Sr Examen
Integral de dx/cos(p/4-x)^2 dx
Solución
Ha introducido
[src]
1 / | | 1 | ----------- dx | 2/p \ | cos |- - x| | \4 / | / 0
$$\int\limits_{0}^{1} \frac{1}{\cos^{2}{\left(\frac{p}{4} - x \right)}}\, dx$$
Integral(1/(cos(p/4 - x)^2), (x, 0, 1))
Respuesta (Indefinida)
[src]
// / x p\ \ / || 2*tan|- - + -| | | || \ 2 8/ | | 1 ||------------------ for p != -2*pi + 4*x| | ----------- dx = C + |< 2/ x p\ | | 2/p \ ||-1 + tan |- - + -| | | cos |- - x| || \ 2 8/ | | \4 / || | | \\ zoo*x otherwise / /
$$\int \frac{1}{\cos^{2}{\left(\frac{p}{4} - x \right)}}\, dx = C + \begin{cases} \frac{2 \tan{\left(\frac{p}{8} - \frac{x}{2} \right)}}{\tan^{2}{\left(\frac{p}{8} - \frac{x}{2} \right)} - 1} & \text{for}\: p \neq 4 x - 2 \pi \\\tilde{\infty} x & \text{otherwise} \end{cases}$$
Respuesta
[src]
/p\ / 1 p\
2*tan|-| 2*tan|- - + -|
\8/ \ 2 8/
- ------------ + ------------------
2/p\ 2/ 1 p\
-1 + tan |-| -1 + tan |- - + -|
\8/ \ 2 8/
$$\frac{2 \tan{\left(\frac{p}{8} - \frac{1}{2} \right)}}{\tan^{2}{\left(\frac{p}{8} - \frac{1}{2} \right)} - 1} - \frac{2 \tan{\left(\frac{p}{8} \right)}}{\tan^{2}{\left(\frac{p}{8} \right)} - 1}$$
=
=
/p\ / 1 p\
2*tan|-| 2*tan|- - + -|
\8/ \ 2 8/
- ------------ + ------------------
2/p\ 2/ 1 p\
-1 + tan |-| -1 + tan |- - + -|
\8/ \ 2 8/
$$\frac{2 \tan{\left(\frac{p}{8} - \frac{1}{2} \right)}}{\tan^{2}{\left(\frac{p}{8} - \frac{1}{2} \right)} - 1} - \frac{2 \tan{\left(\frac{p}{8} \right)}}{\tan^{2}{\left(\frac{p}{8} \right)} - 1}$$
-2*tan(p/8)/(-1 + tan(p/8)^2) + 2*tan(-1/2 + p/8)/(-1 + tan(-1/2 + p/8)^2)
Estos ejemplos se pueden aplicar para introducción de los límites de integración inferior y superior.