MHB Quick Trigonometric Substitution Question

[ardentmed]

View attachment 2797

Hey guys,

I'm really doubting my answer for 4b specifically.

I used x=tan(Ø) and got (-1/4)cot(Ø) - (Ø) + C. I'm really not sure about this one.

Thanks in advance.

 

[MarkFL]

We are given:

$$\int\frac{1}{x^2\sqrt{x^2+4}}\,dx$$

Now, I would use:

$$x=2\tan(\theta)\,\therefore\,dx=2\sec^2(\theta)\,d\theta$$

and we have:

$$\int\frac{2\sec^2(\theta)}{4\tan^2(\theta)2\sec(\theta)}\,d\theta=\frac{1}{4}\int\frac{\cos(\theta)}{\sin^2(\theta)}\,d\theta$$

Can you proceed?

 

[Prove It]

We are given:

$$\int\frac{1}{x^2\sqrt{x^2+4}}\,dx$$

Now, I would use:

$$x=2\tan(\theta)\,\therefore\,dx=2\sec^2(\theta)\,d\theta$$

and we have:

$$\int\frac{2\sec^2(\theta)}{4\tan^2(\theta)\sec(\theta)}\,d\theta=\frac{1}{2}\int\frac{\cos(\theta)}{\sin^2(\theta)}\,d\theta$$

Can you proceed?

I'm sure you mean $\displaystyle \begin{align*} \frac{1}{4} \int{ \frac{\cos{\left( \theta \right) } }{\sin^2{\left( \theta \right) } } \,\mathrm{d}\theta} \end{align*}$ don't you Mark?

 

[MarkFL]

I'm sure you mean $\displaystyle \begin{align*} \frac{1}{4} \int{ \frac{\cos{\left( \theta \right) } }{\sin^2{\left( \theta \right) } } \,\mathrm{d}\theta} \end{align*}$ don't you Mark?

Yes, I corrected my mistake...thanks! :D

 

[ardentmed]

We are given:

$$\int\frac{1}{x^2\sqrt{x^2+4}}\,dx$$

Now, I would use:

$$x=2\tan(\theta)\,\therefore\,dx=2\sec^2(\theta)\,d\theta$$

and we have:

$$\int\frac{2\sec^2(\theta)}{4\tan^2(\theta)2\sec(\theta)}\,d\theta=\frac{1}{4}\int\frac{\cos(\theta)}{\sin^2(\theta)}\,d\theta$$

Can you proceed?

I'm guessing that u=sin $ \theta $

And thus the cos gets substituted out?

Thanks.

Edit: But how would I go about expressing it in terms of "x" again?

 

[Prove It]

I'm guessing that u=sin $ \theta $

And thus the cos gets substituted out?

Thanks.

Edit: But how would I go about expressing it in terms of "x" again?

If by "substituted out" you mean "since $\displaystyle \begin{align*} \frac{\mathrm{d}u}{\mathrm{d}x} = -\cos{(x)} \end{align*}$, that means the integral with respect to u becomes $\displaystyle \begin{align*} -\int{ \frac{1}{u^2}\,\mathrm{d}u} \end{align*}$, then yes...

 

[MarkFL]

If by "substituted out" you mean "since $\displaystyle \begin{align*} \frac{\mathrm{d}u}{\mathrm{d}x} = -\cos{(x)} \end{align*}$, that means the integral with respect to u becomes $\displaystyle \begin{align*} -\int{ \frac{1}{u^2}\,\mathrm{d}u} \end{align*}$, then yes...

I'm certain you mean $$\frac{\mathrm{d}u}{\mathrm{d}\theta}=\cos(\theta)$$ right? (Thinking)

...Edit: But how would I go about expressing it in terms of "x" again?

If $x=2\tan(\theta)$, then what is $\theta$ in terms of $x$?

 

[Prove It]

I'm certain you mean $$\frac{\mathrm{d}u}{\mathrm{d}\theta}=\cos(\theta)$$ right? (Thinking)
If $x=2\tan(\theta)$, then what is $\theta$ in terms of $x$?

Yes that is exactly what I meant. I think between us Mark, we have one functioning human being hahaha :P

 

[ardentmed]

Yes that is exactly what I meant. I think between us Mark, we have one functioning human being hahaha :P

True, but how would I go about expressing it in terms of x?

(I'm referring to your other post by the way).

Thanks in advance.

 

[MarkFL]

Since:

$$x=2\tan(\theta)$$

then:

$$\theta=\tan^{-1}\left(\frac{x}{2}\right)$$

 

[ardentmed]

Wait, so doesn't -1/(4sinarctan(x/2)) +c seem a bit off? Or is it just me?

 

[MarkFL]

You have the correct solution, which I would choose to write as:

$$I=-\frac{1}{4}\csc\left(\tan^{-1}\left(\frac{x}{2}\right)\right)+C$$

So, this angle $\theta$ is such that its tangent is $$\frac{x}{2}$$.

This means we may draw a right triangle where the side opposite $\theta$ is $x$ and the side adjacent is $2$. In order to determine its cosecant, we need to know the hypotenuse, which we can obtain via Pythagoras, and then we can use the definition:

$$\csc(\theta)=\frac{\text{hypotenuse}}{\text{opposite}}$$

So, what do you get?