Question

\[evaluate \int\limits\limits \frac{dx}{4x \sqrt{x ^{2}-4^{2}}}\]

Answer 1

you can take out 4

\[\frac{1}{4} \int \frac{dx}{x\sqrt{x^2 -4^2}}\]
i believe this is arcsec yes?

Answer 2

trig sub

Answer 3

yup it is an arcsec.

Answer 4

then problem solved! :D

Answer 5

\(x=4sec\theta\) => \(dx = 4tan\theta sec\theta\)

Answer 6

the answer must \[\frac{1}{16}\sec ^{-1}\frac{1}{4}x+C\] but how???

Answer 7

sounds right...

Answer 8

\[\int \frac{dx}{x\sqrt{x^2 - 4^2}} \implies \frac{1}{4} \sec^{-1} \frac{1}{4}x\]

agree?

Answer 9

go back to the theorem of \[\int\limits \frac{du}{u \sqrt{u ^{2}-a ^{2}}} ==> \frac{1}{a}\sec ^{-1}\frac{u}{a}+C\]

Answer 10

@lgbasallote the frac in first term must \[\frac{1}{16}\]

Answer 11

that's because there is still a 1/4 out side

Answer 12

\[\frac{1}{4} [ \frac{1}{4} \sec^{-1} \frac 14 x \implies \frac{1}{16} \sec^{-1} \frac 14 x + C\]

Answer 13

yah your right. Thanks.

Answer 14

<tips hat>