Hi, Im having a little bit of trouble with the following: [tex]\int sintsec^2(cost)dt[/tex] heres what I have so far [tex]u=cost[/tex] [tex]du=-sintdt[/tex] [tex]-\int sec^2(u)du[/tex] [tex]-2tan(u) + C[/tex] is this right?
I have another..i dont know where to start..can someone point out what I should sub U for? [tex]\int_{1/2}^{1/6}csc \pi t cot \pi t dt[/tex]
ok..so i did.. [tex] \int_{1/2}^{1/6} \frac{1}{sin\pi t) \frac{cos \pi t}{sin \pi t}dt[/tex] ended up with... [tex] \int_{1/2}^{1/6} cot \pi t[/tex] im guessing its wrong haha
It's almost there [tex]\frac{1}{\sin \pi t} \cdot \frac{\cos \pi t}{\sin \pi t} = \frac{\cos \pi t}{\sin^2 \pi t}[/tex] Now, see what you can do with the identity [itex]\sin^2 \theta + \cos^2 \theta = 1[/itex]. Btw, I think this can be done without a substitution. ~H
As others have pointed out, the anti-derivative of sec^{2}(u) is tan(u), not -tan(u). Also, the original problem does not say anything about "u"! That was your "invention". To properly answer the question, you need to go back to t: [tex]\int sintsec^2(cost)dt= 2tan(cos(t)+ C[/tex]
Actually, the original poster (with respect to the original question) is correct, save for the two. [tex] \int sin(t) sec (cos(x))^2 dx [/tex] [tex] u = cos(x) du = -sin(x) dx [/tex] This is right. Substituting back yields [tex] - \int sec(u)^2 du [/tex] Just as the OP said. The negative sign is the result of the du = -sin(x) dx part. Now, the antiderivative of sec(x)^2 = tan(x) + C (as everyone as stated) - ( tan(u) + C) -tan(cos(x) + C. The original poster's only error is the two infront. The minus sign is correct . Check with differentiatoin (or use the "Integrator").