Jul 24, 2024
\sin^2(x) + \cos^2(x) = 11 + \tan^2(x) = \sec^2(x)1 + \cot^2(x) = \csc^2(x)\sin(2x) = 2 \sin(x) \cos(x)\cos(2x) has three forms:
\cos^2(x) - \sin^2(x)1 - 2 \sin^2(x)2 \cos^2(x) - 1\sin^2(x) = \frac{1}{2} (1 - \cos(2x))\cos^2(x) = \frac{1}{2} (1 + \cos(2x))\cos^3(x)\cos^3(x) = \cos^2(x) \cdot \cos(x)\cos^2(x) = 1 - \sin^2(x)u = \sin(x),du = \cos(x) dx\int (1 - u^2) duu - \frac{u^3}{3} + C\sin(x) - \frac{\sin^3(x)}{3} + C\cos^5(x)\cos^5(x) = \cos^4(x) \cdot \cos(x)\cos^4(x) = (\cos^2(x))^2 = (1 - \sin^2(x))^2u = \sin(x), du = \cos(x) dx(1 - u^2)^2 = 1 - 2u^2 + u^4\int (1 - 2u^2 + u^4) duu - \frac{2u^3}{3} + \frac{u^5}{5} + C\sin(x) - \frac{2\sin^3(x)}{3} + \frac{\sin^5(x)}{5} + C\cos^5(x) \cdot \sin(x)u = \cos(x), du = -\sin(x) dx-\int u^5 du-\frac{u^6}{6} + C-\frac{\cos^6(x)}{6} + C\sin^5(x) \cos^2(x)\sin^5(x) = (\sin^4(x))\sin(x), \cos^2(x) is left.\sin^4(x) = (\sin^2(x))^2 and \sin^2(x) = 1 - \cos^2(x)u = \cos(x), du = -\sin(x) dx\int (1 - u^2)^2 u^2 du- (\frac{u^3}{3} - \frac{2u^5}{5} + \frac{u^7}{7}) + C- (\frac{\cos^3(x)}{3} - \frac{2\cos^5(x)}{5} + \frac{\cos^7(x)}{7}) + C\sin^2(x)\sin^2(x) = \frac{1}{2}(1 - \cos(2x))\frac{1}{2} \int (1 - \cos(2x)) dx\frac{1}{2} (x - \frac{\sin(2x)}{2}) + C\cos^2(3x)\cos^2(x) = \frac{1}{2}(1 + \cos(2x))\cos^2(3x) = \frac{1}{2}(1 + \cos(6x))\frac{1}{2} \int (1 + \cos(6x)) dx\frac{1}{2} (x + \frac{\sin(6x)}{6}) + C\sin^4(x)\sin^2(x) = \frac{1}{2}(1 - \cos(2x)) and square it.\sin^2(x) \cos^2(x)\frac{1}{4} \int (1 - \cos(2x))(1 + \cos(2x)) dx\cos^2(x) \tan^3(x)