Evaluate the integral sinx/cos^2x dx from 0 to pi/3.
So I've gotten it to tanxcosx, but I don't know the indefinite integral of these, and I couldn't find it in my book. Can anyone help?
$\displaystyle \displaystyle \int _0 ^{\frac {\pi}{3} } \frac {\sin{x}}{\cos^2{x}} \; dx = \int _0 ^{\frac {\pi}{3} } \sec{x} \tan{x} \; dx =$
$\displaystyle \displaystyle u = \sec {x} $ and now the $\displaystyle \displaystyle du= \tan{x}\sec{x}\; dx$
after that ... i just got
$\displaystyle \displaystyle \int _0 ^{\frac {\pi}{3} }\;du = u \big{|} _0 ^{\frac {\pi}{3} }= \sec{x} \big{|} _0 ^{\frac {\pi}{3} }$
and with limits you get "1"
Just use a $\displaystyle u$ substitution...
$\displaystyle \int{\frac{\sin{x}}{\cos^2{x}}\,dx} = -\int{\cos^{-2}{x}(-\sin{x})\,dx}$
Let $\displaystyle u=\cos{x}$ so that $\displaystyle du = -\sin{x}\,dx$ and the integral becomes
$\displaystyle -\int{u^{-2}\,du}$
$\displaystyle = -\left(\frac{u^{-1}}{-1}\right) + C$
$\displaystyle = \frac{1}{u} + C$
$\displaystyle = \frac{1}{\cos{x}} + C$.
Therefore $\displaystyle \int_0^{\frac{\pi}{3}}{\frac{\sin{x}}{\cos^2{x}}\, dx} = \left[\frac{1}{\cos{x}}\right]_0^{\frac{\pi}{3}}$
$\displaystyle = \left[\frac{1}{\cos{\left(\frac{\pi}{3}\right)}}\right] - \left[\frac{1}{\cos{(0)}}\right]$
$\displaystyle = \frac{1}{\frac{1}{2}} - \frac{1}{1}$
$\displaystyle = 2 - 1$
$\displaystyle = 1$.
You forgot to use the chain rule when finding du.
$\displaystyle \begin{aligned}
u &= \sec 2x \\
du &= 2\sec 2x\, \tan 2x \, dx \\
\frac{1}{2}du &= \sec 2x\, \tan 2x \, dx
\end{aligned}$
$\displaystyle \displaystyle \int \sec 2x\,\tan 2x\, dx$
$\displaystyle = \displaystyle \int \frac{1}{2} du$
Can you take it from here?