Find the value of the definite integral

[chwala]

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 $du=dt$

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

 

[pasmith]

Completing the square is the right idea, but don't treat the general case. The substitution depends on what is left after making a linear change of variable and extracting a positive common factor. That leaves three cases to consider: <br /> \begin{split}<br /> \int \sqrt{X^2 + C^2}\,dX &amp; \Rightarrow X = Cf_1(u) \\<br /> \int \sqrt{X^2 - C^2}\,dX &amp; \Rightarrow X = Cf_2(u) \\<br /> \int \sqrt{C^2 - X^2}\,dX &amp; \Rightarrow X = Cf_3(u)<br /> \end{split}

 

[Mark44]

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

Let $u=t+1$ then $du=dt$
$$\sqrt8\int_0^2[\sqrt{u^2+1} ]du$$

This integral is ripe for a trig substitution.

 

[chwala]

aaaaaaahahaha...seen it @Mark44

Thanks to our physics forum in particular;... recent study of hyperbolic functions... I'll post my working later...we shall integrate $\sinh^2x$wrt $x$ ...arrive at solution $12.64$...

 

[pasmith]

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 $du=dt$

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

If t is between 0 and 2, then u = 1 + t is between 1 and 3.

 

[chwala]

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 $du=dt$

$$\sqrt8\int_1^3[\sqrt{u^2+1} ]du$$

Let $u=\sinh x ⇒du=\cosh x dx$

$$\sqrt8\int_1^3[\sqrt{u^2+1} ]du=\sqrt8\int_{0.881}^{1.8184}\sqrt{\cosh^2 x} ⋅\cosh x dx=\sqrt8\int_{0.881}^{1.8184}\cosh^2 x dx$$

$$=\sqrt8\left[\frac{1}{2}\left[x+\frac{1}{2} \sinh (2x)\right]\right]_{0.881}^{1.8184}=\sqrt8\left[5.652-1.147\right]=\sqrt8×4.505=12.742$$

Bingo!!

 

[chwala]

Bringing me to my next question, How do we handle cubic or quartic integration under square roots or even cube roots?...should i come up with a question on a new thread?

 

[pasmith]

Note that if u = \sinh x then x = \ln\left(u + \sqrt{1 + u^2}\right) and <br /> \sinh 2x = 2\sinh x \cosh x = 2u\sqrt{1 + u^2}. This leads to the exact result <br /> \sqrt{2} \ln\left(-3 + 3\sqrt{2} + 2\sqrt{5} - \sqrt{10} \right) + 6\sqrt{5} - 2.

 

[dextercioby]

Bringing me to my next question, How do we handle cubic or quartic integration under square roots or even cube roots?...should i come up with a question on a new thread?

These are the elliptic integrals. Their value could only be determined numerically for definite integrals (i.e. with limits).