Question

Trig substitution hard question help

Answer 1

\[\int\limits_{}^{}\sqrt{9x ^{2}-361}/x\]

Answer 2

i got \[\sqrt{9x ^{2}-361}-arcsec(3x/19) + C\]

Answer 3

wolfram alpha got that also but they simplified it to:
\[\sqrt{9x ^{2}-361}+19\arctan(19/\sqrt{9x ^{2}-361})+C\]

Answer 4

becuase of restricted domain.

Answer 5

both are wrong according to webworks..

Answer 6

ops forgot to put a 19 behind arcsecant(3x/19)

Answer 7

\[\int\limits_{}^{}\frac{\sqrt{(3x)^{2}-19^{2}}}{x}dx\]
\[3x=-19\sec \theta\]
\[x=\frac{-19\sec \theta}{3}\]
\[dx=-(19/3)\tan \theta \sec \theta d \theta\]
\[\huge \int\limits_{}^{}\frac{\sqrt{(-19\sec \theta)^{2}-19^{2})}}{\frac{-19\sec \theta}{3}}(\frac{-19}{3}\tan \theta \sec \theta d \theta)\]
\[\int\limits_{}^{}19\tan^2 \theta d \theta\]
\[=19(\tan \theta -\theta)\]

Answer 8

that looks like exactly what i had earlier

Answer 9

but just made changed it in terms of x

Answer 10

nvm i got it!

Answer 11

thanks!! just had a typo

Answer 12

\[\int\limits_{}^{}19\tan^2θ-19\tan \theta dθ\]
should be like this

Answer 13

thank you! and you were right. Your work made me see the typo

Answer 14

ok yw