Solve G ( x ) = ∫ (from cos x to sin x) of t^5dt

$$G ( x ) = \int _ { \cos x } ^ { \sin x } t ^ { 5 } d t$$

Evaluate

$$-\frac{\cos(2x)\left(\cos(4x)+7\right)}{48}$$

Solution Steps

Evaluate the indefinite integral first.

$$\int t^{5}\mathrm{d}t$$

Since \int t^{k}\mathrm{d}t=\frac{t^{k+1}}{k+1} for k\neq -1, replace \int t^{5}\mathrm{d}t with \frac{t^{6}}{6}.

$$\frac{t^{6}}{6}$$

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.

$$\frac{1}{6}\left(\sin(x)\right)^{6}-\frac{1}{6}\left(\cos(x)\right)^{6}$$

Simplify.

$$-\frac{1}{48}\cos(2x)\left(7+48\times \frac{\cos(4x)}{48}\right)$$

Differentiate w.r.t. x

$$\frac{\sin(2x)\left(\cos(4x)+7\right)+2\sin(4x)\cos(2x)}{24}$$