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I am having trouble with a question really need help please.

$$ \int {\sqrt{x^2 + 81} \over 2} \,dx $$

I thought about taking the square root off and turning the question into $\frac 12 \int (x^2 +81)^{1/2}\, dx$ but then wondered if I could use the quotient rule.

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Unfortunately, there is no such thing as a "quotient rule" for integration. What other techniques do you know? If this is HW for a course, what section are you currently studying? There is a particular technical method that works, based on the fact that "$x^2 + a^2$" shows up inside a radical. – Shaun Ault Mar 7 '12 at 13:48

Hint: Put $x = 9 \tan{t}$, then $x^{2}+81 = 81(\tan^{2}(t)+1) = 81 \cdot \sec^{2}(t)$.

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+1, very good substitution. – Américo Tavares Mar 7 '12 at 22:33

$$ \int \sqrt{x^2+81} dx $$

Integration by parts formula

$$ \int u(x)v^{'}(x) dx = u(x)v(x) - \int v(x)u^{'}(x) dx$$

Therefore assume that $v^{'}(x) = 1$ in this case

Denote the integral

$$ I = \int \sqrt{x^2+81} dx$$

$$ \begin{align*} I &= x \sqrt{x^2+81} - \int \frac{x^2}{\sqrt{x^2+81}} dx\\ &= x \sqrt{x^2+81} - \int \frac{x^2+81-81}{\sqrt{x^2+81}} dx\\ &= x \sqrt{x^2+81} - I + 81 \int \frac{1}{\sqrt{x^2+81}} dx \end{align*} $$


$$ 2I = x \sqrt{x^2+81} +81 \int \frac{1}{\sqrt{x^2+81}} dx$$

the rest you should do it yourself.

I was told that we are not supposed to give complete solution for homework questions.


$u(x) = \sqrt{x^2+81}$ and

$v(x) = x$, therefore $v^{'}(x) = 1$

$$ u^{'}(x) = \frac{1}{2}\left(x^2+81\right)^{-\frac{1}{2}} \times 2x = \frac{x}{\sqrt{x^2+81}}$$

$$ \begin{align*} \int \sqrt{x^2+81} dx &= \int u(x) v^{'}(x) \\ &= u(x)v(x) - \int u^{'}(x) v(x) dx\\ &= x \sqrt{x^2+81} - \int \frac{x^2+81-81}{\sqrt{x^2+81}} dx\\ &= x \sqrt{x^2+81} - \int \sqrt{x^2+81} \hspace{3pt} dx + \int \frac{81}{\sqrt{x^2+81}} dx\\ \end{align*} $$

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In your final expression, the variable $x$ appears outside the integral, without any proper definition there. You should change this notation. – shuhalo Mar 7 '12 at 15:12

The simplest method I see here is to replace x by something else. We know that $a^2+x^2 = x^2+81 = 9^2+x^2$ therefore we can put $x = 9\tan(\theta)$ and get

$dx = 9\sec^2\theta d\theta$

Therefore $$1/2 \int \sqrt{9^2+9^2\tan^2(\theta)} 9\sec^2 \theta d\theta =$$

$$27/2 \int \sqrt{1+\tan^2(\theta)} \sec^2 \theta d\theta =$$

$$27/2 \int \sec^3 \theta d\theta $$

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you don't need to delete your own answer if someone answers at the same time as you. – Harry Stern Mar 7 '12 at 18:56
@HarryStern: It is usually a good idea to try to reduce noise, though. – Aryabhata Mar 7 '12 at 21:14
okay! thanks. PS sec^3 is a lot of fun to integrate – Sam Creamer Mar 9 '12 at 16:58
I am just new here, but I wonder how people vote here because this elaborates first answer (by Chandrasekhar), but that got 11 votes and this has 0 (I am voting this one)- Weird voting system – Jeremy Carlos Mar 17 '12 at 15:02

Added. Euler substitution

  • The Euler substitution $$\begin{equation*} \sqrt{x^{2}+81}=t-x\Leftrightarrow x=\frac{1}{2}\frac{t^{2}-81}{t}, dx=\frac{1}{2}\frac{t^{2}+81}{t^{2}}\;\mathrm{d}t \end{equation*}$$ reduces the given integral to an integral of a rational function in $t$ $$\begin{eqnarray*} \frac{1}{2}\int \sqrt{x^{2}+81}dx &=&\frac{1}{2}\int \left( t-\frac{1}{2}\frac{t^{2}-81}{t}\right) \cdot \frac{1}{2}\frac{t^{2}+81}{t^{2}}dt \\ &=&\frac{1}{2}\int \frac{6561+162t^{2}+t^{4}}{4t^{3}}dt=\frac{1}{16}t^{2}-\frac{6561}{ 16t^{2}}+\frac{81}{4}\ln t. \end{eqnarray*}$$

Added 2. Hyperbolic substitution

  • The hyperbolic substitution $$\begin{equation*} x=9\sinh t\Leftrightarrow t=\operatorname{arcsinh}\frac{x}{9},dx=9\cosh tdt,\end{equation*}$$ gives $$\begin{eqnarray*} \frac{1}{2}\int \sqrt{x^{2}+81}dx &=&\frac{1}{2}\int 9\sqrt{81\sinh ^{2}t+81} \cosh t\,dt \\ &=&\frac{81}{2}\int \cosh ^{2}t\,dt \\ &=&\frac{81}{2}\int \left( \frac{e^{2t}}{4}+\frac{1}{2}+\frac{e^{-2t}}{4} \right) \,dt \\ &=&\frac{81}{2}\left( \frac{1}{8}e^{2t}+\frac{1}{2}t-\frac{1}{8} e^{-2t}\right). \end{eqnarray*}$$ Since $t=\operatorname{arcsinh}\frac{x}{9}=\ln \left( \frac{x}{9}+\frac{1}{9}\sqrt{x^{2}+81}\right) $, we obtain $$\begin{eqnarray*} \frac{1}{2}\int \sqrt{x^{2}+81}dx&=&\frac{81}{2}\left( \frac{1}{8}e^{2\operatorname{arcsinh}\frac{x}{9}}+\frac{1}{2}\operatorname{arcsinh}\frac{x}{9}-\frac{1}{8}e^{-2\operatorname{arcsinh}\frac{x}{9}}\right) \\ &=&\frac{81}{16}\left( \frac{1}{9}x+\frac{1}{9}\sqrt{x^{2}+81}\right) ^{2}+ \frac{81}{4}\ln \left( \frac{1}{9}x+\frac{1}{9}\sqrt{x^{2}+81}\right) \\ &&-\frac{81}{16}\left( \frac{1}{9}x+\frac{1}{9}\sqrt{x^{2}+81}\right) ^{-2} \\ &=&\frac{81}{4}\ln \left( \frac{1}{9}x+\frac{1}{9}\sqrt{x^{2}+81}\right) +\frac{1}{4}x\sqrt{x^{2}+81}+\text{Constant}. \end{eqnarray*}$$

There is a general method for integrating a function of the form $P(x)/\sqrt{ax^{2}+bx+c}$ I have already posted here.

If $P(x)$ is a polynomial of degree $n\geq 2$, we can find a polynomial $Q(x) $ of degree $n-1$ and a constant $C$ such that$^1$

$$\int \frac{P(x)}{\sqrt{ax^{2}+bx+c}}\;\mathrm{d}x=Q(x)\sqrt{ax^{2}+bx+c}+\int \frac{C}{\sqrt{ax^{2}+bx+c}}\;\mathrm{d}x.$$

The given integral is $1/2$ of $$ \begin{equation*} I(x):=\int \sqrt{x^{2}+81}dx=\int \frac{x^{2}+81}{\sqrt{x^{2}+81}}\mathrm{d}x=Q(x) \sqrt{x^{2}+81}+\int \frac{C}{\sqrt{x^{2}+81}}\mathrm{d}x, \end{equation*}$$ with $Q(x)=Ax+B$. To find the constants $A$ and $B$ differentiate both sides $$\begin{eqnarray*} \frac{x^{2}+81}{\sqrt{x^{2}+81}} &=&A\sqrt{x^{2}+81}+\frac{2x\cdot \left( Ax+B\right) }{2\sqrt{x^{2}+81}}+\frac{C}{\sqrt{x^{2}+81}} \\ &=&\frac{2Ax^{2}+Bx+81A+C}{\sqrt{x^{2}+81}} \end{eqnarray*}$$ and equate the coefficients in the numerators $$\begin{eqnarray*} 2A &=&1\Leftrightarrow A=\frac{1}{2} \\ B &=&0 \\ 81A+C &=&81\Leftrightarrow C=\frac{81}{2}. \end{eqnarray*}$$ Consequently $$ \begin{equation*} I(x)=\frac{1}{2}x\sqrt{x^{2}+81}+\frac{81}{2}\int \frac{1}{\sqrt{x^{2}+81}}\mathrm{d}x.\end{equation*}$$ Note: This result is essentially the same as Kirthi Raman's. The integral on the right can be easily evaluated using the substitution $x=9u$, because then becomes a direct integral $$\begin{eqnarray*} \int \frac{1}{\sqrt{x^{2}+81}}dx &=&\int \frac{1}{\sqrt{u^{2}+1}}\mathrm{d}u \\ &=&\operatorname{arcsinh} u=\operatorname{arcsinh}\frac{x}{9}+\text{Constant}. \end{eqnarray*}$$

To write it in terms of the natural logarithmic use the identity $$\begin{equation*} \operatorname{arcsinh}u=\ln \left( u+\sqrt{u^{2}+1}\right) \end{equation*}.$$

$^1$ Described in Cálculo Integral em $\mathbb{R}$ by M. Olga Baptista.

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There exists a simple formula for the integrals of the type $\int{\sqrt{x^2 + a^2}} dx$ i.e. $$\int{\sqrt{x^2 + a^2}} dx=\space\frac{x}{2}\sqrt{x^2 + a^2}+\frac{a^2}{2}\arctan{\frac{x}{a}}+ C$$ You can prove this by putting $x= a\tan\theta$, differantiating both sides to get $dx=a \ \sec^2\theta \space d\theta$, and evaluating the simple integral that follows. Using this formula, your problem is given by $$\int {\sqrt{x^2 + 81} \over 2} \,dx= \int \sqrt{x^2 + 9^2} \,dx=\frac{x}{2}\sqrt{x^2 + 9^2}+\frac{81}{2} \arctan{\frac{x}{9}+C}$$ Hope this helps.

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