Exercise
$\int_{-\frac{\pi}{6}}^0sin^4\left(w\right)\left[2-tan^2\left(w\right)\right]cos^3\left(w\right)dw$
Step-by-step Solution
Final answer to the exercise
$2\cdot \left(\left(-\frac{1}{7}\cdot \cos\left(-\frac{\pi }{6}\right)^2-1\right)\cdot \sin\left(-\frac{\pi }{6}\right)^2+\frac{3\cdot \cos\left(-\frac{\pi }{6}\right)^{4}\sin\left(-\frac{\pi }{6}\right)}{35}- \sin\left(-\frac{\pi }{6}\right)^2\cdot \left(-\frac{4}{35}\cdot \cos\left(-\frac{\pi }{6}\right)^2-\frac{8}{35}\right)\right)+\frac{2}{7}\cdot \sin\left(-\frac{\pi }{6}\right)^{3}\cdot \cos\left(-\frac{\pi }{6}\right)^{4}+\frac{- \left(- \sin\left(-\frac{\pi }{6}\right)^{7}\right)}{7}$