How do you integrate #int x / sqrt(-x^2+8x) dx# using trigonometric substitution?

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Answer 1 · 2016-09-25T13:59:46

#4arcsin((x-4)/4)-sqrt(-x^2+8x)+C#

Complete the square in the denominator:

#intx/sqrt(-((x-4)^2-16))dx=intx/sqrt(16-(x-4)^2)dx#

Let #x-4=4sintheta#. Thus, #dx=4costhetad theta#. Also note that #x=4sintheta+4#. Substituting in:

#=int(4sintheta+4)/sqrt(16-16sin^2theta)(4costhetad theta)#

Factoring out both of the #4# terms and the #sqrt16# from the denominator:

#16/sqrt16int((sintheta+1)costheta)/sqrt(1-sin^2theta)d theta#

Note that #cos^2theta+sin^2theta=1#, so #sqrt(1-sin^2theta)=costheta#:

#=4int(sintheta+1)d theta#

#=-4costheta+4theta+C#

From #x-4=4sintheta#, we see that #theta=arcsin((x-4)/4)#. Furthermore, we see that #sintheta=(x-4)/4#. Now, use #costheta=sqrt(1-sin^2theta)#:

#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#