How do you integrate #int x^3 sqrt(-x^2 - 8x-41)dx# using trigonometric substitution?

1 Answer
Aug 7, 2018

#intx^3sqrt(-x^2-8x-41)dx=5^5i*(1/5sec(theta)^5−1/3sec(theta)^3)-5^4*3/2i(2sec(theta)^3tan(theta)-sec(theta)tan(theta)-ln(|sec(theta)+tan(theta)|))+16isec(theta)^3-5^2*32i(sec(theta)tan(theta)+ln(|sec(theta)+tan(theta)|)+C#,

#C in RR#,

#theta=tan^(-1)((x+4)/5)#,

#sec(tan^(-1)(u))=sqrt(u^2+1)#, here #u=(x+4)/5#

See explanations below.

Explanation:

#I=intx^3sqrt(-x^2-8x-41)dx#

#=iintx^3sqrt(x^2+8x+41)dx#

Complete the square:

#I=iintx^3sqrt((x+4)^2+25)dx#

#=iintx^3sqrt((x+4)^2+5^2)dx#

Now let #x+4=5tan(theta)#

#dx=5sec(theta)^2d theta#

So:

#I=iint(5tan(theta)-4)^3sqrt(5^2(tan(theta)^2+1))*5sec(theta)^2d theta#

Because #tan(theta)^2+1=sec(theta)^2#

#=5iint(5tan(theta)-4)^3*5sec(theta)^3d theta#

#=25iint(5tan(theta)-4)^3sec(theta)^3d theta#

Now, #(a-b)^3=a^3-3a^2b+3ab^2-b^3#, and here, #a=5tan(theta)#, #b=4#

So:

#I=5^2iint(5^3tan(theta)^3-5^2*12tan(theta)^2+5*48tan(theta)-64)sec(theta)^3d theta#

#=5^5i##inttan(theta)^3sec(theta)^3d theta##-5^4*12i##inttan(theta)^2sec(theta)^3d theta##+5^3*48i##inttan(theta)sec(theta)^3d theta##-5^2*64i##intsec(theta)^3d theta#

Well, at this point, you have a sum of defined integral, you just have to sum
( click on each integral to see explanations associated ).

#I=5^5i*(1/5sec(theta)^5−1/3sec(theta)^3)-5^4*3/2i(2sec(theta)^3tan(theta)-sec(theta)tan(theta)-ln(|sec(theta)+tan(theta)|))+16isec(theta)^3-5^2*32i(sec(theta)tan(theta)+ln(|sec(theta)+tan(theta)|)+C#, #C in RR#

Finally, #theta=tan^(-1)((x+4)/5)#, and #sec(tan^(-1)(u))=sqrt(u^2+1)#, here #u=(x+4)/5#

I let you find what is the writing of #I# using #x# and not #theta#, it's very long but easy.

\0/ Here's our answer !