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# Integrate the inverse sine of x ---- by parts ???

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I have to integrate the inverse sine of x but I have to do it by parts ???

How do I do that?

Marcop  Oct 18, 2014

#1
+91454
+15

the formula is

$$\\\boxed{\int f(x)g'(x)\;dx=f(x)g(x)-\int f'(x)g(x)\;dx}\\\\ or\\\\ \boxed{\int uv'\;dx=uv-\int u'v\;dx}\\\\$$

$$\\\int sin^{-1}\;dx=\int sin^{-1}*1\;dx\\ let\;u=sin^{-1}\;\;and\;\;v'=1\\\\ u=sin^{-1}x\\ x=sin\;u\\ \frac{dx}{du}=cos\;u\\ \frac{du}{dx}=\frac{1}{cos\;u}\\ \frac{du}{dx}=\frac{1}{\sqrt{cos^2 u}}\\ \frac{du}{dx}=\frac{1}{\sqrt{1-sin^2 u}}\\ \frac{du}{dx}=\frac{1}{\sqrt{1-x^2}}\\$$

$$so\\ u=sin^{-1}x\\ u'=\frac{1}{\sqrt{1-x^2}}\\ v'=1\\ v=x\\\\ \int sin^{-1}x\;dx\\ =\int sin^{-1}*1\;dx\\ =uv-\int u'v\;dx\\ =sin^{-1}x*x-\int \frac{x}{\sqrt{1-x^2}}\;dx\\ =xsin^{-1}x-\int \frac{x}{\sqrt{1-x^2}}\;dx\\\\ let\;\;t=1-x^2\\\\ \frac{dt}{dx}=-2x\\\\ dx=\frac{dt}{-2x}=\frac{dt}{-2\sqrt{1-t}}\\\\ \sqrt{1-t}=x\\\\$$

$$so\\\\ xsin^{-1}x-\int \frac{x}{\sqrt{1-x^2}}\;dx\\\\ =xsin^{-1}x-\int \frac{\sqrt{1-t}}{\sqrt{t}}\;\frac{dt}{-2\sqrt{1-t}}\\\\ =xsin^{-1}x-\int \frac{\sqrt{1-t}}{\sqrt{t}(-2\sqrt{1-t})}\;dt\\\\ =xsin^{-1}x-\int \frac{\sqrt{1-t}}{-2\sqrt{t}\sqrt{1-t}}\;dt\\\\ =xsin^{-1}x-\int \frac{1}{-2\sqrt{t}}\;dt\\\\ =xsin^{-1}x+\frac{1}{2}\int t^{-0.5}\;dt\\\\ =xsin^{-1}x+\frac{1}{2}\times \frac{t^{0.5}}{0.5}+c\\\\ =xsin^{-1}x+ \sqrt{t}+c\\\\ =xsin^{-1}x+ \sqrt{1-x^2}+c\\\\$$

Melody  Oct 18, 2014
Sort:

#1
+91454
+15

the formula is

$$\\\boxed{\int f(x)g'(x)\;dx=f(x)g(x)-\int f'(x)g(x)\;dx}\\\\ or\\\\ \boxed{\int uv'\;dx=uv-\int u'v\;dx}\\\\$$

$$\\\int sin^{-1}\;dx=\int sin^{-1}*1\;dx\\ let\;u=sin^{-1}\;\;and\;\;v'=1\\\\ u=sin^{-1}x\\ x=sin\;u\\ \frac{dx}{du}=cos\;u\\ \frac{du}{dx}=\frac{1}{cos\;u}\\ \frac{du}{dx}=\frac{1}{\sqrt{cos^2 u}}\\ \frac{du}{dx}=\frac{1}{\sqrt{1-sin^2 u}}\\ \frac{du}{dx}=\frac{1}{\sqrt{1-x^2}}\\$$

$$so\\ u=sin^{-1}x\\ u'=\frac{1}{\sqrt{1-x^2}}\\ v'=1\\ v=x\\\\ \int sin^{-1}x\;dx\\ =\int sin^{-1}*1\;dx\\ =uv-\int u'v\;dx\\ =sin^{-1}x*x-\int \frac{x}{\sqrt{1-x^2}}\;dx\\ =xsin^{-1}x-\int \frac{x}{\sqrt{1-x^2}}\;dx\\\\ let\;\;t=1-x^2\\\\ \frac{dt}{dx}=-2x\\\\ dx=\frac{dt}{-2x}=\frac{dt}{-2\sqrt{1-t}}\\\\ \sqrt{1-t}=x\\\\$$

$$so\\\\ xsin^{-1}x-\int \frac{x}{\sqrt{1-x^2}}\;dx\\\\ =xsin^{-1}x-\int \frac{\sqrt{1-t}}{\sqrt{t}}\;\frac{dt}{-2\sqrt{1-t}}\\\\ =xsin^{-1}x-\int \frac{\sqrt{1-t}}{\sqrt{t}(-2\sqrt{1-t})}\;dt\\\\ =xsin^{-1}x-\int \frac{\sqrt{1-t}}{-2\sqrt{t}\sqrt{1-t}}\;dt\\\\ =xsin^{-1}x-\int \frac{1}{-2\sqrt{t}}\;dt\\\\ =xsin^{-1}x+\frac{1}{2}\int t^{-0.5}\;dt\\\\ =xsin^{-1}x+\frac{1}{2}\times \frac{t^{0.5}}{0.5}+c\\\\ =xsin^{-1}x+ \sqrt{t}+c\\\\ =xsin^{-1}x+ \sqrt{1-x^2}+c\\\\$$

Melody  Oct 18, 2014
#2
+81022
+5

Very nice, Melody.

CPhill  Oct 18, 2014
#3
+91454
0

Thanks Chris

Melody  Oct 18, 2014

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