Re: Integral problem
Hello, grapz!
This is a messy one . . .
∫ arcsin ( 2 x 1 + x 2 ) d x \displaystyle \int \arcsin\left(\frac{2x}{1+x^2}\right)\,dx ∫ arcsin ( 1 + x 2 2 x ) d x
I'm going to change the inverse trig function . . .
Let: θ = arcsin ( 2 x 1 + x 2 ) ⇒ sin θ = 2 x 1 + x 2 = o p p h y p \displaystyle \text{Let: }\:\theta \:=\:\arcsin\left(\frac{2x}{1+x^2}\right) \quad\Rightarrow\quad \sin\theta \:=\:\frac{2x}{1+x^2} \:=\:\frac{opp}{hyp} Let: θ = arcsin ( 1 + x 2 2 x ) ⇒ sin θ = 1 + x 2 2 x = h y p o p p
\(\displaystyle \theta\text{ is in a right triangle with: }\
pp \,= \,2x,\;hyp \,= \,1+x^2\)
\(\displaystyle \text{Hence: }\:adj \,=\,1-x^2\quad\hdots\;\;\text{and: }\:\tan\theta \:= \:\frac{2x}{1-x^2} \quad\Rightarrow\quad\theta\;=\;\arctan\left(\frac{2x}{1-x^2}\right)\)
The integral becomes: ∫ arctan ( 2 x 1 − x 2 ) d x \displaystyle \text{The integral becomes: }\;\int\arctan\left(\frac{2x}{1-x^2}\right)dx The integral becomes: ∫ arctan ( 1 − x 2 2 x ) d x
Integrate by parts . . .
. . ** . u = arctan ( 2 x 1 − x 2 ) d v = d x d u = 2 d x 1 + x 2 v = x \displaystyle \begin{array}{ccccccc}u &=&\arctan\left(\dfrac{2x}{1-x^2}\right) & & dv &=& dx \\ du &=&\dfrac{2\,dx}{1+x^2} & & v &=& x \end{array} u d u = = arctan ( 1 − x 2 2 x ) 1 + x 2 2 d x d v v = = d x x Then we have: x ⋅ arctan ( 2 x 1 − x 2 ) − ∫ 2 x 1 + x 2 d x \displaystyle \text{Then we have: }\;x\cdot\arctan\left(\frac{2x}{1-x^2}\right) - \int\frac{2x}{1+x^2}\,dx Then we have: x ⋅ arctan ( 1 − x 2 2 x ) − ∫ 1 + x 2 2 x d x . . . . . . = x ⋅ arctan ( 2 x 1 − x 2 ) − ln ( 1 + x 2 ) + C \displaystyle =\;\boxed{x\cdot\arctan\left(\frac{2x}{1-x^2}\right) - \ln(1 + x^2) + C} = x ⋅ arctan ( 1 − x 2 2 x ) − ln ( 1 + x 2 ) + C
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** We have: u = arctan ( 2 x 1 − x 2 ) \displaystyle \text{We have: }\;u \;=\;\arctan\left(\frac{2x}{1-x^2}\right) We have: u = arctan ( 1 − x 2 2 x ) Differentiate: \displaystyle \text{Differentiate:} Differentiate: d u d x = 1 1 + ( 2 x 1 − x 2 ) 2 ⋅ ( 1 − x 2 ) ⋅ 2 − ( 2 x ) ( − 2 x ) ( 1 − x 2 ) 2 = ( 1 − x 2 ) 2 ( 1 − x 2 ) 2 + 4 x 2 ⋅ 2 − 2 x 2 + 4 x 2 ( 1 − x 2 ) 2 \displaystyle \frac{du}{dx} \;=\;\frac{1}{1 + \left(\frac{2x}{1-x^2}\right)^2} \cdot\frac{(1-x^2)\cdot2 - (2x)(-2x)}{(1-x^2)^2} \;= \;\frac{(1-x^2)^2}{(1-x^2)^2 + 4x^2}\cdot\frac{2-2x^2+4x^2}{(1-x^2)^2} d x d u = 1 + ( 1 − x 2 2 x ) 2 1 ⋅ ( 1 − x 2 ) 2 ( 1 − x 2 ) ⋅ 2 − ( 2 x ) ( − 2 x ) = ( 1 − x 2 ) 2 + 4 x 2 ( 1 − x 2 ) 2 ⋅ ( 1 − x 2 ) 2 2 − 2 x 2 + 4 x 2 . . . = 1 1 − 2 x 2 + x 4 + 4 x 2 ⋅ 2 + 2 x 2 1 = 1 1 + 2 x 2 + x 4 ⋅ 2 ( 1 + x 2 ) 1 \displaystyle =\;\frac{1}{1 - 2x^2 + x^4 + 4x^2}\cdot\frac{2+ 2x^2}{1} \;=\;\frac{1}{1+2x^2+x^4}\cdot\frac{2(1+x^2)}{1} = 1 − 2 x 2 + x 4 + 4 x 2 1 ⋅ 1 2 + 2 x 2 = 1 + 2 x 2 + x 4 1 ⋅ 1 2 ( 1 + x 2 ) . . \(\displaystyle = \;\frac{1}{(1+x^2)^2}\cdot\frac{2(1+x^2)}{1} \;=\;\frac{2}{1+x^2}\quad\hdots\quad \text{Isn't that amazing?}\)
Last edited by a moderator: Apr 13, 2016
pp \,= \,2x,\;hyp \,= \,1+x^2\)