How do you integrate #int sqrt(13+25x^2)# using trig substitutions?

Question

Emma Aldridge · Accepted Answer

To integrate an irrational function with a sum of squares under the root, you use the substitution with the tangent.

First reduce the radical to a sum of squares: #int(sqrt(13+25x^2))dx = 1/5int(sqrt((sqrt(13)/5)^2+x^2))dx# Pose #x=sqrt(13)/5tan(t)# and as #(d tan(t))/dt = sec^2(t)# #dx=sqrt(13)/5sec^2(t)dt# #int(sqrt(13+25x^2))dx = 1/5int(sqrt(13/25+13/25tan^2(t))sqrt(13)/5sec^2(t)dt = 13/125int(sqrt(1+tan^2(t))sec^2(t)dt# Use the trigonometric identity: #1+tan^2t=sec^2t# #int(sqrt(13+25x^2))dx = 13/125int(sqrt(sec^2(t))sec^2(t)dt =# #= 13/125int|sec(t)|^3dt#. Now, to integrate this last integral let's limit ourselves in the interval #t in [-pi/2,pi/2]# where #sec(t) # is positive and use integration by parts: #intsec^3(t)dt = int (sect*sec^2t)dt = intsect * d(tant) = sect tant -int tan t* d(sec t) = sect tant -int tan ^2t*sec t dt = # but #tan^2t = sec^2t-1#, so #intsec^3(t)dt = sect tant + int (sect -sec^3t)dt= sect tant +int sect dt -int sec^3(t)tdt# or #2intsec^3(t)dt = sect tant +int sect dt # Finally: #intsec^3(t)dt = 1/2sect tant +1/2 ln(|sec t + tan t|)# Now reverse the substitution: #tan t = 5/sqrt(13)x# and #sect = sqrt(1+tan^2(t)) = sqrt(1+25/13x^2)# So: #int(sqrt(13+25x^2))dx = 5/(2sqrt(13))xsqrt(1+25/13x^2)+1/2ln(|5/sqrt(13)x+sqrt(1+25/13x^2)|)#

Elijah Abram · Answer

undefined To integrate $\int \sqrt{13 + 25x^2} \, dx$ using trigonometric substitution, follow these steps:

1. **Identify the substitution**: Recognize that the integrand has the form $\sqrt{a^2 + u^2}$. Here, $a^2 = 13$ and $u^2 = 25x^2$, which suggests $a = \sqrt{13}$ and $u = 5x$. We use the trigonometric substitution $u = a\tan(\theta)$, or in this case, $5x = \sqrt{13}\tan(\theta)$.

2. **Substitute and simplify**: From $5x = \sqrt{13}\tan(\theta)$, we have $x = \frac{\sqrt{13}}{5}\tan(\theta)$. The differential $dx$ is then $dx = \frac{\sqrt{13}}{5}\sec^2(\theta) d\theta$. Substituting $x$ and $dx$ into the integral gives
   $$
   \int \sqrt{13 + 25\left(\frac{\sqrt{13}}{5}\tan(\theta)\right)^2} \, \frac{\sqrt{13}}{5}\sec^2(\theta) d\theta.
   $$

3. **Evaluate the integral**: Simplifying the integrand, we have
   $$
   \int \sqrt{13 + 13\tan^2(\theta)} \, \frac{\sqrt{13}}{5}\sec^2(\theta) d\theta = \int \frac{\sqrt{13}}{5} \sec^3(\theta) d\theta.
   $$
   The integral of $\sec^3(\theta)$ can be found using integration techniques (e.g., integration by parts or using a reduction formula), yielding
   $$
   \frac{\sqrt{13}}{5} \left( \frac{1}{2} \sec(\theta)\tan(\theta) + \frac{1}{2} \ln|\sec(\theta) + \tan(\theta)| \right) + C,
   $$
   where $C$ is the constant of integration.

4. **Back-substitute $\theta$**: Recall that $5x = \sqrt{13}\tan(\theta)$, so $\tan(\theta) = \frac{5x}{\sqrt{13}}$. Using trigonometric identities, we can express $\sec(\theta)$ in terms of $x$:
   $$
   \sec(\theta) = \sqrt{1 + \tan^2(\theta)} = \sqrt{1 + \left(\frac{5x}{\sqrt{13}}\right)^2}.
   $$
   Thus, the integral in terms of $x$ is
   $$
   \frac{\sqrt{13}}{5} \left( \frac{1}{2} \sqrt{1 + \left(\frac{5x}{\sqrt{13}}\right)^2}\frac{5x}{\sqrt{13}} + \frac{1}{2} \ln\left|\sqrt{1 + \left(\frac{5x}{\sqrt{13}}\right)^2} + \frac{5x}{\sqrt{13}}\right| \right) + C.
   $$
   Simplifying further will give the final answer. Let's proceed with these calculations for precise results.After simplifying and back-substituting $\theta$ with the expressions in terms of $x$, the integral $\int \sqrt{13 + 25x^2} \, dx$ results in:
$$0.5x\sec(\theta) + 0.1\sqrt{13}\log\left|\frac{5\sqrt{13}x + 13\sec(\theta)}{13}\right| + C,$$
where $\sec(\theta) = \sqrt{1 + \left(\frac{5x}{\sqrt{13}}\right)^2}$.

This expression includes a placeholder $\sec(\theta)$ which should be replaced with its expression in terms of $x$ for the complete solution. Thus, the correct interpretation in terms of $x$ should include substituting $\sec(\theta)$ with $\sqrt{1 + \left(\frac{5x}{\sqrt{13}}\right)^2}$, yielding the final result in terms of $x$ only. This step concludes the trigonometric substitution process for the given integral.