Question

Please help!!! Integration

Answer 1

How do I solve the following?
\[\int\limits_{}^{}\frac{ dx }{ 16 +9x ^{2} }\]

Answer 2

I know the answer is meant to be \[\frac{ 1 }{ 12} \tan^{-1} (\frac{ 3x }{ 4 })+C\] but I'm not sure how to get there :/ !

Answer 3

this is a standard tan substitution, and uses identity tan^ ø + 1 = sec^2 ø as a substitution in the denominator.

to make the denominator fit this pattern, you want 9x^2 to be 16 tan^ ø so that you have 16(tan^1 + 1) in the denominator. that means you sub 9x^2 to = 16 tan^ ø --->> 3x = 4 tanø --->> x = 4/3 tanø.

the reason this sub is so handy is that x = 4/3 tanø -->> dx = 4/3 sec^2 ø dø (!!) so you have sec^ in the numerator and denominator.

Answer 4

thank you!

Answer 5

no worries, but a lot of confusing typos in my post so i will just correct the worst in case i get my head chewed off by others! the identity is:
\[\tan^{2}ø + 1 = \sec^{2}ø\]