Question

Help with evaluating an indefinite integral?

Answer 1

\[\int\limits_{}^{}\frac{ (\sec^2\theta*\tan^2\theta) }{ \sqrt(9-\tan^2\theta)}\]

Answer 2

Take the integral:
integral (tan^2(x) sec^2(x))/sqrt(9-tan^2(x)) dx

For the integrand (tan^2(x) sec^2(x))/sqrt(9-tan^2(x)),
substitute u = tan(x) and du = sec^2(x) dx:
= integral u^2/sqrt(9-u^2) du

For the integrand u^2/sqrt(9-u^2),
substitute u = 3 sin(s) and du = 3 cos(s) ds. Then sqrt(9-u^2) = sqrt(9-9 sin^2(s)) = 3 cos(s) and s = sin^(-1)(u/3):
= integral 9 sin^2(s) ds

Factor out constants:
= 9 integral sin^2(s) ds

Write sin^2(s) as 1/2-1/2 cos(2 s):
= 9 integral (1/2-1/2 cos(2 s)) ds

Integrate the sum term by term and factor out constants:
= 9/2 integral 1 ds-9/2 integral cos(2 s) ds

For the integrand cos(2 s), substitute p = 2 s and dp = 2 ds:
= 9/2 integral 1 ds-9/4 integral cos(p) dp

The integral of cos(p) is sin(p):
= 9/2 integral 1 ds-(9 sin(p))/4

The integral of 1 is s:
= (9 s)/2-(9 sin(p))/4+constant

Substitute back for p = 2 s:
= (9 s)/2-9/4 sin(2 s)+constant

Apply the double angle formula sin(2 s) = 2 cos(s) sin(s):
= (9 s)/2-9/2 sin(s) cos(s)+constant

Express cos(s) in terms of sin(s) using cos^2(s) = 1-sin^2(s):
= (9 s)/2-9/2 sin(s) sqrt(1-sin^2(s))+constant

Substitute back for s = sin^(-1)(u/3):
= 9/2 sin^(-1)(u/3)-1/2 u sqrt(9-u^2)+constant

Substitute back for u = tan(x):
= 9/2 csc^(-1)(3 cot(x))-1/2 tan(x) sqrt(9-tan^2(x))+constant

Factor the answer a different way:
= 1/2 (9 csc^(-1)(3 cot(x))-tan(x) sqrt(9-tan^2(x)))+constant

Which is equivalent for restricted x values to:
Answer: = (sec(x) (9 sqrt(5 cos(2 x)+4) tan^(-1)((sin(x))/sqrt(5 cos(2 x)+4))-(5 cos(2 x)+4) tan(x) sec(x)))/(2 sqrt(9-tan^2(x)))+constant

Answer 3

Hope this helps.