integration by substitution help.

Hi,
im struggling to integrate this using the given substitution, Can someone please help.

Thanks.

$\int_{0}^{2\pi}\frac{1}{7-5sin(x)}$

using $u=tan(\frac{x}{2})$

You can't "simply" do the t-sub (which is just the tan x/2 sub you're doing) in this case because of the limits.

However, note that sin(pi-x) = sin(x) so that $\displaystyle \int_0^{2\pi}\dfrac{1}{7-5\sin x}\,dx = 2 \int_0^{\pi} \dfrac{1}{7 - 5\sin x}\, dx$ (and in fact it also equals $\displaystyle 4 \int_0^{\pi/2} \dfrac{1}{7 - 5\sin x}\, dx$. Both of the latter integrals have "nice" limits after doing the substitution.

Thanks very much for your help. However, we are still struggling with the integration. Please could you show us how to do this?

ermm..we?

Post your working.

Also try googling 't-substitution'

e.g. http://www.ucl.ac.uk/Mathematics/geomath/level2/fint/fisub4.html

Since u = tan(x/2), the limits will change. (I'm also unconvinced you'd done the manipulation right, but since you haven't posted working I can't check it easily).

N.B. Amended version of incorrect earlier post. Sorry about that! I think this is now correct.

You can't "simply" do the t-sub (which is just the tan x/2 sub you're doing) in this case because of the limits.

However, note that sin(pi-x) = sin(x) so that $\displaystyle \int_0^{2\pi}\dfrac{1}{7-5\sin x}\,dx = 2 \int_0^{\pi/2} \dfrac{1}{7 - 5\sin x}\, dx + 2 \int_{3\pi/2}^{2\pi} \dfrac{1}{7 - 5\sin x}\, dx$ and that this also equals $\displaystyle 2 \int_{-\pi/2}^{\pi/2}\dfrac{1}{7 - 5\sin x}\, dx$. This final integral has "nice" limits after doing the substitution.

You can't "simply" do the t-sub (which is just the tan x/2 sub you're doing) in this case because of the limits.

However, note that sin(pi-x) = sin(x) so that $\displaystyle \int_0^{2\pi}\dfrac{1}{7-5\sin x}\,dx = 2 \int_0^{\pi} \dfrac{1}{7 - 5\sin x}\, dx$ (and in fact it also equals $\displaystyle 4 \int_0^{\pi/2} \dfrac{1}{7 - 5\sin x}\, dx$. Both of the latter integrals have "nice" limits after doing the substitution.

Edit: It has been pointed out this is incorrect. See post #18.