Integral de sin^2(2x-1) dx
Solución
Respuesta (Indefinida)
[src]
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| 3 4 2
| 2 x tan (-1/2 + x) tan(-1/2 + x) x*tan (-1/2 + x) 2*x*tan (-1/2 + x)
| sin (2*x - 1) dx = C + --------------------------------------- + --------------------------------------- - --------------------------------------- + --------------------------------------- + ---------------------------------------
| 4 2 4 2 4 2 4 2 4 2
/ 2 + 2*tan (-1/2 + x) + 4*tan (-1/2 + x) 2 + 2*tan (-1/2 + x) + 4*tan (-1/2 + x) 2 + 2*tan (-1/2 + x) + 4*tan (-1/2 + x) 2 + 2*tan (-1/2 + x) + 4*tan (-1/2 + x) 2 + 2*tan (-1/2 + x) + 4*tan (-1/2 + x)
$$\int \sin^{2}{\left(2 x - 1 \right)}\, dx = C + \frac{x \tan^{4}{\left(x - \frac{1}{2} \right)}}{2 \tan^{4}{\left(x - \frac{1}{2} \right)} + 4 \tan^{2}{\left(x - \frac{1}{2} \right)} + 2} + \frac{2 x \tan^{2}{\left(x - \frac{1}{2} \right)}}{2 \tan^{4}{\left(x - \frac{1}{2} \right)} + 4 \tan^{2}{\left(x - \frac{1}{2} \right)} + 2} + \frac{x}{2 \tan^{4}{\left(x - \frac{1}{2} \right)} + 4 \tan^{2}{\left(x - \frac{1}{2} \right)} + 2} + \frac{\tan^{3}{\left(x - \frac{1}{2} \right)}}{2 \tan^{4}{\left(x - \frac{1}{2} \right)} + 4 \tan^{2}{\left(x - \frac{1}{2} \right)} + 2} - \frac{\tan{\left(x - \frac{1}{2} \right)}}{2 \tan^{4}{\left(x - \frac{1}{2} \right)} + 4 \tan^{2}{\left(x - \frac{1}{2} \right)} + 2}$$
2 2
cos (1) sin (1) cos(1)*sin(1)
------- + ------- - -------------
2 2 2
$$- \frac{\sin{\left(1 \right)} \cos{\left(1 \right)}}{2} + \frac{\cos^{2}{\left(1 \right)}}{2} + \frac{\sin^{2}{\left(1 \right)}}{2}$$
=
2 2
cos (1) sin (1) cos(1)*sin(1)
------- + ------- - -------------
2 2 2
$$- \frac{\sin{\left(1 \right)} \cos{\left(1 \right)}}{2} + \frac{\cos^{2}{\left(1 \right)}}{2} + \frac{\sin^{2}{\left(1 \right)}}{2}$$
cos(1)^2/2 + sin(1)^2/2 - cos(1)*sin(1)/2
Estos ejemplos se pueden aplicar para introducción de los límites de integración inferior y superior.