My Math Forum solution of integral

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 December 2nd, 2012, 03:27 PM #1 Member   Joined: Apr 2012 Posts: 72 Thanks: 3 solution of integral cot(cos^-1 x)
 December 2nd, 2012, 06:36 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 465 Math Focus: Calculus/ODEs Re: solution of integral Rewrite the integrand using the identity: $\cot$$\cos^{-1}(x)$$=\frac{x}{\sqrt{1-x^2}}$ then use the substitution: $u=1-x^2\,\therefore\,du=-2x\,dx$ to get: $-\frac{1}{2}\int_0\,^1 u^{-\frac{1}{2}}\,du$
 December 2nd, 2012, 09:38 PM #3 Member   Joined: Apr 2012 Posts: 72 Thanks: 3 Re: solution of integral Thanks.
 December 2nd, 2012, 10:38 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 465 Math Focus: Calculus/ODEs Re: solution of integral Use the power-rule for integration: $\int u^r\,du=\frac{u^{r+1}}{r+1}+C$ where $r\not=-1$ along with the anti-derivative form of the fundamental theorem of calculus: $\int_a\,^b f(u)\,du=F(b)-F(a)$ where $\frac{dF}{du}=f(u)$.
 December 3rd, 2012, 12:33 AM #5 Member   Joined: Apr 2012 Posts: 72 Thanks: 3 Re: solution of integral Can someone please post the way from start to the end?
 December 3rd, 2012, 12:40 AM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 465 Math Focus: Calculus/ODEs Re: solution of integral You cannot evaluate the resulting definite integral using the rules I posted?
 December 3rd, 2012, 12:51 AM #7 Member   Joined: Apr 2012 Posts: 72 Thanks: 3 Re: solution of integral I tried, but I have an error somewhere. I got a division by zero.
 December 3rd, 2012, 01:05 AM #8 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 465 Math Focus: Calculus/ODEs Re: solution of integral $-\frac{1}{2}\int_0\,^1 u^{-\frac{1}{2}}\,du=-\frac{1}{2}$\frac{u^{\frac{1}{2}}}{\frac{1}{2}}$_0^1=-1$$1^{\frac{1}{2}}-0^{\frac{1}{2}}$$=-1$
 December 3rd, 2012, 01:23 AM #9 Member   Joined: Apr 2012 Posts: 72 Thanks: 3 Re: solution of integral How did you do this first step? $\cot$$\cos^{-1}(x)$$=\frac{x}{\sqrt{1-x^2}}$
 December 3rd, 2012, 01:29 AM #10 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 465 Math Focus: Calculus/ODEs Re: solution of integral $\cos^{-1}(x)$ is an angle in a right triangle where the ratio of the side adjacent to that angle and the hypotenuse is $x$, so we may say that the adjacent side is $x$ and the hypotenuse is 1. By the Pythagorean theorem, we may then say the side opposite the angle is $\sqrt{1-x^2}$. Since the cotangent function by definition is the ratio of the adjacent to the opposite, we then have: $\cot$$\cos^{-1}(x)$$=\frac{x}{\sqrt{1-x^2}}$ When I was first learning this, I would draw a right triangle so that I could clearly see what I needed to find.

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