# Answer in Calculus for Carlos #205232

Solve the integral of √(49-x²)÷x from ln1 to ln2

“int_{ln1}^{ln2} {frac{sqrt{49-xu00b2}}{x}}dx”

Let x = 7sin“theta”

dx = 7cos“theta” d“theta”

So “int {frac{sqrt{49-xu00b2}}{x}}dx” =

“int {frac{sqrt{49-49sinu00b2theta}}{7sintheta}}7cos{theta}dtheta”

= “int {frac{7cosu00b2{theta}}{sintheta}}dtheta”

= “7int {frac{1-sinu00b2{theta}}{sintheta}}dtheta”

= “7int [{cosectheta+ sintheta}]dtheta”

Let ln(cosec“theta” + cot“theta”) = t

So “frac{-cosectheta cottheta-cosecu00b2theta)}{cosecthetau03b8 + cotthetau03b8)}dtheta”= dt

=> cosec“theta” d“theta = -dt”

= “-7int dt -7int sintheta dtheta”

= -7t + 7cos“theta” + C

= -7 ln|cosec“theta” + cot“theta” | + 7cos“theta”

= 7 ln| “frac{sintheta}{1+costheta}” | + 7cos“theta” + C

= “ln[frac{2sintheta/2costheta/2}{2cosu00b2theta/2}]+ 7costheta + C”

= ln[tan “theta/2]+ 7costheta+C”

Now as x–> ln1 , “theta” –> 0 and as x–> ln2 , “theta” –> sin ^{-1}(ln2/7)

So

“int_{ln1}^{ln2} {frac{sqrt{49-xu00b2}}{x}}dx” = “[ln(tantheta/2)+7costheta]]_{0}^{sin^{-1}(ln2/7)}”

= “[ln(tan(frac{1}{2}sin^{-1}(ln(2)/7)+”

“7cos((sin^{-1}(ln(2/7)]-“

“[ln(tan(0)+” “7cos(0)]”

= “[ln(tan(frac{1}{2}sin^{-1}(ln(2)/7)+”

“7cos((sin^{-1}(ln(2/7)] -7 – ln(0)”

= A finite quantity + ∞ as ln(0) –>-∞

= ∞