Integral of 1/sqrt(1-x^2) dx
The solution
Detail solution
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The integral of a constant times a function is the constant times the integral of the function:
∫t(1−x2)2dx=t∫(1−x2)2dx
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Rewrite the integrand:
(1−x2)2=x4−2x2+1
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Integrate term-by-term:
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The integral of xn is n+1xn+1 when n=−1:
∫x4dx=5x5
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The integral of a constant times a function is the constant times the integral of the function:
∫(−2x2)dx=−2∫x2dx
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The integral of xn is n+1xn+1 when n=−1:
∫x2dx=3x3
So, the result is: −32x3
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The integral of a constant is the constant times the variable of integration:
∫1dx=x
The result is: 5x5−32x3+x
So, the result is: t5x5−32x3+x
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Now simplify:
15tx(3x4−10x2+15)
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Add the constant of integration:
15tx(3x4−10x2+15)+constant
The answer is:
15tx(3x4−10x2+15)+constant
The answer (Indefinite)
[src]
/
| 3 5
| 2 2*x x
| / 2\ x - ---- + --
| \1 - x / 3 5
| --------- dx = C + -------------
| t t
|
/
∫t(1−x2)2dx=C+t5x5−32x3+x
Use the examples entering the upper and lower limits of integration.