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I'm working on a question that involves using trigonometric substitution on a definite integral that will later use u substitution but I am not sure how to go ahead with this.

$$\int_1^2\frac1{x^2\sqrt{4x^2+9}}dx$$

My first step was to use $\sqrt{a^2+x^2}$ as $x=a\tan\theta$ to get...

$$2x=3\tan\theta :x=\frac32\tan\theta$$ $$dx=\frac32\sec^2\theta$$

Substituting:

$$\int\frac{\frac32\sec^2\theta}{\frac94\tan^2\theta\sqrt{9\tan^2\theta+9}}$$

The problem here is how do I change the limit it goes to?

$$\frac43=\tan\theta$$ and $$\frac23=\tan\theta$$

Following DR.MV's answer so far..

$$\frac29\int_{\tan^-1(\frac23)}^{\tan^-1(\frac43)}\frac{\sec^2\theta}{\tan^2\theta\sqrt{9\sec^2\theta}}d\theta$$

$$=\frac29\int_{\tan^-1(\frac23)}^{\tan^-1(\frac43)}\frac{\sec\theta}{\tan^2\theta}d\theta$$

$$=\frac29\int_{\tan^-1(\frac23)}^{\tan^-1(\frac43)}\frac{\cos\theta}{\sin^2\theta}d\theta$$

Now $u=\sin\theta$ so $du=\cos\theta d\theta$

$$=\frac29\int_{?}^{?}\frac{1}{u^2}du$$

This is where I am stuck now...

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  • $\begingroup$ Do you need to know those values? Maybe you can just live with them as values of inverse trig functions. After all, your answer is likely to involve other trig functions evaluated at $\theta$. $\endgroup$ – lulu Jul 21 '15 at 22:19
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    $\begingroup$ No problem! Denote by $\theta_1$ and $\theta_2$ the angle such that $\tan\theta_1=\frac{4}{3}$ and $\tan\theta_2=\frac{2}{3}$. In the end you will find expressions involving $\tan$ and $\sec$. $\endgroup$ – DiegoMath Jul 21 '15 at 22:20
  • $\begingroup$ Sorry could you explain this a bit more? Do you mean for me to continue my integration from $\arctan\frac23$ to $\arctan\frac43$ $\endgroup$ – Panthy Jul 21 '15 at 22:22
  • $\begingroup$ Compute the indefinite integral and express the result in function of $\tan\theta$. $\endgroup$ – Bernard Jul 21 '15 at 22:24
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    $\begingroup$ @Panthy Yes. Exactly. If you know the value of one trig function at an angle it is generally pretty easy to find the value of the others. $\endgroup$ – lulu Jul 21 '15 at 22:30
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There was a typo in the current post. After enforcing the substitution $2x=3\tan \theta$, the integral ought to read

$$\begin{align}I&=\int_{\arctan(2/3)}^{\arctan(4/3)}\frac{\frac32 \sec^2\theta}{\frac94 \tan^2\theta\sqrt{9\tan^2\theta+9}}d\theta\\\\ &=\frac29 \int_{\arctan(2/3)}^{\arctan(4/3)}\frac{ \sec^2\theta}{ \tan^2\theta\,\sec \theta}d\theta\\\\ &=\frac29 \int_{\arctan(2/3)}^{\arctan(4/3)}\frac{ \sec^2\theta}{ \tan^2\theta\,\sec \theta}d\theta\\\\ &=\frac29 \int_{\arctan(2/3)}^{\arctan(4/3)}\cot \theta \csc \theta d\theta\\\\ &=\frac29 \left.(-\csc \theta)\right|_{\arctan(2/3)}^{\arctan(4/3)}\\\\ &=\frac{\sqrt{13}}{9}-\frac{5}{18}\end{align}$$


NOTES:

Remark 1: When making a substitution of variables in a definite integral, the limits of integration change accordingly. In this example, the substitution was $x=\frac32 \tan \theta$. When $x=1$ at the lower limit, $\tan \theta =\frac23\implies \theta =\arctan(2/3)$. Similarly, when $x=2$ at the upper limit, $\tan \theta =\frac43\implies \theta =\arctan(4/3)$.

Remark 2:
To evaluate $\sin (\arctan(2/3))$, we recall that the arctangent is an angle whose tangent is $2/3$. A picture sometimes facilitates the analysis wherein we draw a right triangle with vertical side of length $2$ and horizontal side of length $3$ forming a right angle.

Note that the angle the hypotenuse makes with the horizontal side is $\arctan(2/3)$. Inasmcuh as the hypotenuse is of length $\sqrt{2^2+3^2}=\sqrt{13}$, we see $\sin(\arctan(2/3))=\frac{2}{\sqrt{13}}$ and thus $\csc (\arctan(2/3))=\frac{\sqrt{13}}{2}$.

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  • $\begingroup$ will continue from this point and update you! thanks $\endgroup$ – Panthy Jul 21 '15 at 22:32
  • $\begingroup$ i updated my answer but still confused $\endgroup$ – Panthy Jul 21 '15 at 22:43
  • $\begingroup$ OK ... posted the full solution to assist. Please let me know how I can improve the answer. I just want to give you the best answer I can. $\endgroup$ – Mark Viola Jul 21 '15 at 22:50
  • $\begingroup$ is the way I did it wrong? im not so confident dealing with csc sec and stuff but either way im reallly not sure how you evaluate $(-csc\theta)_{arctan(\frac23)}^{arctan(\frac43)}$ $\endgroup$ – Panthy Jul 21 '15 at 22:52
  • $\begingroup$ The arctangent gives an angle whose tangent is the argument. So, if an angle has a tangent of $2/3$ then its sine is $\sqrt{2^2+3^2}=\sqrt{13}$. Likewise, the sine of an angle whose tangent is $4/3$ is $\sqrt{4^2+3^2}=5$. How else can I improve the answer? $\endgroup$ – Mark Viola Jul 21 '15 at 23:00
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\begin{align*}\int\frac{\frac32\sec^2\theta\,\mathrm d\mkern1.5mu\theta}{\frac94\tan^2\theta\sqrt{9\tan^2\theta+9}}&=\frac29\int\frac{\mathrm d\mkern1.5mu\theta}{\sin^2\theta\sqrt{1+\tan^2\theta}}=\frac29\int\frac{\lvert\cos\theta\rvert\,\mathrm d\mkern1.5mu\theta}{\sin^2\theta}\\[1ex] &=\frac29\int\frac{\cos\theta\,\mathrm d\mkern1.5mu\theta}{\sin^2\theta}\qquad\text{since}\enspace 0\le\theta<\dfrac\pi2\\[1ex] &=-\frac2{9\sin\theta} \end{align*} Some trigonometry will let you determine the bounds for $\sin\theta\;$ from the bounds for $\tan\theta$: since $0\le\theta<\dfrac\pi2$, we have: $\cos\theta=\dfrac1{\sqrt{1+\tan^2\theta}}$, hence $$\sin\theta=\tan\theta\cos\theta=\frac{\tan\theta}{\sqrt{1+\tan^2\theta}}=\frac{\cfrac{2x}3}{\sqrt{1+\cfrac{4x^2}9}}=\frac{2x}{\sqrt{4x^2+9}}$$ so that the indefinite integral is: $$-\frac{\sqrt{4x^2+9}}{9x}.$$

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  • $\begingroup$ There appears to be an error of omission. The secant squares term disappeared in the second integral. $\endgroup$ – Mark Viola Jul 21 '15 at 22:52
  • $\begingroup$ It's been replaced with $\cos^2\theta$ in the denominator; $\endgroup$ – Bernard Jul 21 '15 at 22:54
  • $\begingroup$ Why the downvote? $\endgroup$ – Bernard Jul 21 '15 at 22:55
  • $\begingroup$ OK. Yes. But the integrand is missing the term in the denominator, namely $x^2=\frac94 \tan^2 \theta$. $\endgroup$ – Mark Viola Jul 21 '15 at 22:56
  • $\begingroup$ By the way, I was not the down voter. In fact, I very very rarely down vote. Rather, I like to neutralize down votes by giving an up vote. $\endgroup$ – Mark Viola Jul 21 '15 at 22:57
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$$\int_1^2\frac1{x^2\sqrt{4x^2+9}}dx = \int_a^b\frac{\frac{2}{3}\sec^2\theta}{\frac{9}{4}\tan^2\theta\sqrt{9\tan^2\theta+9}}d\theta$$

where $a = \tan^{-1}\frac{2}{3}$ and $b = \tan^{-1}\frac{4}{3}$

$$ \int_a^b\frac{2\sec\theta}{9\tan^2\theta}d\theta = \left(-\frac{2\text{cosec}\theta}{9} \right)_a^b$$

Hint: $\text{cosec}\left(\tan^{-1}\frac{4}{3}\right) = \frac{5}{4}$, $\text{cosec}\left(\tan^{-1}\frac{2}{3}\right) = \frac{\sqrt{13}}{2}$

EDIT: let $\tan^{-1}\frac{4}{3}=d $

$$ \frac{4}{3} = \tan d$$ $$\frac{4}{3} = \frac{\sin d}{\cos d} $$ $$ \frac{1}{\sin d} = \text{cosec}\ d = \frac{3}{4}\sec d$$

$$ \sec^2d = \tan^2 d + 1 = 1+ \left(\frac{4}{3}\right)^2 = \frac{25}{9}$$ $$\sec d =\frac{5}{3}$$

$$ \text{cosec}\ d = \frac{3}{4}\times \frac{5}{3} =\frac{5}{4}$$

You can do the same for the second one.

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  • $\begingroup$ how do you know that? $\endgroup$ – Panthy Jul 21 '15 at 22:47
  • $\begingroup$ could you please elaborate im new to that kind of thing... sorry if i sound stupid $\endgroup$ – Panthy Jul 21 '15 at 22:54
  • $\begingroup$ I'll explain further in my solution. $\endgroup$ – John_dydx Jul 21 '15 at 22:54
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    $\begingroup$ @Panthy, I hope the explanation helps. $\endgroup$ – John_dydx Jul 21 '15 at 23:10

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