Integral de sinx/(cos^2)x+1 dx
Solución
Solución detallada
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Integramos término a término:
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No puedo encontrar los pasos en la búsqueda de esta integral.
Pero la integral
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La integral de las constantes tienen esta constante multiplicada por la variable de integración:
El resultado es:
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Ahora simplificar:
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Añadimos la constante de integración:
Respuesta:
Respuesta (Indefinida)
[src]
/ / /x\\ / /x\\ 2/x\ / /x\\ 2/x\ 2/x\ / /x\\
| log|1 + tan|-|| log|-1 + tan|-|| tan |-|*log|-1 + tan|-|| x*tan |-| tan |-|*log|1 + tan|-||
| / sin(x) \ \ \2// x \ \2// \2/ \ \2// \2/ \2/ \ \2//
| |-------*x + 1| dx = C + x + --------------- - ------------ - ---------------- + ------------------------ - ------------ - -----------------------
| | 2 | 2/x\ 2/x\ 2/x\ 2/x\ 2/x\ 2/x\
| \cos (x) / -1 + tan |-| -1 + tan |-| -1 + tan |-| -1 + tan |-| -1 + tan |-| -1 + tan |-|
| \2/ \2/ \2/ \2/ \2/ \2/
/
$$\int \left(x \frac{\sin{\left(x \right)}}{\cos^{2}{\left(x \right)}} + 1\right)\, dx = C + x - \frac{x \tan^{2}{\left(\frac{x}{2} \right)}}{\tan^{2}{\left(\frac{x}{2} \right)} - 1} - \frac{x}{\tan^{2}{\left(\frac{x}{2} \right)} - 1} + \frac{\log{\left(\tan{\left(\frac{x}{2} \right)} - 1 \right)} \tan^{2}{\left(\frac{x}{2} \right)}}{\tan^{2}{\left(\frac{x}{2} \right)} - 1} - \frac{\log{\left(\tan{\left(\frac{x}{2} \right)} - 1 \right)}}{\tan^{2}{\left(\frac{x}{2} \right)} - 1} - \frac{\log{\left(\tan{\left(\frac{x}{2} \right)} + 1 \right)} \tan^{2}{\left(\frac{x}{2} \right)}}{\tan^{2}{\left(\frac{x}{2} \right)} - 1} + \frac{\log{\left(\tan{\left(\frac{x}{2} \right)} + 1 \right)}}{\tan^{2}{\left(\frac{x}{2} \right)} - 1}$$
2 2 2
1 log(1 + tan(1/2)) tan (1/2) pi*I + log(1 - tan(1/2)) tan (1/2)*(pi*I + log(1 - tan(1/2))) tan (1/2)*log(1 + tan(1/2))
1 - -------------- + ----------------- - pi*I - -------------- - ------------------------ + ------------------------------------ - ---------------------------
2 2 2 2 2 2
-1 + tan (1/2) -1 + tan (1/2) -1 + tan (1/2) -1 + tan (1/2) -1 + tan (1/2) -1 + tan (1/2)
$$\frac{\log{\left(\tan{\left(\frac{1}{2} \right)} + 1 \right)}}{-1 + \tan^{2}{\left(\frac{1}{2} \right)}} - \frac{\log{\left(\tan{\left(\frac{1}{2} \right)} + 1 \right)} \tan^{2}{\left(\frac{1}{2} \right)}}{-1 + \tan^{2}{\left(\frac{1}{2} \right)}} - \frac{\tan^{2}{\left(\frac{1}{2} \right)}}{-1 + \tan^{2}{\left(\frac{1}{2} \right)}} + 1 - \frac{1}{-1 + \tan^{2}{\left(\frac{1}{2} \right)}} - i \pi + \frac{\left(\log{\left(1 - \tan{\left(\frac{1}{2} \right)} \right)} + i \pi\right) \tan^{2}{\left(\frac{1}{2} \right)}}{-1 + \tan^{2}{\left(\frac{1}{2} \right)}} - \frac{\log{\left(1 - \tan{\left(\frac{1}{2} \right)} \right)} + i \pi}{-1 + \tan^{2}{\left(\frac{1}{2} \right)}}$$
=
2 2 2
1 log(1 + tan(1/2)) tan (1/2) pi*I + log(1 - tan(1/2)) tan (1/2)*(pi*I + log(1 - tan(1/2))) tan (1/2)*log(1 + tan(1/2))
1 - -------------- + ----------------- - pi*I - -------------- - ------------------------ + ------------------------------------ - ---------------------------
2 2 2 2 2 2
-1 + tan (1/2) -1 + tan (1/2) -1 + tan (1/2) -1 + tan (1/2) -1 + tan (1/2) -1 + tan (1/2)
$$\frac{\log{\left(\tan{\left(\frac{1}{2} \right)} + 1 \right)}}{-1 + \tan^{2}{\left(\frac{1}{2} \right)}} - \frac{\log{\left(\tan{\left(\frac{1}{2} \right)} + 1 \right)} \tan^{2}{\left(\frac{1}{2} \right)}}{-1 + \tan^{2}{\left(\frac{1}{2} \right)}} - \frac{\tan^{2}{\left(\frac{1}{2} \right)}}{-1 + \tan^{2}{\left(\frac{1}{2} \right)}} + 1 - \frac{1}{-1 + \tan^{2}{\left(\frac{1}{2} \right)}} - i \pi + \frac{\left(\log{\left(1 - \tan{\left(\frac{1}{2} \right)} \right)} + i \pi\right) \tan^{2}{\left(\frac{1}{2} \right)}}{-1 + \tan^{2}{\left(\frac{1}{2} \right)}} - \frac{\log{\left(1 - \tan{\left(\frac{1}{2} \right)} \right)} + i \pi}{-1 + \tan^{2}{\left(\frac{1}{2} \right)}}$$
1 - 1/(-1 + tan(1/2)^2) + log(1 + tan(1/2))/(-1 + tan(1/2)^2) - pi*i - tan(1/2)^2/(-1 + tan(1/2)^2) - (pi*i + log(1 - tan(1/2)))/(-1 + tan(1/2)^2) + tan(1/2)^2*(pi*i + log(1 - tan(1/2)))/(-1 + tan(1/2)^2) - tan(1/2)^2*log(1 + tan(1/2))/(-1 + tan(1/2)^2)
Estos ejemplos se pueden aplicar para introducción de los límites de integración inferior y superior.