Question

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

Answer 1

`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 !