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[1] Generally, if the function sin x {\displaystyle \sin x} is any trigonometric function, and cos x {\displaystyle \cos x} is its derivative, ∫ a cos n x d x = a n sin n x + C {\displaystyle \int a\cos nx\,dx={\frac {a}{n}}\sin nx+C}
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
Case I: Integrands containing a 2 − x 2 [ edit ] Let x = a sin θ , {\displaystyle x=a\sin \theta ,} and use the identity 1 − sin 2 θ = cos 2 θ . {\displaystyle 1-\sin ^{2}\theta =\cos ^{2}\theta .}
The real part of the other side is a polynomial in cos x and sin x, in which all powers of sin x are even and thus replaceable through the identity cos 2 x + sin 2 x = 1. By the same reasoning, sin nx is the imaginary part of the polynomial, in which all powers of sin x are odd and thus, if one factor of sin x is factored out, the remaining ...
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 ...
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 = = .
The notations sin −1, cos −1, etc. are often used for arcsin and arccos, etc. When this notation is used, inverse functions could be confused with multiplicative inverses. The notation with the "arc" prefix avoids such a confusion, though "arcsec" for arcsecant can be confused with "arcsecond".