Definite Integration with Trigonometric Substitution
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...
$\endgroup$ 53 Answers
$\begingroup$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 $\csc (\arctan(2/3))=\frac{\sqrt{13}}{2}$.
$\endgroup$ 15 $\begingroup$\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: $$-\frac{\sqrt{4x^2+9}}{9x}.$$
$\endgroup$ 14 $\begingroup$$$\int_1^2\frac1{x^2\sqrt{4x^2+9}}dx = \int_a^b\frac{\frac{2}{3}\sec^2\theta}{\frac{9}{4}\tan^2\theta\sqrt{9\tan^2\theta+9}}d\theta$$
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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