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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]
A formula for computing the trigonometric identities for the one-third angle exists, but it requires finding the zeroes of the cubic equation 4x 3 − 3x + d = 0, where is the value of the cosine function at the one-third angle and d is the known value of the cosine function at the full angle.
Plot of Si(x) for 0 ≤ x ≤ 8π. Plot of the cosine integral function Ci(z) in the complex plane from −2 − 2i to 2 + 2i. The different sine integral definitions are = = .
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 ...
3.1 Integrals of hyperbolic tangent, cotangent, secant, cosecant functions. 3.2 Integrals involving hyperbolic sine and cosine functions.
For the sine function, we can handle other values. If θ > π /2, then θ > 1. But sin θ ≤ 1 (because of the Pythagorean identity), so sin θ < θ. So we have < <. For negative values of θ we have, by the symmetry of the sine function
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.
As t goes from 0 to 1, the point follows the part of the circle in the first quadrant from (1, 0) to (0, 1). Finally, as t goes from 1 to +∞, the point follows the part of the circle in the second quadrant from (0, 1) to (−1, 0). Here is another geometric point of view. Draw the unit circle, and let P be the point (−1, 0).