evaluate the integral $\\int_{0}^{\\pi / 6} \\int_{3}^{9}(y \\cos x-5) d y d x$.
evaluate the integral $\\int_{0}^{\\pi / 6} \\int_{3}^{9}(y \\cos x-5) d y d x$.
Answer
Explanation:
Step1: Integrate with respect to ( y )
Use the power rule (\int y^n dy=\frac{y^{n + 1}}{n+1}) ((n\neq - 1)) and (\int a dy=ay) ((a) is a constant). [ \begin{align*} \int_{3}^{9}(y\cos x-5)dy&=\cos x\int_{3}^{9}y dy-5\int_{3}^{9}dy\ &=\cos x\left[\frac{y^{2}}{2}\right]{3}^{9}-5\left[y\right]{3}^{9}\ &=\cos x\left(\frac{9^{2}}{2}-\frac{3^{2}}{2}\right)-5(9 - 3)\ &=\cos x\left(\frac{81 - 9}{2}\right)-5\times6\ &=36\cos x-30 \end{align*} ]
Step2: Integrate the result with respect to ( x )
Use (\int\cos xdx=\sin x+C) and (\int a dx=ax + C) ((a) is a constant). [ \begin{align*} \int_{0}^{\frac{\pi}{6}}(36\cos x-30)dx&=36\int_{0}^{\frac{\pi}{6}}\cos xdx-30\int_{0}^{\frac{\pi}{6}}dx\ &=36\left[\sin x\right]{0}^{\frac{\pi}{6}}-30\left[x\right]{0}^{\frac{\pi}{6}}\ &=36\left(\sin\frac{\pi}{6}-\sin0\right)-30\left(\frac{\pi}{6}-0\right)\ &=36\times\frac{1}{2}-30\times\frac{\pi}{6}\ &=18 - 5\pi \end{align*} ]
Answer:
(18-5\pi)