Integration help please!!?

Question

integral (tan^3x)(sec^4x)dx

Answer 1

Integral (tan^3(x) sec^4(x) dx )

First, decompose sec^4(x) into sec^2(x) and sec^2(x).

Integral (tan^3(x) sec^2(x) sec^2(x) dx )

Use the identity sec^2(x) = tan^2(x) + 1.

Integral (tan^3(x) (tan^2(x) + 1) sec^2(x) dx )

At this point, we use substitution.

Let u = tan(x). Then
du = sec^2(x) dx. {Note: this is the tail end of our integral, and we replace as appropriate.}

Integral (u^3 (u^2 + 1) du )

Distributing the u^3,

Integral (u^5 + u^3) du

Now, we integrate normally using the reverse power rule.

(1/6)u^6 + (1/4)u^4 + C

Replacing u = tan(x), we have

(1/6) tan^6(x) + (1/4) tan^4(x) + C

Answer 2

Integrate (tan³x)(sec^4(x)) with respect to x.

First rearrange the terms to make it easier to integrate.

(tan³x)(sec^4(x)) = (tanx)(tan²x)(sec^4(x))
= (tanx)(sec²x - 1)(sec^4(x))
= (tanx)(sec^6(x) - sec^4(x))
= [(tanx)(secx)][sec^5(x) - sec³x]

Now we can integrate.

∫(tan³x)(sec^4(x)) dx
= ∫[(tanx)(secx)][sec^5(x) - sec³x] dx
= (1/6) sec^6(x) - (1/4) sec^4(x) + C