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Me again, probably someone is going to blame me. I have this:

$$\int{\frac{dx}{\sqrt{16-9x^2}}}$$

I have asked an old teacher of mine an he said me I should let $x=\frac{4}{3}\sin{\big(\frac{3}{4}x\big)}$ but I don't how he realise that. I have tried $u-substitution$ and Integration by Parts but the square root is making the whole thing much harder. I tried multiplying both terms by $\sqrt{16-9x^2}$ but nothing. Also tried to factor to eliminate square root but nothing. I don't want that anyone solve my integral, just give me a hint about how do it, and how I can realise what to do in this cases :)

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5 Answers 5

up vote 3 down vote accepted

In general, computing an integral is a guessing process. We usually rely on the fundamental theorem of calculus to provide us the initial guess i.e. if you want to compute $\displaystyle \int f(x) dx$, we need to first guess $F(x)$ such that $F'(x) = f(x)$. Then we make use of the fundamental theorem of calculus to conclude that $\displaystyle \int f(x) dx = F(x) + c$.

If you have an integral of the form $\displaystyle \int \dfrac{dx}{\sqrt{b^2 - a^2x^2}}$, recall that you would seen something similar before. You would have learnt that the derivative of $\arcsin(x)$ is $\dfrac1{\sqrt{1-x^2}}$. But now you the $a$ and $b$ hanging around. The goal now is to convert $\sqrt{b^2 - a^2x^2}$ into something like $\sqrt{1-u^2}$, so that we can then recognize that to be the derivative of $\arcsin(u)$.

Consider $\sqrt{b^2 - a^2x^2}$. The first thing is to pull out the $b$. This gives us $$\sqrt{b^2 - a^2x^2} = b \sqrt{1 - \dfrac{a^2}{b^2}x^2}$$ Now it looks very similar to a scaled version of $\sqrt{1-u^2}$ except for the coefficient in front of $x^2$. This gives us the motivation to make the substitution. $u^2 = \dfrac{a^2}{b^2}x^2$ i.e. $u = \dfrac{a}b x$.

In our case $a = 3$ and $b=4$. Hence, substitute $u = \dfrac34x$. We then get $du = \dfrac34 dx$. Plug this into the integral and recall that the derivative of $\arcsin(u)$ is $\dfrac1{\sqrt{1-u^2}}$ to get the integral.

Move you mouse over the gray area for the answer.

$$ I = \int \dfrac{dx}{\sqrt{16-9x^2}} = \int \dfrac43 \dfrac{du}{\sqrt{16 - 9 \times \dfrac{16}9 u^2}} = \int \dfrac43 \dfrac{du}{4\sqrt{1-u^2}} = \dfrac13 \arcsin(u) + c = \dfrac13 \arcsin \left( \dfrac34x\right) + c$$

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Nice, so it is possible to do whitout trigonometric substitution –  Andres Jun 24 '12 at 16:08
    
@Andres To do the last step $$\int \dfrac{du}{\sqrt{1-u^2}}$$ you can either recall from before or make use of the trigonometric substitution $u = \sin(\theta)$. –  user17762 Jun 24 '12 at 16:09
    
Thank you I fully understand it now :) If only my teachers were like you :) –  Andres Jun 24 '12 at 17:00

Let $x=\frac{4}{3}u$.${}{}{}{}$ After some algebra, we end up with the familiar integral $$\int\frac{k}{\sqrt{1-u^2}}\,du$$ for a not difficult to compute constant $k$.

Motivation The bottom looks like $\sqrt{1-x^2}$, except that the constants are wrong. One way to think about it is to note that the bottom is equal to $4\sqrt{1-\frac{9x^2}{16}}$, which is closer to the desired shape. Now we would like to have $\frac{9x^2}{16}=u^2$, which is achieved by setting $u=\frac{3x}{4}$. We did it a little more efficiently, by deciding we wanted a $16$ where there is a $9$.

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It is easy calculation just you have to continue $x=\frac{3}{4}u$..... –  Babak Miraftab Jun 24 '12 at 15:49

The substitution is "trigonometric substitution": one of the standard tools one learns in an introductory calculus course. It is usually one of the first things to try on any integral involving the square root of a quadratic function.

To work out the detail, it may help to draw a right triangle, where the integrand is the length of one the three sides, one of the three sides has constant length, and the length of the remaining side involves the variable. This problem is particularly simple because it's easy to rewrite the integrand to involve $\sqrt{a^2 - (b x)^2}$. The substitution is to rewrite the integral in terms of an angle of the triangle.

Messier problems would involve completing the square before you could do this.

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Mmmm... My teacher didn't teach me that. Is any rule to do that substituion like $u-substitution$ ? –  Andres Jun 24 '12 at 15:49
    
@Andres: Yes, trigonometric substitution is just an ordinary substitution. It has a special name because it comes up frequently enough to deserve one. –  Hurkyl Jun 24 '12 at 15:50

All you need here is to apply the well-known formula:

$$\int\frac{dx}{\sqrt{a^2-x^2}}=\arcsin{\frac{x}{a}}+C.$$

Therefore we get that:

$$\int\frac{dx}{\sqrt{16-9x^2}}=\frac{1}{3}\int\frac{dx}{\sqrt{(\frac{4}{3})^2-x^2}}=\frac{1}{3}\arcsin\left(\frac{3}{4}x\right)+C.$$

NOTE: it's important to recognize the elementary integrals because in some situations when dealing with more complex integrals, we know how to successfully split it into elementary
integrals, when this is possible.

Q.E.D.

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The given integral is: $\int{\frac{dx}{\sqrt{16-9x^2}}}$.

This integral can remark as: $\int{\frac{dx}{\sqrt{4^2-(3x)^2}}}$.

By received substition 3x with $4\sin t$ it is $3x=4\sin t$ have:

$3x=4\sin t$ $\Rightarrow$ $x=\frac{4}{3}\sin t$.

Hence we: $dx=\frac{4}{3}\cos t$

If this data replacement the intial integral we have:

$\int{\frac{dx}{\sqrt{16-9x^2}}}$=$\int{\frac{dx}{\sqrt{4^2-(3x)^2}}}$=$\int\frac{4\cos t dt}{3\sqrt{16-(4\sin t)^2}}$=$\frac{4}{3}\int\frac{cos t dt}{\sqrt{16-16\sin^2 t}}$=$\frac{4}{3}\int\frac{cos t dt}{\sqrt{16(1-\sin^2 t)}}$=$\frac{4}{3}\int\frac{cos t dt}{\sqrt{16}\cdot\sqrt{1-\sin^2 t}}$=$\frac{4}{3}\int\frac{cos t dt}{4\cdot\cos t}$=$\frac{4}{3}\cdot\frac{1}{4}\int{dt}$=$\frac{1}{3}t$

By substition of the above have:

$3x=4\sin t$ $\Rightarrow$ $\sin t=\frac{3}{4}\cdot x$ $\Rightarrow$ $ t=\arcsin \frac{3}{4}\cdot x $.

Theres definitly the given integral is:

$\int{\frac{dx}{\sqrt{16-9x^2}}}$=$\frac{1}{3}t$=$\frac{1}{3}\arcsin \frac {3}{4}\cdot x+C $.

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