# How do you use the second fundamental theorem of Calculus to find the derivative of given int cos 2t dt from x to pi/4?

Jan 26, 2016

$\frac{1}{4} \cos \left(\frac{x}{2}\right) - 2 \cos 2 x$.

#### Explanation:

Using the substitution technique, together with appropriate changing of limits for the substituted variable, we get :

${\int}_{x}^{\frac{\pi}{4}} \cos 2 t \mathrm{dt} = \frac{1}{2} {\int}_{2 x}^{\frac{\pi}{2}} \cos u \mathrm{du}$, where $u = 2 t$

$= \frac{1}{2} {\left[\sin u\right]}_{2 x}^{\frac{x}{2}}$

$= \frac{1}{2} \left(\sin \left(\frac{x}{2}\right) - \sin 2 x\right)$.

We may now differentiate this result to obtain :

d/dx[1/2(sin(x/2)-sin2x]=1/4cos(x/2)-2cos2x.