Someone please help me! I don't get this problem...
Use substitution to find: the integral from 0 to pi/3 of sinx/cos(squared)x dx
Hello juicysharpie,
iam just learning mathematics ,but i think i can answer this question
let cos x = t
differentiating this
-sin x dx = dt
so the numerator can be replaced by -dt and the denominator by tē.
the limits are changed to t
when x= 0,t becomes cos 0 and hence 1
when x= pi/3 ,t becomes cos pi/3 and hence 1/2 .these are the limits of integration.
hope you can do this from here.
there are some very good mathematicians here ,but i just tried because i thought i could answer this.
$\displaystyle \int\limits_{0}^{\frac{\pi}{3}} \frac{sinx}{(cosx)^2} dx$
$\displaystyle u = cos x$
$\displaystyle du = - sinx dx$
$\displaystyle -du = sinx dx$
The substitution gets rid of the sinx and dx in your integral which leaves you: ** don't forget to put that negative out in front of your integral!!
$\displaystyle - \int\limits_{0}^{\frac{\pi}{3}} \frac{1}{(u)^2} du$
You could plug cos x back in for u and solve using the same limits on your integral OR like Krizalid said you can integrate with the u in there and change your limits based on using u instead of cos x.
If you are just learning to do u-substitution it might be better to plug cos x back in until you are comfortable with the concept of u-substitution.
FYI, if you did want to change your limits so you could integrate leaving u in there all you do is plug your old limits into the "x" of your u = cos x equation. So
$\displaystyle u = cos (0) = 1$ (would be your new lower limit)
$\displaystyle u = cos(\frac{\pi}{3} = \frac{1}{2} $(would be your new upper limit)