MHB Transforming Trigonometric Integrals with Substitution

[karush]

mnt{7.3} nmh{2000}
$$\displaystyle
\int\frac{1}{{t}^{2}\sqrt{1+{t}^{2}}} \ dt = \frac{-\sqrt{t^2+1}}{t}+C$$

$\displaystyle t=\tan\left({u}\right)$
$\displaystyle dt=\sec^2(u) \ du $

$$\int\frac{\sec^2\left({u}\right)}
{\tan^2\left({u}\right)sec\left({u}\right)} \ du
\implies\int\frac{\sec\left({u}\right)}{\tan^2\left({u}\right)} \ du $$

Then ??

 

[MarkFL]

What did you get after applying the substitution and simplifying?

 

[karush]

I added more to the OP
Thinking a reply would come after
Sorry

 

[karush]

$$\int\frac{\sec^2\left({u}\right)}
{\tan^2\left({u}\right)sec\left({u}\right)} \ du
\implies\int\frac{\sec\left({u}\right)}{\tan^2\left({u}\right)} \ du
\implies\int\frac{\cos\left({u}\right)}{\sin^2\left({u}\right)} \ du
$$

 

[MarkFL]

I added more to the OP
Thinking a reply would come after
Sorry

Okay...good...now think of the definitions:

$$\tan(u)\equiv\frac{\sin(u)}{\cos(u)}$$

$$\sec(u)\equiv\frac{1}{\cos(u)}$$

And write everything in terms of sine and cosine in a simple fraction...what do you get?

 

[karush]

$$ \int\frac{cos\left({u}\right)}{\sin^2\left({u}\right)} \ du
$$

$$w=\sin\left({u}\right) \ \ dw=\cos\left({u}\right) \ du $$

Then

$$\int \frac{dw}{{w}^{2}}$$

Well ?

 

[MarkFL]

What you actually now have is:

$$I=\int w^{-2}\,dw$$

Use the power rule, then back-substitute for $w$, and then for $u$...:)

 

[karush]

$$I=\int w^{-2}\,dw=\frac{1}{w}+C $$

$w=\sin\left({u}\right)$
So
$\frac{1}{\sin\left({u}\right)}+C$
Or
$\sec(u)+C$

Wait I'm lost...

 

[MarkFL]

You would have:

$$I=-w^{-1}+C=-\csc(u)+C=-\csc(\arctan(t))+C=...$$?

 

[karush]

Well I think then that
$$\csc(u)=-\sqrt{1+{t}^{2}}/t $$
So
$$ I=-\frac{\sqrt{1+{t}^{2}}} {t}+C $$
The substitutions of $u$ and $w$ kinda ###

 

[MarkFL]

Don't forget the negative sign in front of the cosecant function. :D

 

[MarkFL]

We could also approach this integral using a hyperbolic trig. substitution...we are given:

$$I=\int \frac{1}{t^2\sqrt{t^2+1}}\,dt$$

Observing that we have:

$$\cosh^2(u)=\sinh^2(u)+1$$

We could then let:

$$t=\sinh(u)\,\therefore\,dt=\cosh(u)\,du$$

And we get:

$$I=\int \frac{1}{\sinh^2(u)\sqrt{\sinh^2(u)+1}}\,\cosh(u)\,du=\int \csch^2(u)\,du$$

Using the fact that:

$$\frac{d}{du}(-\coth(u))=\csch^2(u)$$

We now have:

$$I=-\coth(u)+C$$

Now, using the identity:

$$\coth(u)=\sqrt{1+\csch^2(u)}$$

We then may write:

$$I=-\sqrt{1+t^{-2}}+C=-\frac{\sqrt{t^2+1}}{t}+C$$

 

[Greg]

$$-\sqrt{1+t^{-2}}=-\dfrac{\sqrt{1+t^2}}{|t\,|}$$