Solve ∫ _{-pi/2}^frac{pi{2}}6cosθsin^2θ=

$$\int _ { - \frac { \pi } { 2 } } ^ { \frac { \pi } { 2 } } 6 \cos \theta \sin ^ { 2 } \theta =$$

Evaluate

$$3\pi \sin(\theta )\sin(2\theta )$$

Solution Steps

Evaluate the indefinite integral first.

$$\int 6\cos(\theta )\left(\sin(\theta )\right)^{2}\mathrm{d}x$$

Find the integral of 6\cos(\theta )\left(\sin(\theta )\right)^{2} using the table of common integrals rule \int a\mathrm{d}x=ax.

$$6\cos(\theta )\left(\sin(\theta )\right)^{2}x$$

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.

$$\frac{1}{2}\times 6\cos(\theta )\left(\sin(\theta )\right)^{2}\pi +\frac{1}{2}\times 6\cos(\theta )\left(\sin(\theta )\right)^{2}\pi$$

Simplify.

$$3\pi \sin(2\theta )\sin(\theta )$$

Differentiate w.r.t. θ

$$3\pi \left(\sin(2\theta )\cos(\theta )+2\sin(\theta )\cos(2\theta )\right)$$