Integration involving substitutions II

[paulmdrdo1]

\begin{align*}\displaystyle \int\frac{du}{u\sqrt{16u^2-9}}\\& let\,\acute{u} =\,4u\\&d\acute{u}=4du\\&du=\frac{1}{4}(d\acute{u})\\ &\frac{1}{4}\int\frac{d\acute{u}}{\acute{u}\sqrt{(\acute{u})^2-a^2}}\\ & =\frac{1}{12}sec^{-1}\frac{4}{3}u+c -->> is\, this\, correct?\end{align*}

 

[Prove It]

Re: Integration Inverse trig II

\begin{align*}\displaystyle \int\frac{du}{u\sqrt{16u^2-9}}\\& let\,\acute{u} =\,4u\\&d\acute{u}=4du\\&du=\frac{1}{4}(d\acute{u})\\ &\frac{1}{4}\int\frac{d\acute{u}}{\acute{u}\sqrt{(\acute{u})^2-a^2}}\\ & =\frac{1}{12}sec^{-1}\frac{4}{3}u+c -->> is\, this\, correct?\end{align*}

It's better to use a different letter for a new variable, I know you've changed the symbol slightly but it's difficult to read.

Anyway, it's not correct, as the u outside the square root should become [math]\displaystyle \begin{align*} \frac{u'}{4} \end{align*}[/math].

 

[Prove It]

Re: Integration Inverse trig II

\begin{align*}\displaystyle \int\frac{du}{u\sqrt{16u^2-9}}\\& let\,\acute{u} =\,4u\\&d\acute{u}=4du\\&du=\frac{1}{4}(d\acute{u})\\ &\frac{1}{4}\int\frac{d\acute{u}}{\acute{u}\sqrt{(\acute{u})^2-a^2}}\\ & =\frac{1}{12}sec^{-1}\frac{4}{3}u+c -->> is\, this\, correct?\end{align*}

My approach for this question would be to use hyperbolic and trigonometric substitution. Let [math]\displaystyle \begin{align*} 4u = 3\cosh{(t)} \implies du = \frac{3}{4}\sinh{(t)}\,dt \end{align*}[/math] and the integral becomes

[math]\displaystyle \begin{align*} \int{\frac{du}{u \,\sqrt{ \left( 4u \right) ^2 - 9 } } } &= \int{\frac{\frac{3}{4}\sinh{(t)}\,dt}{\frac{3}{4} \cosh{(t)}\sqrt{ \left[ 3\cosh{(t)} \right] ^2 - 9 }}} \\ &= \int{ \frac{\sinh{(t)}\,dt}{\cosh{(t)}\sqrt{9\cosh^2{(t)} - 9}} } \\ &= \int{ \frac{\sinh{(t)}\,dt}{\cosh{(t)}\sqrt{9 \left[ \cosh^2{(t)} - 1 \right] }} } \\ &= \int{\frac{\sinh{(t)}\,dt}{\cosh{(t)}\sqrt{9\sinh^2{(t)}}}} \\ &= \int{\frac{\sinh{(t)}\,dt}{3\cosh{(t)}\sinh{(t)}}} \\ &= \frac{1}{3} \int{ \frac{dt}{\cosh{(t)}} } \\ &= \frac{1}{3} \int{ \frac{\cosh{(t)}\,dt}{\cosh^2{(t)}} } \\ &= \frac{1}{3} \int{ \frac{\cosh{(t)}\,dt}{1 + \sinh^2{(t)}} } \end{align*}[/math]

Now let [math]\displaystyle \begin{align*} \sinh{(t)} = \tan{(\theta)} \implies \cosh{(t)}\,dt = \sec^2{(\theta)} \,d\theta \end{align*}[/math] and the integral becomes

[math]\displaystyle \begin{align*} \frac{1}{3} \int{ \frac{\cosh{(t)}\,dt}{1 + \sinh^2{(t)}} } &= \frac{1}{3} \int{ \frac{\sec^2{(\theta)} \,d\theta}{1 + \tan^2{(\theta)}} } \\ &= \frac{1}{3} \int{ \frac{\sec^2{(\theta)}\,d\theta}{\sec^2{(\theta)}} } \\ &= \frac{1}{3} \int{ 1\,d\theta} \\ &= \frac{1}{3} \theta + C \\ &= \frac{1}{3}\arctan{ \left[ \sinh{(t)} \right] } + C \\ &= \frac{1}{3} \arctan{ \left[ \sqrt{ \cosh^2{(t)} - 1 } \right] } + C \\ &= \frac{1}{3} \arctan{ \left[ \sqrt{ \left( \frac{4u}{3} \right) ^2 - 1 } \right] } + C \\ &= \frac{1}{3} \arctan{ \left[ \frac{\sqrt{16u^2 - 9}}{3} \right] } + C \end{align*}[/math]

 

[paulmdrdo1]

Re: Integration Inverse trig II

It's better to use a different letter for a new variable, I know you've changed the symbol slightly but it's difficult to read.

Anyway, it's not correct, as the u outside the square root should become [math]\displaystyle \begin{align*} \frac{u'}{4} \end{align*}[/math].

\begin{align*}\displaystyle \int\frac{du}{u\sqrt{16u^2-9}}\\& let\,\acute{u} =\,4u\\&d\acute{u}=4du\\& a = 3 \\& du=\frac{1}{4}(d\acute{u} )\\ &\frac{1}{4}\int\frac{d\acute{u}}{\acute{u}\sqrt{({u'})^2-a^2}}\\ & =\frac{1}{12}sec^{-1}\frac{4}{3}u+c\end{align*}

this is the other part of my solution

\begin{align*}\displaystyle\frac{1}{4}(\frac{1}{a}sec^{-1}\frac{u'}{a}) + c\end{align*}

since a = 3 and u' = 4u i substituted it to have

\begin{align*}\displaystyle\frac{1}{4}(\frac{1}{3}sec^{-1}\frac{4u}{3}) + c\\ \frac{1}{12}sec^{-1}\frac{4}{3}u+c\end{align*}
can you tell me what did i do wrong. please.

 

[MarkFL]

It appears to me that you are using a table of integrals, such as:

$$\int\frac{dx}{x\sqrt{x^2-a^2}}=\frac{1}{a}\sec^{-1}\left(\frac{x}{a} \right)+C$$

Applying this to your problem, you should write:

$$\int\frac{du}{u\sqrt{16u^2-9}}=\int\frac{4\,du}{4u\sqrt{(4u)^2-3^2}}=\frac{1}{3}\sec^{-1}\left(\frac{4u}{3} \right)+C$$

You should be able to demonstrate that this is equivalent to the result posted by Prove It.

 

[MarkFL]

I will try to walk you through how I would approach this problem using a trigonometric substitution. We are given to evaluate:

$$\int\frac{1}{u\sqrt{16u^2-9}}\,du=\int\frac{1}{u\sqrt{(4u)^2-3^2}}\,du$$

Now, when I look at the radicand in the denominator, I think of the Pythagorean identity:

$$\tan^2(\theta)=\sec^2(\theta)-1$$

So, this leads me to make the substitution:

$$4u=3\sec(\theta)\,\therefore\,4\,du=3\sec(\theta) \tan(\theta)\,d\theta$$

and this transforms the indefinite integral to:

$$\int\frac{\frac{3}{4}\sec(\theta)\tan(\theta)}{ \frac{3}{4}\sec(\theta)\sqrt{3^2\sec^2(\theta)-3^2}}\,d\theta=\int\frac{\sec(\theta)\tan(\theta)}{3\sec(\theta)\tan(\theta)}\,d\theta= \frac{1}{3}\int\,d\theta=\frac{1}{3}\theta+C$$

Now, using our substitution, we may solve for $\theta$:

$$4u=3\sec(\theta)$$

$$\theta=\sec^{-1}\left(\frac{4u}{3} \right)$$

and so we may conclude:

$$\int\frac{1}{u\sqrt{16u^2-9}}\,du=\frac{1}{3}\sec^{-1}\left(\frac{4u}{3} \right)+C$$

 

[paulmdrdo1]

I will try to walk you through how I would approach this problem using a trigonometric substitution. We are given to evaluate:

$$\int\frac{1}{u\sqrt{16u^2-9}}\,du=\int\frac{1}{u\sqrt{(4u)^2-3^2}}\,du$$

Now, when I look at the radicand in the denominator, I think of the Pythagorean identity:

$$\tan^2(\theta)=\sec^2(\theta)-1$$

So, this leads me to make the substitution:

$$4u=3\sec(\theta)\,\therefore\,4\,du=3\sec(\theta) \tan(\theta)\,d\theta$$

and this transforms the indefinite integral to:

$$\int\frac{\frac{3}{4}\sec(\theta)\tan(\theta)}{ \frac{3}{4}\sec(\theta)\sqrt{3^2\sec^2(\theta)-3^2}}\,d\theta=\int\frac{\sec(\theta)\tan(\theta)}{3\sec(\theta)\tan(\theta)}\,d\theta= \frac{1}{3}\int\,d\theta=\frac{1}{3}\theta+C$$

Now, using our substitution, we may solve for $\theta$:

$$4u=3\sec(\theta)$$

$$\theta=\sec^{-1}\left(\frac{4u}{3} \right)$$

and so we may conclude:

$$\int\frac{1}{u\sqrt{16u^2-9}}\,du=\frac{1}{3}\sec^{-1}\left(\frac{4u}{3} \right)+C$$

i'm still confused about trigonometric substitution. i only use table of integrals leading to inverse trig functions. please enlighten me with trig substitution. what are the rules for that?

 

[MarkFL]

I am trying my best to do just that...at what point in my post do I lose you?

 

[paulmdrdo1]

I am trying my best to do just that...at what point in my post do I lose you?

\begin{align*}\displaystyle tan^2(\theta)=\sec^2(\theta)-1\end{align*} you used this identities.

and you \begin{align*}\displaystyle let\, 4u=3sec(\theta)\, where\, did\, tan^2(\theta)+1\, go? why\, only\, use\, sec(\theta)\end{align*}

 

[MarkFL]

\begin{align*}\displaystyle \tan^2(\theta)=\sec^2(\theta)-1\end{align*} you used this identities.

and you let 4u=3sec where did tan²-1 go? why only use sec?

We see that we want to transform:

$$(4u)^2-3^2$$

and given that:

$$3^2\tan^2(\theta)=3^2\sec^2(\theta)-3^2$$

we then see that if we let:

$$4u=3\sec(\theta)$$

then we will have:

$$(4u)^2-3^2=(3\sec(\theta))^2-3^2=3^2\sec^2(\theta)-3^2=3^2\tan^2(\theta)=(3\tan(\theta))^2$$

Now we have transformed this expression into a square, and since it is under a square root, we can get rid of the radical.

 

[Prove It]

Mark already did in one of your http://www.mathhelpboards.com/f10/integration-involving-trigonometric-substitutions-5545/.

 

[paulmdrdo1]

We see that we want to transform:

$$(4u)^2-3^2$$

and given that:

$$3^2\tan^2(\theta)=3^2\sec^2(\theta)-3^2$$

we then see that if we let:

$$4u=3\sec(\theta)$$

then we will have:

$$(4u)^2-3^2=(3\sec(\theta))^2-3^2=3^2\sec^2(\theta)-3^2=3^2\tan^2(\theta)=(3\tan(\theta))^2$$

Now we have transformed this expression into a square, and since it is under a square root, we can get rid of the radical.

now it's clear! thanks! this is the explanation I'm waiting for.