What is the area enclosed by #r=5cos^3((5theta)/6-(pi)/2)+theta^2/2 # between #theta in [0,pi/2]#?

To find the area enclosed by the curve ( r = 5\cos^3\left(\frac{5\theta}{6} - \frac{\pi}{2}\right) + \frac{\theta^2}{2} ) between ( \theta ) in ( [0, \frac{\pi}{2}] ), we use the formula for the area ( A ) enclosed by a polar curve ( r = f(\theta) ) between ( \theta = \alpha ) and ( \theta = \beta ), which is given by:

[ A = \frac{1}{2} \int_{\alpha}^{\beta} [f(\theta)]^2 , d\theta ]

In this case, the given curve is ( r = 5\cos^3\left(\frac{5\theta}{6} - \frac{\pi}{2}\right) + \frac{\theta^2}{2} ), and ( \theta ) varies from 0 to ( \frac{\pi}{2} ). So, the area is:

[ A = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \left[5\cos^3\left(\frac{5\theta}{6} - \frac{\pi}{2}\right) + \frac{\theta^2}{2}\right]^2 , d\theta ]

We need to simplify the integrand before integrating:

[ \left[5\cos^3\left(\frac{5\theta}{6} - \frac{\pi}{2}\right) + \frac{\theta^2}{2}\right]^2 ] [ = \left[5\cos\left(\frac{5\theta}{6} - \frac{\pi}{2}\right)\right]^2 + 2 \cdot 5\cos\left(\frac{5\theta}{6} - \frac{\pi}{2}\right) \cdot \frac{\theta^2}{2} + \left(\frac{\theta^2}{2}\right)^2 ] [ = 25\cos^2\left(\frac{5\theta}{6} - \frac{\pi}{2}\right) + 25\cos^2\left(\frac{5\theta}{6} - \frac{\pi}{2}\right)\frac{\theta^2}{2} + \frac{\theta^4}{4} ] [ = 25\cos^2\left(\frac{5\theta}{6} - \frac{\pi}{2}\right) + \frac{25}{2}\theta^2\cos^2\left(\frac{5\theta}{6} - \frac{\pi}{2}\right) + \frac{\theta^4}{4} ]

Now, we can integrate this expression from 0 to ( \frac{\pi}{2} ) to find the area. However, the integration of this expression is quite involved and may require numerical methods due to the presence of trigonometric functions and powers.