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I'm trying to use the substitution $x=\frac{1}{4}\sec{\theta}$.

This turns $\sqrt{16x^2-1}$ into $\tan{x}$.

Implicit differentiation gives us $dx = \frac{1}{4} \tan{\theta}\sec{\theta} d\theta$.

Then, $\int\sqrt{16x^2-1} dx = \frac{1}{4}∫\tan^2{\theta}\sec{\theta}d\theta$.

I'm attempting to use Integration by Parts to integrate, but its just taking me around in loops...

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This is a fun one. Let $$I = \int \tan^2\theta \sec\theta d\theta$$ Using integration by parts with $u=\tan\theta, dv=\sec\theta\tan\theta$, $$I = \sec\theta\tan\theta - \int\sec^3\theta d\theta = \sec\theta\tan\theta - \int(1+\tan^2\theta)\sec\theta d\theta = $$$$\sec\theta\tan\theta-I - \int \sec\theta d\theta = \sec\theta\tan\theta-\ln(\sec\theta+\tan\theta)-I$$ Thus, solving for $I$, $$I = \frac{1}{2}(\sec\theta\tan\theta - \ln(\sec\theta+\tan\theta))+C$$ FYI, taking the integral of $\sin(x)e^x$ is similar to this.

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Hint: Let $u=\tan\theta$ and do a u-substitution.

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Hint

Put $$x=\frac{\cosh(t)}{4}$$ and use

$$\cosh^2(t)-1=\sinh^2(t)=\frac{\cosh(2t)-1}{2}.$$

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  • $\begingroup$ Wait, what is h(t)? $\endgroup$ – Saketh Malyala Dec 26 '16 at 21:40
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    $\begingroup$ @SakethMalyala It is not $h(t)$ , it is $cosh(t)$. $\endgroup$ – hamam_Abdallah Dec 26 '16 at 21:41
  • $\begingroup$ why is there an h at the end? $\endgroup$ – Saketh Malyala Dec 26 '16 at 21:42
  • $\begingroup$ He means cosinus and sinus hyperbolicus :) $\endgroup$ – Yaddle Dec 26 '16 at 21:43
  • $\begingroup$ @SakethMalyala It's the "hyperbolic cosine function" $\endgroup$ – Akiva Weinberger Dec 26 '16 at 22:51
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$\frac{1}{4}∫\tan^2{\theta}\sec{\theta}d\theta$ $= \frac{1}{4}∫(sec^2\theta-1){\theta}\sec{\theta}d\theta$ = $\frac{1}{4}∫sec^3\theta d\theta$ - $\frac{1}{4}∫sec\theta d\theta$

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  • $\begingroup$ Yes, I was able to get there, but as I integrated, that put me in a loop. $\endgroup$ – Saketh Malyala Dec 26 '16 at 21:43
  • $\begingroup$ You can use the reduction formula for $sec^n(x)$ $\endgroup$ – idk Dec 26 '16 at 21:50
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What you can also do is using the substitiuion $4x=\cosh t$:

$$\int \sqrt{16x^2-1}dx=\int\sqrt{\cosh^2{t}-1}\cdot\sinh{t} dt=$$ $$\int\sinh^2{t}dt=\int\frac{\cosh{2t}-1}{2}dt=\frac{\sinh{2t}}{4}-\frac{t}{2}+c$$

Substitung it back and using the formula for $\cosh^{-1}t=\ln(x+\frac{x^2-1}{2})$ you get what you wanted

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The easiest way to work out the related integral $\int \sqrt{1 - x^2} \, \mathrm{d}x$ (which differs from yours by a factor of $4i$) is probably geometrically, as the area under a circle:

integral

which you can calculate by adding the area of the triangle and of the circle sector to be $$\int_0^b \sqrt{1-x^2} \, \mathrm{d}x = \frac{\theta}{2} + \frac{\mathrm{area} \, \mathrm{of} \, \mathrm{triangle}}{2} = \frac{1}{2}\arcsin(b) + \frac{1}{2}b \sqrt{1-b^2},$$ with $b$ being the base of the triangle, $\sqrt{1-b^2}$ its height, and $\theta = \arcsin(b)$ the angle of the sector in red.

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