$$\int\limits_{}^{} \arccos(x)dx = \arccos(x)x + \int\limits_{}^{} xdx/\sqrt{1-x^{2}}$$
so I used substitution

u=x^(2)
du/2 = xdx

so I have

$$\arccos(x) + (1/2)\int\limits_{}^{} du/\sqrt{1-u}$$

Then I used trig substitution

u = sin^(2)(s)
(u)^(1/2) = sin(s)
s = arcsin(u^(1/2))

du/ds = 2cos(s)sin(s)
du = cos(s)sin(s)ds

so I make the substitution

$$(2/2)\int\limits_{}^{} \cos(s)\sin(s)ds/\sqrt{1-\sin^{2}} = \int\limits_{}^{} \cos(s)\sin(s)ds/\sqrt{\cos^{2}(s)} = $$

$$\int\limits_{}^{} \cos(s)\sin(s)ds/\cos(s) = \int\limits_{}^{} \sin(s)ds = -\cos(s) + c$$

Question

$$\int\limits_{}^{} \arccos(x)dx  = \arccos(x)x + \int\limits_{}^{} xdx/\sqrt{1-x^{2}}$$
so I used substitution

u=x^(2)
du/2 = xdx

so I have

$$\arccos(x) + (1/2)\int\limits_{}^{} du/\sqrt{1-u}$$

Then I used trig substitution

u = sin^(2)(s)
(u)^(1/2) = sin(s)
s = arcsin(u^(1/2))

du/ds = 2cos(s)sin(s)
du = cos(s)sin(s)ds

so I make the substitution

$$(2/2)\int\limits_{}^{}  \cos(s)\sin(s)ds/\sqrt{1-\sin^{2}} = \int\limits_{}^{}  \cos(s)\sin(s)ds/\sqrt{\cos^{2}(s)} = $$

$$\int\limits_{}^{}  \cos(s)\sin(s)ds/\cos(s) = \int\limits_{}^{}  \sin(s)ds = -\cos(s) + c$$

anonymous · Answer

$$\arccos(x) - \cos(s) + c = \arccos(x) - \cos(\arcsin(x)) + c$$

Can you please check my answer?

experimentx · Answer

the answer is right ... checked from wolfram http://www.wolframalpha.com/input/?i=integrate+arccos+x+dx

anonymous · Answer

thanks :) I forgot the identities so I didnt k now how to convert it over plus im trying to get the method stuck in my head exams are soon :)

experimentx · Answer

which plus??

anonymous · Answer

I think you missinterpreted what I said dont worry about it. Thanks for confirming for me :) I really appreciated it