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

Question

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

1 Answer

Related questions

How do you find the integral $\int \frac{1}{x^{2} \cdot \sqrt{x^{2} - 9}} d x$ $int1/(x^2*sqrt(x^2-9))dx$ ?
How do you find the integral $\int \frac{x^{3}}{\sqrt{x^{2} + 9}} d x$ $intx^3/(sqrt(x^2+9))dx$ ?
How do you find the integral $\int x^{3} \cdot \sqrt{9 - x^{2}} d x$ $intx^3*sqrt(9-x^2)dx$ ?
How do you find the integral $\int \frac{x^{3}}{\sqrt{16 - x^{2}}} d x$ $intx^3/(sqrt(16-x^2))dx$ ?
How do you find the integral $\int \frac{\sqrt{x^{2} - 1}}{x} d x$ $intsqrt(x^2-1)/xdx$ ?
How do you find the integral $\int \frac{\sqrt{x^{2} - 9}}{x^{3}} d x$ $intsqrt(x^2-9)/x^3dx$ ?
How do you find the integral $\int \frac{x}{\sqrt{x^{2} + x + 1}} d x$ $intx/(sqrt(x^2+x+1))dx$ ?
How do you find the integral $\int \frac{d t}{\sqrt{t^{2} - 6 t + 13}}$ $intdt/(sqrt(t^2-6t+13))$ ?
How do you find the integral $\int x \cdot \sqrt{1 - x^{4}} d x$ $intx*sqrt(1-x^4)dx$ ?
How do you prove the integral formula $\int \frac{d x}{\sqrt{x^{2} + a^{2}}} = ln (x + \sqrt{x^{2} + a^{2}}) + C$ $intdx/(sqrt(x^2+a^2)) = ln(x+sqrt(x^2+a^2))+ C$ ?

Impact of this question

2394 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-22T19:44:51

See below

Explanation:

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

$\sqrt{- x^{2} + 8 x - 16} = \sqrt{(- 1) (x^{2} - 8 x + 16)} = \sqrt{- 1} \sqrt{x^{2} - 8 x + 16} = i \sqrt{{(x - 4)}^{2}} = i (x - 4)$ $sqrt(-x^2+8x-16)=sqrt((-1)(x^2-8x+16))=sqrt(-1)sqrt(x^2-8x+16)=isqrt((x-4)^2)=i(x-4)$

We use $i$ $i$ as the imaginary unit $i = \sqrt{- 1}$ $i=sqrt(-1)$

Now in the integral we have

$\int x^{3} \cdot i \cdot (x - 4) d x = i \cdot \int x^{3} (x - 4) d x = i \cdot \int x^{4} d x - 4 i \cdot \int x^{3} d x =$ $intx^3·i·(x-4)dx=i·intx^3(x-4)dx=i·intx^4dx-4i·intx^3dx=$

$= \frac{i}{5} x^{5} - i \cdot x^{4} + C$ $=i/5x^5-i·x^4+C$

Science

Math

Humanities

... and beyond

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

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you integrate int x^3 sqrt(-x^2 + 8x-16)dx∫x3√−x2+8x−16dxint x^3 sqrt(-x^2 + 8x-16)dx using trigonometric substitution?

1 Answer

Explanation:

Related questions

Impact of this question

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