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# Integrate (cos^-1x)^2 /(1-x^2)^1/2

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3147
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Integrate (cos^-1x)^2 /(1-x^2)^1/2

Jun 19, 2015

#2
+117762
+13

Thanks Syllogist.  You're right of course, I just want to try and explain.

$$\\\int\;\frac{(cos^{-1}x)^2 }{(1-x^2)^{1/2}}\;dx\\\\ What I notice is that \;\; \frac{d}{dx}cos^{-1}x=\frac{-1}{\sqrt{1-x^2}}\\\\ So I think the answer will be  some constant* (cos^{-1}x)^3 \;\;+c\\\\ Now if i differentiated this (without the constant multiple) I get\\\\ \frac{d}{dx}(cos^{-1})^3x=3(cos^{-1})^2x\times \frac{-1}{\sqrt{1-x^2}}\\\\ I need to get rid of the - and the 3 so I divide by -3 (that will be the constant multiple\\\\ \int\;\frac{(cos^{-1}x)^2 }{(1-x^2)^{1/2}}\;dx=\frac{-1}{3}cos^{-1}x+c$$

Just like Syllogist said:))

NOTE: I can never remember that inverse trig differential, I have to work it out from scratch every time.

(It doesn't take long to work out)

Jun 19, 2015

#1
+125
+8
-1/3 cos^(-1)(x)^3 + constant
Jun 19, 2015
#2
+117762
+13

Thanks Syllogist.  You're right of course, I just want to try and explain.

$$\\\int\;\frac{(cos^{-1}x)^2 }{(1-x^2)^{1/2}}\;dx\\\\ What I notice is that \;\; \frac{d}{dx}cos^{-1}x=\frac{-1}{\sqrt{1-x^2}}\\\\ So I think the answer will be  some constant* (cos^{-1}x)^3 \;\;+c\\\\ Now if i differentiated this (without the constant multiple) I get\\\\ \frac{d}{dx}(cos^{-1})^3x=3(cos^{-1})^2x\times \frac{-1}{\sqrt{1-x^2}}\\\\ I need to get rid of the - and the 3 so I divide by -3 (that will be the constant multiple\\\\ \int\;\frac{(cos^{-1}x)^2 }{(1-x^2)^{1/2}}\;dx=\frac{-1}{3}cos^{-1}x+c$$

Just like Syllogist said:))

NOTE: I can never remember that inverse trig differential, I have to work it out from scratch every time.

(It doesn't take long to work out)

Melody Jun 19, 2015
#3
+26319
+10

Integrate (cos^-1x)^2 /(1-x^2)^1/2

$$\small{\text{ \begin{array}{lrcl} & \int{ \dfrac{[ \cos^{-1}{(x)} ]^2 } { (1-x^2)^{\frac{1}{2}} } \ dx } = \mathbf{ \int{ [\arccos{(x)}]^2 \cdot \dfrac{ 1 }{ \sqrt{ (1-x^2) } } \ dx } }\\\\ \end{array} }}\\\\$$

$$\small{\text{ \boxed{ \begin{array}{lrcl} \mathrm{Formula:~} & y &=& \dfrac{ f(x)^{n+1} } {n+1} \\\\ & y' &=& \left(\dfrac{n+1}{n+1}\right) \cdot f(x)^{n+1-1}\cdot f'(x) \\\\ & y' &=& f(x)^{n}\cdot f'(x) \\\\ &\mathbf{ \int{ f(x)^{n}\cdot f'(x) \ dx } } & \mathbf{=} & \mathbf{ \dfrac{ f(x)^{n+1} } {n+1} } \end{array}} }}$$

$$\small{\text{ \begin{array}{lrcl} &\mathbf{ \int{ f(x)^{n}\cdot f'(x) \ dx } } & \mathbf{=} & \mathbf{ \dfrac{ f(x)^{n+1} } {n+1} } \qquad f(x) = \arccos{(x)} \qquad f'(x)= -\dfrac{ 1 }{ \sqrt{ (1-x^2) } }\\\\ & \int{ \left[ \arccos{(x)} \right]^2 \cdot \left( -\dfrac{ 1 }{ \sqrt{ (1-x^2) } } \right) \ dx } & = & \dfrac{ \left[ \arccos{(x)} \right]^{3} } {3} \\\\ & -\int{ \left[ \arccos{(x)} \right]^2 \cdot \left( \dfrac{ 1 }{ \sqrt{ (1-x^2) } } \right) \ dx } & = & \dfrac{ \left[ \arccos{(x)} \right]^{3} } {3} \\\\ & \int{ \left[ \arccos{(x)} \right]^2 \cdot \left( \dfrac{ 1 }{ \sqrt{ (1-x^2) } } \right) \ dx } & = & -\dfrac{ \left[ \arccos{(x)} \right]^{3} } {3} \\\\ & \int{ \left[ \arccos{(x)} \right] ^2 \cdot \left( \dfrac{ 1 }{ \sqrt{ (1-x^2) } } \right) \ dx } & = & -\dfrac{1}{3} \cdot \left[ \arccos{(x)} \right]^{3} \\\\ & \int{ \left[ \arccos{(x)} \right]^2 \cdot \left( \dfrac{ 1 }{ \sqrt{ (1-x^2) } } \right) \ dx } & = & -\dfrac{1}{3} \cdot \left[\cos^{-1}{(x)}\right]^3 \\\\ \end{array} }}$$

Jun 19, 2015
#4
+33053
+10

Another approach:

$$\\ \text{Let }\theta=\cos^{-1}x\\\\ x=\cos\theta\quad\text{so }dx=-\sin\theta.d\theta\\\\ \sqrt{1-x^2}=\sqrt{1-\cos^2\theta}=\sin\theta\\\\ \text{So }\\\\ \int\frac{(\cos^{-1}x)^2}{\sqrt{1-x^2}}dx=-\int\frac{\theta^2.\sin\theta}{\sin\theta}d\theta=-\int\theta^2d\theta=-\frac{\theta^3}{3}=-\frac{(\cos^{-1}x)^3}{3}$$

There is a constant to be added of course.

.

Jun 19, 2015
#5
+124524
+10

Here's one more way using  u-du substitution.......

∫ (arccos(x))^2 / √(1 - x^2)   dx

Let  u   = arccos(x)      du = -1/ √(1 - x^2)   dx    implies that.....-du = 1/ √(1 - x^2)   dx

So we have......

-∫ u^2 du =     -(1/3)u^3  + C    =  -(1/3)[arccos(x)] + C

Jun 19, 2015
#6
+117762
+5

Thanks Heureka and Chris,

I really like Alan's solution here.

It is so elegant.

And, once placed in front of me, it is so 'obvious'    LOL

What is the chances of it being obvious (to me) next time without Alan answering first.