Methods of Integration: Powers of Tangent and Secant - Indefinite Integral

  When the integrands are Tangent and Secant together that are raised with a particular power it is better to know the techniques on simplifying it.

Here’s how the given problem should look like,

General form:

    ∫ tan^mu secⁿudu

Where m and n are exponents (constant)

Here are the rules that will help us:

Rule I. If m is any number and n is even.

    Use:
      ∫ tan^mu sec^n-2u sec²udu

Rule II. If m is a positive odd integer and n is any number.

    Use:
        ∫ tan^m-1u sec^n-1u secu tanudu

Rule III. If m is a positive integer and n is zero.
 

    Use:
        ∫ tan^m-2u tan²udu

Examples:

Rule I

1. ∫ tan³x sec⁴xdx = ∫ tan³x sec^4-2x sec²xdx

= ∫ tan³x sec²x sec²xdx

= ∫ tan³x (tan²x + 1) sec²xdx

= ∫ [tan⁵xsec²x + tan³xsec²xdx]dx

= ∫ tan⁵xsec²xdx + ∫ tan³xsec²xdx

Let:

u = tanx

du = sec²xdx

= tan⁶x/6 + tan⁴x/4 + C

Rule II

2. ∫ tan³x sec⁵xdx

= ∫ tan^3-1x sec^5-1x secxtanx dx

= ∫ tan²x sec⁴x secxtanx dx

= ∫ (sec²x - 1) sec⁴x secxtanx dx

= ∫ [sec⁶x secxtanx - sec⁴x secxtanx]dx

= ∫ sec⁶x secxtanxdx - ∫ sec⁴x secxtanxdx

Let:

u = secx
du =secxtanxdx

= sec⁷x/7 - sec⁵x/5 + C

Rule III

3. ∫ tan³xdx

= ∫ tan^3-2x tan²xdx

= ∫ tanx tan²xdx

= ∫ tanx (sec²x - 1)dx

= ∫ [tanxsec²x - tanx]dx

= ∫ tanxsec²xdx - ∫ tanxdx

= tan²x/2 - ln(secx) + C

Note:
    - These rules can also be used in a problem having cosecant and cotangent together as an integrand.