# Find the value of the definite integral

• chwala
In summary: However, as long as the limits are known, the formulas for the elliptic integrals are quite simple.In summary, the author is thinking about integration and is going slow because he has other things to do. He is thinking about integration and is considering the following:-The equation for the sqrt function-The substitution u = t + 1 for the general case-The three cases for when t is between 0 and 2-The elliptic integral for u = sinh x
chwala
Gold Member
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##

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

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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: $$\begin{split} \int \sqrt{X^2 + C^2}\,dX & \Rightarrow X = Cf_1(u) \\ \int \sqrt{X^2 - C^2}\,dX & \Rightarrow X = Cf_2(u) \\ \int \sqrt{C^2 - X^2}\,dX & \Rightarrow X = Cf_3(u) \end{split}$$

chwala
chwala said:
$$\int_0^2 \sqrt{(8t^2+16t+16)} dt=\sqrt8\int_0^2[\sqrt{[(t+1)^2+1^2} ]dt$$
chwala said:
Let ##u=t+1## then ##du=dt##
$$\sqrt8\int_0^2[\sqrt{u^2+1} ]du$$
This integral is ripe for a trig substitution.

PhDeezNutz and 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##...

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chwala said:
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!!

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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?

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

SammyS and chwala
chwala said:
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).

chwala

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