MHB Integral of sqrt(1+x^2)/x: Help Solving w/ Trig Substitution

[MarkFL]

Here is the question:

Integral of sqrt(1+x^2)/x using trigonometric substitution?


Hi, I keep getting the answer wrong on this problem, and I was wondering if someone could please help me figure out how to solve it? I understand the basic concept, and I know that x=tan(θ), but I'm having trouble figuring out what trig identities to use. Thank you for your help!

I have posted a link there to this question so the OP can view my work.

 

[MarkFL]

Hello KTRavenclaw,

We are given to integrate:

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

I agree with your choice of substitution:

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

Using the Pythagorean identity $$\tan^2(u)+1=\sec^2(u)$$ we obtain:

$$\int\frac{\sec^3(\theta)}{\tan(\theta)}\, d\theta=\int\csc(\theta)\sec^2(\theta) \,d\theta$$

Using the above mentioned Pythagorean identity, we may write:

$$\int\csc(\theta)\left(\tan^2(\theta)+1 \right) \,d\theta=\int \sec(\theta)\tan(\theta)+\frac{1}{\sin(\theta)}\, d\theta$$

Now, using the following:

$$\frac{d}{du}\left(\sec(u) \right)=\sec(u)\tan(u)$$

$$\sin(2u)=2\sin(u)\cos(u)$$

We may write the integral as:

$$\int\,d\left(\sec(\theta) \right)+\frac{1}{2}\int\frac{1}{ \sin\left(\frac{\theta}{2} \right) \cos\left(\frac{\theta}{2} \right)}$$

If we multiply the second integrand by $$1=\frac{\sec^2\left(\frac{\theta}{2} \right)}{\sec^2\left(\frac{\theta}{2} \right)}$$ and put the constant with the numerator of the integrand, we obtain:

$$\int\,d\left(\sec(\theta) \right)+\int\frac{\dfrac{1}{2}\sec^2\left( \frac{\theta}{2} \right)}{\tan\left(\frac{\theta}{2} \right)}\, d\theta$$

Now, for the second integral, apply the substitution:

$$u=\tan\left(\frac{\theta}{2} \right)\,\therefore\,du=\frac{1}{2}\sec^2\left( \frac{\theta}{2} \right)\, d\theta$$

Thus, we now have:

$$\int\,d\left(\sec(\theta) \right)+\int\frac{1}{u}\,du$$

And so using the rules of integration, we obtain:

$$\sec(\theta)+\ln|u|+C$$

Back-substitute for $u$:

$$\sec(\theta)+\ln\left|\tan\left(\frac{\theta}{2} \right) \right|+C$$

Using the half-angle identity for tangent:

$$\tan\left(\frac{u}{2} \right)=\csc(u)-\cot(u)$$ we have:

$$\sec(\theta)+\ln\left|\csc(\theta)-\cot(\theta) \right|+C$$

Back-substitute for $\theta$:

$$\sqrt{1+x^2}+\ln\left|\frac{\sqrt{1+x^2}-1}{x} \right|+C$$

And so we may conclude:

$$\int\frac{\sqrt{1+x^2}}{x}\,dx= \sqrt{1+x^2}+\ln\left|\frac{\sqrt{1+x^2}-1}{x} \right|+C$$

 

[tmt1]

Using the Pythagorean identity $$\tan^2(u)+1=\sec^2(u)$$ we obtain:

$$\int\frac{\sec^3(\theta)}{\tan(\theta)}\, d\theta=\int\csc(\theta)\sec^2(\theta) \,d\theta$$

I'm not sure how you make this step. Is this just a simple trig identity?

 

[MarkFL]

I'm not sure how you make this step. Is this just a simple trig identity?

Think of the tangent function like this:

$$\tan(u)=\frac{\sin(u)}{\cos(u)}=\sin(u)\sec(u)$$

 

[Prove It]

In my opinion a more straightforward substitution is $\displaystyle \begin{align*} x = \sinh{(t)} \implies \mathrm{d}x = \cosh{(t)}\,\mathrm{d}t \end{align*}$, giving

$\displaystyle \begin{align*} \int{ \frac{\sqrt{1 + x^2}}{x}\,\mathrm{d}x} &= \int{ \frac{\sqrt{1 + \sinh^2{(t)}}}{\sinh{(t)}}\,\cosh{(t)}\,\mathrm{d}t } \\ &= \int{ \frac{\cosh^2{(t)}}{\sinh{(t)}}\,\mathrm{d}t } \\ &= \int{ \frac{1 + \sinh^2{(t)}}{\sinh{(t)}}\,\mathrm{d}t } \\ &= \int{ \left[ \frac{1}{\sinh{(t)}} + \sinh{(t)} \right] \,\mathrm{d}t } \\ &= \int{ \frac{\sinh{(t)}}{\sinh^2{(t)}}\,\mathrm{d}t } + \int{ \sinh{(t)}\,\mathrm{d}t } \\ &= \int{ \frac{\sinh{(t)}}{\cosh^2{(t)} - 1} \,\mathrm{d}t } + \cosh{(t)} + C_1 \\ &= \int{ \frac{\sinh{(t)}}{\left[ \cosh{(t)} - 1 \right] \left[ \cosh{(t)} + 1 \right] } \,\mathrm{d}t } + \cosh{(t)} + C_1 \end{align*}$

You can solve that resulting integral with $\displaystyle \begin{align*} u = \cosh{(t)} \implies \mathrm{d}u = \sinh{(t)} \,\mathrm{d}t \end{align*}$ giving $\displaystyle \begin{align*} \int{\frac{1}{\left( u - 1 \right) \left( u + 1 \right) } \,\mathrm{d}u } + \cosh{(t)} + C_1 \end{align*}$.

Apply Partial Fractions

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

Let $\displaystyle \begin{align*} u = -1 \end{align*}$ to find $\displaystyle \begin{align*} -2\,A = 1 \implies A = -\frac{1}{2} \end{align*}$.

Let $\displaystyle \begin{align*} u = 1 \end{align*}$ to find $\displaystyle \begin{align*} 2\,B = 1 \implies B = \frac{1}{2} \end{align*}$. Then the integral is

$\displaystyle \begin{align*} \int{ \left[ -\frac{1}{2}\,\left( \frac{1}{u - 1} \right) + \frac{1}{2}\,\left( \frac{1}{u + 1} \right) \right] \,\mathrm{d}u } + \cosh{(t)} + C_1 &= \frac{1}{2} \int{ \left( \frac{1}{u + 1} - \frac{1}{u - 1} \right) \,\mathrm{d}u } + \cosh{(t)} + C_1 \\ &= \frac{1}{2} \,\left( \ln{ \left| u + 1 \right| } - \ln{ \left| u - 1 \right| } \right) + C_2 + \cosh{(t)} + C_1 \\ &= \frac{1}{2} \ln{ \left| \frac{u + 1}{u - 1} \right| } + \cosh{(t)} + C \textrm{ where } C = C_2 + C_1 \\ &= \frac{1}{2}\ln{ \left| \frac{\cosh{(t)} + 1}{\cosh{(t)} - 1 } \right| } + \sqrt{ 1 + \sinh^2{(t)} } + C \\ &= \frac{1}{2}\ln{ \left| \frac{\left[ \cosh{(t)} + 1 \right] ^2}{\left[ \cosh{(t)} - 1 \right] \left[ \cosh{(t)} + 1 \right] } \right| } + \sqrt{1 + x^2} + C \\ &= \frac{1}{2}\ln{ \left| \frac{\cosh^2{(t)} + 2\cosh{(t)} + 1}{\cosh^2{(t)} - 1 } \right| } + \sqrt{1 + x^2} + C \\ &= \frac{1}{2}\ln{ \left| \frac{1 + \sinh^2{(t)} + 2\,\sqrt{ 1 + \sinh^2{(t)}} + 1}{\sinh^2{(t)}} \right| } + \sqrt{1 + x^2} + C \\ &= \frac{1}{2}\ln{ \left| \frac{2 + x^2 + 2\,\sqrt{1 + x^2}}{x^2} \right| } + \sqrt{1 + x^2} + C \end{align*}$

 

[soroban]

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

\text{Let }\,x = \tan\theta,\; dx = \sec^2\!\theta\,d\theta

\text{Substitute: }\;\int \frac{\sec\theta}{\tan\theta}\,\sec^2\theta\,d\theta \;=\;\int\frac{\sec^3\theta}{\tan\theta}\,d\theta \;=\;\int\frac{1}{\cos^3\theta}\,\frac{\cos\theta}{\sin\theta}\,d\theta \;=\;\int\frac{d\theta}{\sin\theta\cos^2\theta}

. . =\;\int \frac{\sec^2\theta}{\sin\theta}\,d\theta \;=\; <br /> \int \frac{1}{\sin\theta} (\tan^2\theta +1)\,d\theta \;=\; \int\left(\frac{\sin\theta}{\cos^2\theta} + \frac{1}{\sin\theta}\right)\,d\theta

. . =\;\int (\sec\theta\tan\theta + \csc\theta)\,d\theta \;=\; \sec\theta + \ln|\csc\theta - \cot\theta| + C\text{Back-substitute: }\;\sqrt{1+x^2} + \ln\left|\frac{\sqrt{1+x^2} - 1}{x}\right| + C