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

Question

2409 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-22T19:44:51

See below

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

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

$=i/5x^5-i·x^4+C$

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