Question

How do you integrate `int x^3 sqrt(-x^2 + 8x-16)dx` using trigonometric substitution?

Answer 1

See below

Explanation:

You don't need trigonometric substitution in this case. Note that

`sqrt(-x^2+8x-16)=sqrt((-1)(x^2-8x+16))=sqrt(-1)sqrt(x^2-8x+16)=isqrt((x-4)^2)=i(x-4)`

We use `i` as the imaginary unit `i=sqrt(-1)`

Now in the integral we have

`intx^3·i·(x-4)dx=i·intx^3(x-4)dx=i·intx^4dx-4i·intx^3dx=`

`=i/5x^5-i·x^4+C`