How to solve the given integral?

Question

tan 2x / sqrt tan² 2x-15

Answer 1

I'm going to assume that your question is written as:

tan(2x) / sqrt(tan^2(2x)-15)

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First, let's use trig identities to replace tan^2(2x) with sec^2(2x) - 1, simplifying to:

tan(2x) / sqrt(sec²(2x)-16)

Now we do u-substitution, using the entire function under the sqrt as our 'u':

u = sec²(2x) - 16 [note: sec^2(2x) = u + 16]

du = 2(sec(2x))*sec(2x)tan(2x)*2 dx = 4sec^2(2x)tan(2x) dx

(1/4)du = sec²(2x)tan(2x) dx

Let's now substitute sec²(2x) with u+16:

(1/4)du = (u+16)tan(2x) dx

Now, substitute into integral function:

(1/4) ∫ (u+16) / sqrt(u) du

And now we have a much easier integral function to deal with.

Split fraction into 2 parts:

(1/4) ∫ sqrt(u) + 16sqrt(u) du

Take the integral of both parts, we get:

(1/4) [(2/3)u^3/2 + 32sqrt(u)] + C

Substitute u for sec²(2x)-16, the final solution should be:

(1/4) [(2/3)(sec²(2x)-16)^3/2 + 32sqrt(sec²(2x)-16)] + C

Please let me know if you have any questions or if this is the correct answer or not. Thank you.