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The following is a list of integrals (antiderivative functions) of trigonometric functions. 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.
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.
The following is a list of integrals (anti-derivative functions) of hyperbolic functions. For a complete list of integral functions, see list of integrals. In all formulas the constant a is assumed to be nonzero, and C denotes the constant of integration.
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).
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 ...
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.
For a definite integral, the bounds change once the substitution is performed and are determined using the equation = , with values in the range < <. Alternatively, apply the boundary terms directly to the formula for the antiderivative.
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.