Is This Trig Substitution Correct for This Integral?

[tmt1]

$$\int_{}^{} \frac{1}{x\sqrt{x^2 + 16}} \,dx$$

I can set $x = 4 tan\theta$. Thus $dx = 4 sec^2 \theta d\theta$

So, plug this into the first equation:

$$\int_{}^{} \frac{4 sec^2 \theta }{4 tan\theta \sqrt{16 tan^2\theta + 16}} \,d\theta$$

Then,

$$\int_{}^{} \frac{ sec^2 \theta }{4 tan\theta sec\theta} \,d\theta$$

then:

$$\frac{1}{4}\int_{}^{} \frac{ sec \theta }{ tan\theta} \,d\theta$$

I can simplify this to:

$$\frac{1}{4}\int_{}^{} \frac{ 1}{ sin\theta} \,d\theta$$

which evaluates evaluates to

$$\frac{1}{4} ln|sin \theta| + C$$

I have $x = 4 tan\theta$, therefore $\frac{x}{4} = tan\theta$.

$tan\theta = \frac{opposite}{adjacent}$ and $sin\theta = \frac{opposite}{hypotenuse}$.

So, given $\frac{x}{4} = tan\theta$, then $opposite = x$ and $adjacent = 4$, and using the pythagorean theorem, $hypotenuse = \sqrt{x^2 + 16}$.

Plugging that into $sin\theta = \frac{opposite}{hypotenuse}$, $sin\theta = \frac{x}{\sqrt{x^2 + 16}}$, therefore

$$\frac{1}{4} ln|sin \theta| + C$$

is the same as

$$\frac{1}{4} ln|\frac{x}{\sqrt{x^2 + 16}}| + C$$

Is this right?

 

[MarkFL]

I'm with you up to this point:

\(\displaystyle I=\frac{1}{4}\int \frac{1}{\sin(\theta)}\,d\theta\)

However, you next state:

\(\displaystyle I=\frac{1}{4}\ln\left|\sin(\theta)\right|+C\)

If this is correct, and if we differentiate $I$ w.r.t $\theta$, we should get:

\(\displaystyle \frac{1}{4\sin(\theta)}\)

Do we?

 

[Prove It]

$$\int_{}^{} \frac{1}{x\sqrt{x^2 + 16}} \,dx$$

I can set $x = 4 tan\theta$. Thus $dx = 4 sec^2 \theta d\theta$

So, plug this into the first equation:

$$\int_{}^{} \frac{4 sec^2 \theta }{4 tan\theta \sqrt{16 tan^2\theta + 16}} \,d\theta$$

Then,

$$\int_{}^{} \frac{ sec^2 \theta }{4 tan\theta sec\theta} \,d\theta$$

then:

$$\frac{1}{4}\int_{}^{} \frac{ sec \theta }{ tan\theta} \,d\theta$$

I can simplify this to:

$$\frac{1}{4}\int_{}^{} \frac{ 1}{ sin\theta} \,d\theta$$

which evaluates evaluates to

$$\frac{1}{4} ln|sin \theta| + C$$

I have $x = 4 tan\theta$, therefore $\frac{x}{4} = tan\theta$.

$tan\theta = \frac{opposite}{adjacent}$ and $sin\theta = \frac{opposite}{hypotenuse}$.

So, given $\frac{x}{4} = tan\theta$, then $opposite = x$ and $adjacent = 4$, and using the pythagorean theorem, $hypotenuse = \sqrt{x^2 + 16}$.

Plugging that into $sin\theta = \frac{opposite}{hypotenuse}$, $sin\theta = \frac{x}{\sqrt{x^2 + 16}}$, therefore

$$\frac{1}{4} ln|sin \theta| + C$$

is the same as

$$\frac{1}{4} ln|\frac{x}{\sqrt{x^2 + 16}}| + C$$

Is this right?

$\displaystyle \begin{align*} \int{ \frac{1}{x\,\sqrt{x^2 + 16}} \,\mathrm{d}x } &= \int{ \frac{2\,x}{2\,x^2\,\sqrt{ x^2 + 16 }} \,\mathrm{d}x } \end{align*}$

Let $\displaystyle \begin{align*} u = x^2 + 16 \implies \mathrm{d}u = 2\,x\,\mathrm{d}x \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \int{ \frac{2\,x}{2\,x^2\,\sqrt{x^2 + 16}} \,\mathrm{d}x } &= \int{ \frac{1}{2\,\left( u - 16 \right) \, \sqrt{ u }} \,\mathrm{d}u } \\ &= \int{ \frac{1}{2\,\left[ \left( \sqrt{u} \right) ^2 - 16 \right] \, \sqrt{u}} \,\mathrm{d}u } \end{align*}$

Let $\displaystyle \begin{align*} v = \sqrt{u} \implies \mathrm{d}v = \frac{1}{2\,\sqrt{u}}\,\mathrm{d}u \end{align*}$ and the integral becomes

$\displaystyle \begin{align*} \int{ \frac{1}{2\,\left[ \left( \sqrt{u} \right) ^2 - 16 \right] \,\sqrt{u} } \,\mathrm{d}u } &= \int{ \frac{1}{v^2 - 16}\,\mathrm{d}v } \\ &= \int{ \frac{1}{\left( v - 4 \right) \left( v + 4 \right) } \,\mathrm{d}v } \end{align*}$

Applying Partial Fractions:

$\displaystyle \begin{align*} \frac{A}{v - 4} + \frac{B}{v + 4} &\equiv \frac{1}{ \left( v - 4 \right) \left( v + 4 \right) } \\ A\,\left( v + 4 \right) + B \,\left( v - 4 \right) &\equiv 1 \end{align*}$

Let $\displaystyle \begin{align*} v = 4 \end{align*}$ to find $\displaystyle \begin{align*} 8\,A = 1 \implies A = \frac{1}{8} \end{align*}$.

Let $\displaystyle \begin{align*} v = -4 \end{align*}$ to find $\displaystyle \begin{align*} -8\,B = 1 \implies B = -\frac{1}{8} \end{align*}$. Then

$\displaystyle \begin{align*} \int{ \frac{1}{\left( v - 4 \right) \left( v + 4 \right) } \,\mathrm{d}v } &= \frac{1}{8} \int{ \left( \frac{1}{v - 4} - \frac{1}{v + 4} \right) \,\mathrm{d}v } \\ &= \frac{1}{8} \, \left( \ln{ \left| v - 4 \right| } - \ln{ \left| v + 4 \right| } \right) + C \\ &= \frac{1}{8} \ln{ \left| \frac{v - 4}{v + 4} \right| } + C \\ &= \frac{1}{8} \ln{ \left| \frac{\sqrt{u} - 4}{\sqrt{u} + 4} \right| } + C \\ &= \frac{1}{8} \ln{ \left| \frac{\sqrt{x^2 + 16} - 4}{\sqrt{x^2 + 16} + 4} \right| } + C \end{align*}$