How do you integrate #int x / sqrt(-x^2+8x) dx# using trigonometric substitution?
1 Answer
Explanation:
Complete the square in the denominator:
#intx/sqrt(-((x-4)^2-16))dx=intx/sqrt(16-(x-4)^2)dx#
Let
#=int(4sintheta+4)/sqrt(16-16sin^2theta)(4costhetad theta)#
Factoring out both of the
#16/sqrt16int((sintheta+1)costheta)/sqrt(1-sin^2theta)d theta#
Note that
#=4int(sintheta+1)d theta#
#=-4costheta+4theta+C#
From
#costheta=sqrt(1-((x-4)/4)^2)=sqrt((16-(x-4)^2)/16)=1/4sqrt(-x^2-8x)#
Thus the integral equals:
#=-4(1/4sqrt(-x^2-8x))+4arcsin((x-4)/4)+C#
#=4arcsin((x-4)/4)-sqrt(-x^2+8x)+C#