6.5
Integrals involving powers of tangent and secant can be solved using substitution, and the approach depends on whether the power of secant is even or the power of tangent is odd.
If the power of secant is even, choose tangent as the new variable, because its derivative is secant square, which can be factored out.
The remaining even power of secant can then be completely converted into tangent terms using the identity that relates secant square to tangent square.
This substitution turns the integral into a polynomial, which is then integrated and converted back to the original variable.
If the power of tangent is odd, a similar method is used. Let secant be the new variable, since its derivative secant times tangent can be factored out.
Use a trigonometric identity to rewrite the remaining even power of tangent in terms of secant.
After substitution, the integral becomes a polynomial expression, which can be integrated directly.
Finally, convert the result back to the original trigonometric function to complete the solution.
Integrals involving powers of tangent and secant are commonly evaluated using substitution, with the strategy determined by the parity of the exponent…
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