Cual es la integral dela función

$$\begin{align}&\int_{-\pi/4}^{\pi/3}sen^5{\theta}\,d\theta=\\ &\\ &\int_{-\pi/4}^{\pi/3}sen^2\theta·sen^2\theta·sen\theta\,d\theta=\\ &\\ &\\ &\int_{-\pi/4}^{\pi/3}(1-\cos^2\theta)·(1-\cos^2\theta)·sen\theta\,d\theta=\\ &\\ &\\ &t =\cos\theta\quad dt=-sen\theta\,d\theta\\ &\\ &\theta=-\frac{\pi}{4}\implies t = \frac{\sqrt 2}{2}\\ &\\ &\theta= \frac{\pi}{3}\implies t = \frac 12\\ &\\ &\\ &=-\int_{\sqrt 2/2}^{1/2}(1-t^2)(1-t^2)dt=\\ &\\ &\\ &-\int_{\sqrt 2/2}^{1/2}(1-2t^2+t^4)dt=\\ &\\ &\\ &-\left[t-\frac{2t^3}{3}+\frac{t^5}{5}  \right]_{\sqrt 2/2}^{1/2}=\\ &\\ &\\ &-\left(\frac 12-\frac{2}{24}+\frac {1}{160}-\frac{\sqrt 2}{2}+\frac{4 \sqrt 2}{24}-\frac{4 \sqrt 2}{160}  \right)=\\ &\\ &\\ &-\left(\frac{240-40+3-240 \sqrt 2+80 \sqrt 2-12 \sqrt 2}{480} \right)=\\ &\\ &\\ &-\left(\frac{203-172 \sqrt 2}{480}  \right)= \frac{172 \sqrt 2-203}{480}\\ &\\ &\end{align}$$

Y eso es odo.