hey can anyone help me evaluate the integral: cot(x)/ln[sinx] dx
B br8909 New member Joined Oct 13, 2009 Messages 1 Oct 13, 2009 #1 hey can anyone help me evaluate the integral: cot(x)/ln[sinx] dx
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Oct 13, 2009 #2 This one is actually easier than it looks. ∫cot(x)ln(sin(x))dx\displaystyle \int\frac{cot(x)}{ln(sin(x))}dx∫ln(sin(x))cot(x)dx Let u=ln(sin(x)), du=cot(x)dx\displaystyle u=ln(sin(x)), \;\ du=cot(x)dxu=ln(sin(x)), du=cot(x)dx Now, make the subs and get: ∫1udu\displaystyle \int\frac{1}{u}du∫u1du Integrate and resub. That's it.
This one is actually easier than it looks. ∫cot(x)ln(sin(x))dx\displaystyle \int\frac{cot(x)}{ln(sin(x))}dx∫ln(sin(x))cot(x)dx Let u=ln(sin(x)), du=cot(x)dx\displaystyle u=ln(sin(x)), \;\ du=cot(x)dxu=ln(sin(x)), du=cot(x)dx Now, make the subs and get: ∫1udu\displaystyle \int\frac{1}{u}du∫u1du Integrate and resub. That's it.
B BigGlenntheHeavy Senior Member Joined Mar 8, 2009 Messages 1,577 Oct 13, 2009 #3 ∫cot(x)ln∣sin(x)∣dx = ∫cos(x)dxsin(x)[ln∣sin(x)∣]\displaystyle \int\frac{cot(x)}{ln|sin(x)|}dx \ = \ \int\frac{cos(x)dx}{sin(x)[ln|sin(x)|]}∫ln∣sin(x)∣cot(x)dx = ∫sin(x)[ln∣sin(x)∣]cos(x)dx Let u = sin(x), then du = cos(x)dx\displaystyle Let \ u \ = \ sin(x), \ then \ du \ = \ cos(x)dxLet u = sin(x), then du = cos(x)dx Ergo, ∫duu[ln∣u∣]\displaystyle Ergo, \ \int\frac{du}{u[ln|u|]}Ergo, ∫u[ln∣u∣]du Now, let k = ln∣u∣, then dk = 1udu\displaystyle Now, \ let \ k \ = \ ln|u|, \ then \ dk \ = \ \frac{1}{u}duNow, let k = ln∣u∣, then dk = u1du Hence, we now have ∫dkk = ∫k−1dk =ln∣k∣+C.\displaystyle Hence, \ we \ now \ have \ \int\frac{dk}{k} \ = \ \int k^{-1}dk \ = \ln|k|+C.Hence, we now have ∫kdk = ∫k−1dk =ln∣k∣+C. Resubstituting, we get ln[ln∣u∣]+C = ln[ln∣sin(x)∣] +C QED.\displaystyle Resubstituting, \ we \ get \ ln[ln|u|]+C \ = \ ln[ln|sin(x)|] \ +C \ QED.Resubstituting, we get ln[ln∣u∣]+C = ln[ln∣sin(x)∣] +C QED.
∫cot(x)ln∣sin(x)∣dx = ∫cos(x)dxsin(x)[ln∣sin(x)∣]\displaystyle \int\frac{cot(x)}{ln|sin(x)|}dx \ = \ \int\frac{cos(x)dx}{sin(x)[ln|sin(x)|]}∫ln∣sin(x)∣cot(x)dx = ∫sin(x)[ln∣sin(x)∣]cos(x)dx Let u = sin(x), then du = cos(x)dx\displaystyle Let \ u \ = \ sin(x), \ then \ du \ = \ cos(x)dxLet u = sin(x), then du = cos(x)dx Ergo, ∫duu[ln∣u∣]\displaystyle Ergo, \ \int\frac{du}{u[ln|u|]}Ergo, ∫u[ln∣u∣]du Now, let k = ln∣u∣, then dk = 1udu\displaystyle Now, \ let \ k \ = \ ln|u|, \ then \ dk \ = \ \frac{1}{u}duNow, let k = ln∣u∣, then dk = u1du Hence, we now have ∫dkk = ∫k−1dk =ln∣k∣+C.\displaystyle Hence, \ we \ now \ have \ \int\frac{dk}{k} \ = \ \int k^{-1}dk \ = \ln|k|+C.Hence, we now have ∫kdk = ∫k−1dk =ln∣k∣+C. Resubstituting, we get ln[ln∣u∣]+C = ln[ln∣sin(x)∣] +C QED.\displaystyle Resubstituting, \ we \ get \ ln[ln|u|]+C \ = \ ln[ln|sin(x)|] \ +C \ QED.Resubstituting, we get ln[ln∣u∣]+C = ln[ln∣sin(x)∣] +C QED.
