Question

Evaluate the integral from 0 to pi/4 for [(1/(cos(x))squared) + sin (2x)] dx

Answer 1

\[\frac{1}{\cos^2(x)}=\sec^2(x)\] and an anti derivative for secant squared is
\[\tan(x)\] is a start

Answer 2

\[\sin(2x)=2\sin(x)\cos(x)\] and that has
\[\sin^2(x)\]as an anti-derivative

Answer 3

so you have
\[\tan(x)+\sin^2(x)\] as your anti derivative. evaluate at upper limit, lower limit and subtract

Answer 4

\[\int\limits_{0}^{\pi/4} [(\frac{1}{\cos(x)^{2}} + \sin (2x)] dx= Tan(x)-\frac{1}{2}Cos(2x)\]
Put in the limits of integration and Bingo.....so \[\frac{1}{2}+1=\frac{3}{2}\]

Answer 5

thanks!