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I'm at a loss here. I tried everything I could think of, can't seem to get the correct answer after plugging in the new boundaries (the teacher wants us to use the new boundaries instead of the ones given). How do I find where the radical values of $\sin(\arctan(x/3))$? I am supposed to evaluate the integral using trig substitution for the following:

$$\int_\sqrt3 ^3 \frac{\sqrt{9+x^2}}{x^6}dx $$

I inserted a picture of my work, sorry if you can't read my terrible writing, let me know if you have any questions regarding what I wrote.

Here is what I have so far

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  • $\begingroup$ I just realized, I added it in :) $\endgroup$ – ChrisD93 Oct 1 '17 at 17:02
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To answer your question about the sine of the arctangent.

Since trig functions only depend on ratios of sides, you can pick any triange that would represent the quantities you need. In your case, $\arctan(x/3)$ implies you have a tangent of some angle that is $x/3$, so in $\Delta ABC$, say, angle $B = 90^\circ$ and side $AB = 3$ while side $BC = x$, which would make $\tan A = x/3$.

So you are asked what is the $\sin A$ and from this triangle it is clear that hypotenuse $AC$ is $\sqrt{x^2+3^2}$ and hence $$ \sin(\arctan(x/3)) = \sin A = \frac{x}{\sqrt{x^2+3^2}} = \frac{x}{\sqrt{x^2+9}}. $$

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With substitution $x=\sqrt{3}\tan u$: \begin{align} \int_\sqrt3 ^3 \frac{\sqrt{9+x^2}}{x^6}dx &= \int_{\frac{\pi}{4}}^{\frac{\pi}{3}}\dfrac{\sqrt{3}}{9}\dfrac{1-\sin^2u}{\sin^6u}\cos u\,du \\ &= \dfrac{\sqrt{3}}{9}\left(\dfrac{1}{3\sin^3u}-\dfrac{1}{5\sin^5u}\right)_{\frac{\pi}{4}}^{\frac{\pi}{3}} \\ &= \dfrac{\sqrt{3}}{9}\left(\dfrac{8}{45\sqrt{3}}+\dfrac{12}{60\sqrt{2}}\right) \end{align}

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  • $\begingroup$ Where did the sqrt(3)/9 come from? The equation I got is (1/81)[(-1/5sin^5 <theta>)+(1/3sin^3 <theta>)] $\endgroup$ – ChrisD93 Oct 1 '17 at 17:08
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Set up a triangle with side lengths $x$ and $3,$ then the hypotenuse has length $\sqrt{x^2+9}.$ Choosing the angle opposite one of these legs to be $\theta.$ It doesn't matter which one, but just keep your choice consistent.

I'm choosing $\theta$ opposite $3$. This means that hypotenuse/opposite gives us $\sec(\theta)=\frac{\sqrt{x^2+9}}{3},$ or $3\sec(\theta)=\sqrt{x^2+9}.$ Making this choice also provides us with $\frac{3}{x}=\tan(\theta),$ or $\frac{1}{x}=\frac{\tan(\theta)}{3}.$ Lastly, we have $\cot(\theta)=\frac{x}{3},$ so $\frac{dx}{3}=-\csc^2(\theta)d\theta.$ This makes original integral into the following (not as definite integral)

$$ \int3\sec(\theta)\cdot \left(\frac{\tan(\theta)}{3}\right)^6\cdot(-3)\csc^2(\theta)d\theta.$$

A quick reduction of the integrand using algebra gives us

$$ \frac{-1}{3^4}\int \tan^6(\theta)\cdot\csc(\theta)d\theta.$$

This can be done via a $u-$substitution where $\sec(\theta)=u,$

yielding

$$\frac{\sec^5(\theta)}{5}-\frac{2\sec^3(\theta)}{3}+\sec(\theta)+C$$

Recall that $\sec(\theta)=\frac{\sqrt{x^2+9}}{3},$ and I leave the rest of the computation to you.

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  • $\begingroup$ Weird, I got dx = 3sec^2 <theta> d<theta>, did I do it a harder way or something? $\endgroup$ – ChrisD93 Oct 1 '17 at 17:12
  • $\begingroup$ Nah. Trig hides when two things are the same. $\endgroup$ – Chickenmancer Oct 1 '17 at 17:17

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