- #1

chwala

Gold Member

- 2,691

- 354

- Homework Statement
- $$\int_0^2 \sqrt{(8t^2+16t+16)} dt$$

- Relevant Equations
- Integration

Looking at integration today...i will go slow as i also try finish other errands anyway; i am thinking along these lines;

$$\int \sqrt{(ax^2+bx+c)} dx=\sqrt{a}\int \sqrt{\left[x+\frac{b}{2a}\right]^2+\left[\frac{4ac-b^2}{4a^2}\right]} dx$$

...

Therefore,

$$\int_0^2 \sqrt{(8t^2+16t+16)} dt=\sqrt8\int_0^2[\sqrt{[(t+1)^2+1^2} ]dt$$

Let ##u=t+1## then ##du=dt##

$$\sqrt8\int_0^2[\sqrt{u^2+1} ]du$$

$$\int \sqrt{(ax^2+bx+c)} dx=\sqrt{a}\int \sqrt{\left[x+\frac{b}{2a}\right]^2+\left[\frac{4ac-b^2}{4a^2}\right]} dx$$

...

Therefore,

$$\int_0^2 \sqrt{(8t^2+16t+16)} dt=\sqrt8\int_0^2[\sqrt{[(t+1)^2+1^2} ]dt$$

Let ##u=t+1## then ##du=dt##

$$\sqrt8\int_0^2[\sqrt{u^2+1} ]du$$

Last edited: