Solve = ∫ (from 0 to pi / 2) of (9/2w^2cosu-18vcos^2u)|_0^5du=

$$= \int _ { 0 } ^ { \pi / 2 } ( \frac { 9 } { 2 } w ^ { 2 } \cos u - 18 v \cos ^ { 2 } u ) | _ { 0 } ^ { 5 } d u =$$

Evaluate

$$\frac{9\pi \cos(u)\left(-4v\cos(u)+w^{2}\right)}{4}$$

Solution Steps

Evaluate the indefinite integral first.

$$\int \frac{9w^{2}\cos(u)}{2}-18v\left(\cos(u)\right)^{2}\mathrm{d}x$$

Find the integral of \frac{9w^{2}\cos(u)}{2}-18v\left(\cos(u)\right)^{2} using the table of common integrals rule \int a\mathrm{d}x=ax.

$$\left(\frac{9w^{2}\cos(u)}{2}-18v\left(\cos(u)\right)^{2}\right)x$$

The definite integral is the antiderivative of the expression evaluated at the upper limit of integration minus the antiderivative evaluated at the lower limit of integration.

$$\frac{1}{2}\left(\frac{9}{2}w^{2}\cos(u)-18v\left(\cos(u)\right)^{2}\right)\pi +0\left(\frac{9}{2}w^{2}\cos(u)-18v\left(\cos(u)\right)^{2}\right)$$

Simplify.

$$\frac{9\cos(u)\left(w^{2}-4\cos(u)v\right)\pi }{4}$$

Differentiate w.r.t. w

$$\frac{9\pi w\cos(u)}{2}$$