin 2 θ = 1, the integral can be rewritten as
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 .
In the integral , we may use = , = , = . Then, = = () = = = + = +. The above step requires that > and > We can choose to be the principal root of , and impose the restriction / < < / by using the inverse sine function.
[1] [2] One reason for this is that they can greatly simplify differential equations that do not need to be answered with absolute precision. There are a number of ways to demonstrate the validity of the small-angle approximations. The most direct method is to truncate the Maclaurin series for each of the trigonometric functions.
Equation 2 is differentiated with respect to t on both sides and rearranged to obtain + = (+) = (). Replacing the quotient of the square root with its definition in Eq. 2 , and equating the coefficients of powers of t in the resulting expansion gives Bonnet’s recursion formula ( n + 1 ) P n + 1 ( x ) = ( 2 n + 1 ) x P n ( x ) − n P n − 1 ...