Integration of dx/3sin^2 -4

= 1/2 tan^(-1)(2 cot(x)) + constant

Step-by-step explanation:

Take the integral:

integral1/(3 sin^2(x) - 4) dx

Multiply numerator and denominator of 1/(3 sin^2(x) - 4) by -sec^2(x):

= integral-(sec^2(x))/(4 sec^2(x) - 3 tan^2(x)) dx

Prepare to substitute u = tan(x). Rewrite -(sec^2(x))/(4 sec^2(x) - 3 tan^2(x)) using sec^2(x) = tan^2(x) + 1:

= integral-(sec^2(x))/(tan^2(x) + 4) dx

For the integrand -(sec^2(x))/(tan^2(x) + 4), substitute u = tan(x) and du = sec^2(x) dx:

= integral-1/(u^2 + 4) du

Factor out constants:

= - integral1/(u^2 + 4) du

Factor 4 from the denominator:

= - integral1/(4 (u^2/4 + 1)) du

Factor out constants:

= -1/4 integral1/(u^2/4 + 1) du

For the integrand 1/(u^2/4 + 1), substitute s = u/2 and ds = 1/2 du:

= -1/2 integral1/(s^2 + 1) ds

The integral of 1/(s^2 + 1) is tan^(-1)(s):

= -1/2 tan^(-1)(s) + constant

Substitute back for s = u/2:

= -1/2 tan^(-1)(u/2) + constant

Substitute back for u = tan(x):

= -1/2 tan^(-1)(tan(x)/2) + constant

Which is equivalent for restricted x values to:

= 1/2 tan^(-1)(2 cot(x)) + constant

idk the answer but just put a

step-by-step explanation:

the solution is 12x-6y