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$x = \sec\theta, \mathrm dx = \sec\theta \tan\theta\mathrm d\theta$

$$\begin{align}\int \frac{\mathrm dx}{(x^2-1)^{\frac{3}{2}}}&=\int \frac{\mathrm dx}{(\tan^2\theta)^{\frac{3}{2}}}\\&=\int \frac{\mathrm dx}{\tan^{\frac{7}{2}}\theta}\\&=\int \frac{\sec\theta \tan\theta}{\tan\theta ^{\frac{7}{2}}}\mathrm d\theta\\&=\int \tan^{\frac{-5}{2}}\theta \sec\theta\mathrm d\theta\end{align}$$

Here is where I get stuck...I tried converting $\tan\theta$ and $\sec\theta$ in terms of $\cos\theta$ and $\sin\theta$, but that didn't seem to get me anywhere...What is my next move from here? Did I even start this problem correctly? I can't tell :(

Update with more work after initial answers:

$$\int \frac{\cos\theta}{\sin^2\theta}\mathrm d\theta$$ $u = \sin\theta, \mathrm du = \cos\theta\,\mathrm d\theta$

I found $\sin\theta = \frac{\sqrt{x^2-1}}{x}$

$$\implies \int \frac{\cos\theta}{\sin^2\theta}\mathrm d\theta=\int \frac{\mathrm du}{u^2} = \frac{1}{ \frac{1}{3}u^3}+C= \frac{1}{3\sin^3\theta}+C= 3 \bigg( \frac{x}{\sqrt{x^2-1}} \bigg)^3+C$$

Integreek
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Evan Kim
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    In the denominator you added the exponents instead of multiplying them. – John Wayland Bales Mar 08 '19 at 00:08
  • Why would I multiply them? I guess it is because it is an exponent of an exponent instead of an exponent * an exponent – Evan Kim Mar 08 '19 at 00:28
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    You should never keep the differentials by changing the variable. You should do it always together. The problem starts from very beginning, where you use both $\theta$ and $dx$. – user Mar 08 '19 at 01:00
  • @user it is totally correct to do so, just that it won’t be useful in integration. – Integreek Dec 14 '24 at 16:10

5 Answers5

5

$$\int \frac{dx}{(x^2-1)^{\frac{3}{2}}} = \int \frac{\tan\theta\sec\theta d\theta}{\underbrace{(\tan^2\theta)^{\frac{3}{2}}}_{\tan^3\theta}} = \int \frac{\sec\theta d\theta}{\tan^2\theta}= -\frac1{\sin \theta}=-\frac{x}{\sqrt{x^2-1}}.$$

user
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  • How do you get -$\frac{1}{\sin\theta}$? Please see my update – Evan Kim Mar 08 '19 at 00:54
  • $$\int \frac{\sec\theta d\theta}{\tan^2\theta}=\int \frac{\cos\theta d\theta}{\sin^2\theta}=\int \frac{d\sin\theta}{\sin^2\theta}=-\frac1{\sin^2\theta}$$ – user Mar 08 '19 at 00:57
  • I don't understand, what is the $d\sin\theta$ notation mean? It doesn't look like regular u-substitution? – Evan Kim Mar 08 '19 at 00:58
  • For simplicity imagine that I introduced new variable $u=\sin\theta$. Thus $$\int\frac{d\sin\theta}{\sin^2\theta}=\int\frac{du}{u^2}=-\frac1u=-\frac1{\sin\theta}.$$ – user Mar 08 '19 at 01:04
4

Note that $$(\tan^2 x)^{3/2}=\tan^3 x$$

Thus, yous should instead obtain $$\int\frac1{(x^2-1)^{3/2}}dx =\int\frac{\sec t\tan t}{\tan^3 t}dt $$

which simplifies to $$\int\frac{\cos t}{\sin^2 t}=\int\frac1{\sin ^2 t}d(\sin t)=-\frac1{\sin t}+C=-\csc t+C$$

Now reverse the substitution by the identity $\csc^2 t=\frac1{1-\cos^2 t}=\frac1{1-x^{-2}}$.

Lee
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1

If trigonometric substitution isn’t necessary, the fastest way to evaluate the given integral is to observe that it can be written as $\int\frac{\mathrm dx}{(x^2-1)\sqrt{x^2-1}}$ which is of the form $\int\frac{\mathrm dx}{Q_1\sqrt Q_2}$, the substitution $x=\frac1t$ can be performed:

$$\begin{align}\int\frac{\mathrm dx}{(x^2-1)^\frac32}&=-\text{sgn}(x)\int\frac{t\mathrm dt}{(1-t^2)^\frac32}\\&=-\frac{\text{sgn}(x)}{\sqrt{1-t^2}}+C\\&=-\frac{x}{\sqrt{1-x^2}}+C \end{align}$$

Integreek
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0

If you do $x=\sec\theta\,\mathrm d\theta$ and $\mathrm dx=\sec(\theta)\tan(\theta)\,\mathrm d\theta$, what you get is$$\int\frac{\sec(\theta)\tan(\theta)}{\tan^3\theta}\,\mathrm d\theta=\int\frac{\cos^2\theta}{\sin^3\theta}\,\mathrm d\theta=\int\frac{\cos^2(\theta)\sin(\theta)}{\bigl(1-\cos^2(\theta)\bigr)^2}\,\mathrm d\theta.$$Now, do $\cos\theta=u$ and $\sin\theta\,\mathrm d\theta=\mathrm du$.

José Carlos Santos
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0

If trig sub isn’t necessary, the given integral can be evaluated using Feynman’s technique by considering the function:

$$\mathcal I(a)=\int_a^x\frac{\mathrm dt}{\sqrt{t^2-a^2}}$$

Note that $\mathcal I(a)$ can be simplified as

$$\mathcal I(a)=\ln|x+\sqrt{x^2-a^2}|$$

Hence, by the Leibniz integral rule, and since $x$ is constant w.r.t $a$,

$$\int_a^x\frac{\partial }{\partial a}\left(\frac1{\sqrt{t^2-a^2}}\right)\mathrm dt=\frac{\frac{-a}{\sqrt{x^2-a^2}}}{x+\sqrt{x^2-a^2}}$$ $$\implies \int_a^x\frac{\mathrm dt}{(t^2-a^2)^\frac32}=\frac{-2a}{\sqrt{x^2-a^2}(x+\sqrt{x^2-a^2})}$$

If $f(t)$ be a function such that $f’(t)=\frac1{(t^2-a^2)^\frac32}$, then

$$\require{cancel}\begin{align}f(x)&=f(a)-\frac{2\cancel{a}}{\sqrt{x^2-a^2}}\frac{x-\sqrt{x^2-a^2}}{a\cancel{^2}}\\&=f(a)-\frac{x}{\sqrt{x^2-a^2}}+\frac2{a}\end{align}$$

Since infinitely many such functions $f(x)$ are possible, all of which differ by only a constant,

$$\therefore\int\frac{\mathrm dx}{(x^2-a^2)^\frac32}=-\frac{x}{\sqrt{x^2-a^2}}+C$$ $$\implies\int\frac{\mathrm dx}{(x^2-1)^\frac32}=-\frac{x}{\sqrt{x^2-1}}+C $$

Integreek
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