How I integrate this?
int_0^(1/2)dx/(4-5sqrt(1-x^2)
2 Answers
Rationalize the denominator then use lots of partial fraction decompositions.
Explanation:
Let
I=int_0^(1/2)dx/(4-5sqrt(1-x^2))
Rationalize:
I=int_0^(1/2)dx/(4-5sqrt(1-x^2))*((4+5sqrt(1-x^2))/(4+5sqrt(1-x^2)))
Simplify:
I=int_0^(1/2)(4+5sqrt(1-x^2))/(25x^2-9)dx
Integration is distributive:
I=4int_0^(1/2)1/(25x^2-9)dx+5int_0^(1/2)sqrt(1-x^2)/(25x^2-9)dx
Rearrange:
I=-4int_0^(1/2)1/((3+5x)(3-5x))dx+5int_0^(1/2)(1-x^2)/(25x^2-9)dx/sqrt(1-x^2)
Apply partial fraction decomposition:
I=-2/3int_0^(1/2)(1/(3+5x)+1/(3-5x))dx-1/5int_0^(1/2)(1-16/(25x^2-9))dx/sqrt(1-x^2)
Apply the substitution
I=-2/15[ln|3+5x|-ln|3-5x|]_ 0^(1/2)-1/5int_0^(pi/6)(1-16/(25sin^2theta-9))d theta
Integration is distributive:
I=-2/15ln(11)-1/5int_0^(pi/6)d theta+16/5int_0^(pi/6)1/(25sin^2theta-9)d theta
Rearrange:
I=-2/15ln(11)-pi/30+16/5int_0^(pi/6)sec^2theta/(25tan^2theta-9sec^2theta)d theta
Since
I=-2/15ln(11)-pi/30+16/5int_0^(pi/6)sec^2theta/(16tan^2theta-9)d theta
Apply the difference of squares:
I=-2/15ln(11)-pi/30-16/5int_0^(pi/6)sec^2theta/((3+4tantheta)(3-4tantheta))d theta
Apply partial fraction decomposition:
I=-2/15ln(11)-pi/30-8/15int_0^(pi/6)(1/(3+4tantheta)+1/(3-4tantheta))sec^2thetad theta
Integrate term by term:
I=-2/15ln(11)-pi/30-2/15[ln|3+4tantheta|-ln|3-4tantheta|]_0^(pi/6)
Insert the limits of integration:
I=-2/15ln(11)-pi/30-2/15ln((3sqrt3+4)/(3sqrt3-4))
Rationalize:
I=-2/15ln(11)-pi/30-2/15ln((3sqrt3+4)^2/11)
Simplify:
I=-pi/30-4/15ln(3sqrt3+4)
Explanation:
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