Question

Пожалуйста решите интеграл

Answer 1

`(√(3) )/(72)`

Explanation:

Evaluate the given definite integral.

`\int\limits^6_3 {(√(x^2-9) )/(x^4) } \, dx`

`\hrulefill`

(1) - Apply trig substitution

`\boxed{\left\begin{array}{ccc}\text{\underline{Trig Substitution:}}\\\\√(x^2-a^2) \rightarrow a\tan(\theta)\\\\ x=a\sec(\theta) \end{array}\right}`

`\int\limits^6_3 {(√(x^2-9) )/(x^4) } \, dx \\ \\ \\ \boxed{\left\begin{array}{ccc}\text{Our Substitutions...}\\√(x^2-9) \rightarrow 3 \tan(\theta)\\x=3sec(\theta)\\dx=3\sec(\theta)\tan(\theta)d\theta\end{array}\right}\\\\ \\ \Longrightarrow \int\limits^6_3 {(3\tan(\theta))/((3\sec(\theta))^4) } \, \cdot3\sec(\theta)\tan(\theta)d\theta \\\\\Longrightarrow \int\limits^6_3 {(3\tan(\theta))/(81\sec^4(\theta)) } \, \cdot3\sec(\theta)\tan(\theta)d\theta \\`

`\Longrightarrow (1)/(9) \int\limits^6_3 {(\tan(\theta))/(\sec^4(\theta)) } \, \cdot\sec(\theta)\tan(\theta)d\theta \\\\\Longrightarrow (1)/(9) \int\limits^6_3 {(\tan^2(\theta))/(\sec^3(\theta)) } \, d\theta`

(2) - Adjust the integral boundaries

`x=3sec(\theta)\rightarrow \theta=\sec^(-1)((x)/(3) )\\\\\text{\underline{When x=6}}\\\theta=\sec^(-1)((6)/(3) )\\\\\Longrightarrow \theta=\sec^(-1)(2 )\\\\\Longrightarrow \boxed{\theta=(\pi)/(3) }\\\\\text{\underline{When x=3}}\\\theta=\sec^(-1)((3)/(3) )\\\\\Longrightarrow \theta=\sec^(-1)(1 )\\\\\Longrightarrow \boxed{\theta=0 }\\\\\therefore \ \text{we have} \ \boxed{(1)/(9)\int\limits^{(\pi)/(3) }_(0) {(\tan^2(\theta))/(\sec^3(\theta)) }\, d\theta}}`

(3) - Simplify the integral using trig identities

`\boxed{\left\begin{array}{ccc}\text{\underline{Reciprocal/Quotient Identities:}}\\\\\tan(A)=(\sin(A))/(\cos(A)) \\\\\sec(A)=(1)/(\cos(A)) \end{array}\right}`

`(1)/(9)\int\limits^{(\pi)/(3) }_(0) {(\tan^2(\theta))/(\sec^3(\theta)) }\, d\theta}\\\\\Longrightarrow (1)/(9) \int\limits^(\frac\pi3)_0 {(((\sin(\theta))/(\cos(\theta)) )^2)/(((1)/(\cos(\theta)) )^3) } \, d\theta\\\\\Longrightarrow \boxed{(1)/(9)\int\limits^\frac\pi3_0 {\sin^2(\theta)\cos(\theta)} \, d\theta }`

(4) - Apply u-substitution and evaluate the integral

`u=\sin(\theta)\rightarrow du=\cos(\theta)d\theta\\\\\Longrightarrow (1)/(9)\int\limits^\frac\pi3_0 {u^2} \, du\\\\\Longrightarrow \frac19\Big[\frac13u^3\Big]\limits^(\frac\pi3)_(0)\\\\\Longrightarrow \frac19\Big[\frac13\sin^3(\theta)\Big]\limits^(\frac\pi3)_(0)\\\\\Longrightarrow \frac19\Big[\frac13\sin^3(\pi/3)-\frac13\sin^3(0)\Big]\\\\\Longrightarrow \frac19\Big[\frac13((3√(3) )/(8) )-0\Big]\\\\\Longrightarrow \frac19\Big[(√(3) )/(8)\Big]\\`

`\therefore \int\limits^6_3 {(√(x^2-9) )/(x^4) } \, dx = \boxed{\boxed{(√(3) )/(72) }}`