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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]
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.
In mathematics, the definite integral ()is the area of the region in the xy-plane bounded by the graph of f, the x-axis, and the lines x = a and x = b, such that area above the x-axis adds to the total, and that below the x-axis subtracts from the total.
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 ...
As another example, to find the area of the region bounded by the graph of the function f(x) = between x = 0 and x = 1, one can divide the interval into five pieces (0, 1/5, 2/5, ..., 1), then construct rectangles using the right end height of each piece (thus √ 0, √ 1/5, √ 2/5, ..., √ 1) and sum their areas to get the approximation
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.
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 ...
[1] [2] This can be seen by using Dirichlet's test for improper integrals. It is a good illustration of special techniques for evaluating definite integrals, particularly when it is not useful to directly apply the fundamental theorem of calculus due to the lack of an elementary antiderivative for the integrand, as the sine integral , an ...