Solving for the t in the Integral of 1/sqrt(9+t^2) dt

[karush]

$\int_{0}^{4}\frac{1}{\sqrt{9+t^2}} \,dt$

The book said the and was 2 but the TI said it was $\ln\left({3}\right)$

I tried using $$u=\sqrt{9+t^2}$$ but for$$\int_{}^{}\frac{1}{\sqrt{9+t^2}} \,dt$$
I got $\ln\left({\sqrt{9+t^2}}+t\right)$
Where does the $t$ come from

 

[MarkFL]

According to W|A, this definite integral is:

$$\arsinh\left(\frac{4}{3}\right)=\ln(3)$$

 

[karush]

Sorry I posted before finishing
But on the indefinite integral where does the $t$ come from?

 

[Fantini]

Use the relation $$\cosh^2(x) - \sinh^2(x) = 1$$ by using the substitution $t=3\sinh(x)$. You obtain $dt = 3 \cosh(x) \, dx$ and therefore $$\int_0^4 \frac{1}{\sqrt{9+t^2}} \, dt = \int_0^{\sinh^{-1} (4/3)} \, dx = \operatorname{arcsinh} \left( \frac{4}{3} \right).$$ To see this is equal to $\ln(3)$, let $$\operatorname{arcsinh} \left( \frac{4}{3} \right) = \theta.$$ Then $$\sinh(\theta) = \frac{e^{\theta} - e^{-\theta}}{2} = \frac{4}{3}.$$ Multiply it by $e^{\theta}$ and simplify. You'll get $$e^{2 \theta} - \frac{8}{3} e^{\theta} -1=0.$$ Finding the root of this second degree polynomial you'll find $e^{\theta} = 3$, therefore $\theta = \ln(3)$.

 

[Prove It]

$\int_{0}^{4}\frac{1}{\sqrt{9+t^2}} \,dt$

The book said the and was 2 but the TI said it was $\ln\left({3}\right)$

I tried using $$u=\sqrt{9+t^2}$$ but for$$\int_{}^{}\frac{1}{\sqrt{9+t^2}} \,dt$$
I got $\ln\left({\sqrt{9+t^2}}+t\right)$
Where does the $t$ come from

$\displaystyle \begin{align*} \int{ \frac{1}{\sqrt{9 + x^2}}\,\mathrm{d}x} &= \int{ \frac{1}{\sqrt{9 + \left[ 3\tan{(t)} \right] ^2}}\,3\sec^2{(t)}\,\mathrm{d}t} \textrm{ after substituting } x = 3\tan{(t)} \implies \mathrm{d}x = 3\sec^2{(t)}\,\mathrm{d}t \\ &= \int{ \frac{3\sec^2{(t)}}{\sqrt{9 + 9\tan^2{(t)}}}\,\mathrm{d}t} \\ &= \int{ \frac{3\sec^2{(t)}}{\sqrt{9\sec^2{(t)}}} \,\mathrm{d}t} \\ &= \int{ \frac{3\sec^2{(t)}}{3\sec{(t)}}\,\mathrm{d}t } \\ &= \int{\sec{(t)}\,\mathrm{d}t} \\ &= \int{ \frac{1}{\cos{(t)}}\,\mathrm{d}t} \\ &= \int{\frac{\cos{(t)}}{\cos^2{(t)}}\,\mathrm{d}t} \\ &= \int{ \frac{\cos{(t)}}{1 - \sin^2{(t)}}\,\mathrm{d}t} \\ &= \int{ \frac{1}{1 - u^2}\,\mathrm{d}u} \textrm{ after substituting } u = \sin{(t)} \implies \mathrm{d}u =\cos{(t)}\,\mathrm{d}t \\ &= \int{ \frac{1}{\left( 1 - u \right) \left( 1 + u \right) }\,\mathrm{d}u} \\ &= \int{ \frac{1}{2 \left( u + 1 \right) } - \frac{1}{2\left( u - 1 \right) } \,\mathrm{d}u} \\ &= \frac{1}{2} \ln{ \left| u + 1 \right| } - \frac{1}{2} \ln{ \left| u - 1 \right| } + C \\ &= \frac{1}{2} \ln{ \left| \frac{ u + 1}{u - 1} \right| } + C \\ &= \frac{1}{2} \ln{ \left| \frac{\sin{(t)} + 1}{\sin{(t)} - 1} \right| } + C \\ &= \frac{1}{2} \ln{ \left| \frac{\left[ \sin{(t)} + 1 \right] ^2}{ \left[ \sin{(t)} - 1 \right] \left[ \sin{(t)} + 1 \right] } \right| } + C \\ &= \frac{1}{2}\ln{ \left| \frac{\sin^2{(t)} + 2\sin{(t)}\cos{(t)} + 1}{\sin^2{(t)} - 1} \right| } + C \\ &= \frac{1}{2} \ln{ \left| \frac{\sin^2{(t)} + 2\sin{(t)}\cos{(t)} + 1}{-\cos^2{(t)}} \right| } + C \\ &= \frac{1}{2} \ln{ \left| -\tan^2{(t)} - 2\tan{(t)} - \sec^2{(t)} \right| } + C \\ &= \frac{1}{2} \ln{ \left| -\tan^2{(t)} - 2\tan{(t)} - \left[ 1 + \tan^2{(t)} \right] \right| } +C \\ &= \frac{1}{2}\ln{ \left| -2\tan^2{(t)} - 2\tan{(t)} - 1 \right| } +C \\ &= \frac{1}{2} \ln{ \left| 2\tan^2{(t)} + 2\tan{(t)} + 1 \right| } +C \\ &= \frac{1}{2} \ln{ \left| 2 \left( \frac{x}{3} \right) ^2 + 2 \left( \frac{x}{3} \right) + 1 \right| } +C \\ &= \frac{1}{2} \ln{ \left| \frac{2x^2}{9} + \frac{2x}{3} + 1 \right| } +C \\ &= \frac{1}{2} \ln{ \left| \frac{2x^2 + 6x + 9}{9} \right| } +C \\ &= \ln{ \left| \frac{\sqrt{2x^2 + 6x + 9}}{3} \right| } + C \end{align*}$

PHEW!

 

[Fantini]

Wow. That is truly worthy of the name tour-de-force!

 

[Euge]

$\int_{0}^{4}\frac{1}{\sqrt{9+t^2}} \,dt$

The book said the and was 2 but the TI said it was $\ln\left({3}\right)$

I tried using $$u=\sqrt{9+t^2}$$ but for$$\int_{}^{}\frac{1}{\sqrt{9+t^2}} \,dt$$
I got $\ln\left({\sqrt{9+t^2}}+t\right)$
Where does the $t$ come from

Hi karush,

A main reason that $t$ is present is because $9$ is the product of the conjugates $\sqrt{9 + t^2} + t$ and $\sqrt{9 + t^2} - t$. Let me explain more.

Since $(\sqrt{9 + t^2} + t)(\sqrt{9 + t^2} - t) = (9 + t^2) - t^2 = 9$, $$\int \frac{dt}{\sqrt{9 + t^2}}$$

$$= \frac{1}{9} \int \frac{(\sqrt{9 + t^2} + t)(\sqrt{9 + t^2} - t)}{\sqrt{9 + t^2}} \, dt$$

$$=\frac{1}{9} \int \left(1 + \frac{t}{\sqrt{9 + t^2}}\right) (\sqrt{9 + t^2} - t)\, dt$$

$$= \int \left(1 + \frac{t}{\sqrt{9 + t^2}}\right) \frac{1}{\sqrt{9 + t^2} + t}\, dt$$

Now let $u = t + \sqrt{9 + t^2}$. Then $du = (1 + t/\sqrt{9 + t^2})\, dt$. So the last integral becomes $\int \frac{du}{u} = \ln|u| + C = \ln|t + \sqrt{9 + t^2}| + C$. Thus

$$\int \frac{1}{\sqrt{9 + t^2}}\, dt = \ln|t + \sqrt{9 + t^2}| + C.$$

Using this method, you may deduce more generally

$$\int \frac{1}{\sqrt{\lambda + t^2}}\, dt = \ln|t + \sqrt{\lambda + t^2}| + C.$$

 

[karush]

Wow, that was a LOT of help