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integrating products of secants and tangents

Integrating products of secants and tangents

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When integrating products of secants and tangents, we are integrating expressions that are in the form:

$$\int tan^m(x)sec^n(x) $$

secant and tangent are both trigonometric functions that relate the angles of a right triangle to the ratios of its sides.

There are several cases to consider in this type of integration.

case 1

This is a case where m is an odd positive integer . In this case we split off the sec(x)tan(x) to form a differential sec(x)tan(x) of sec(x) along with dx.

We then use the identity ‘sec2x = 1-tan2x‘ to convert the remaining power into powers of sec(x). This way, we prepares the integrand for the substitution of u = sec(x). consider the following:

$$ \int tan^3x sec^3x dx = \int tan^2(x)sec^2(x)sec(x)tan(x)dx$$

u = sec(x)

$$\frac{du}{dx} = sec(x) tan(x) $$
$$dx = \frac{du}{sec(x)tan(x) }$$

hence

$$\int(sec^2x-1)sec^2(x)sec(x)tan(x)dx $$

substituting;

$$\int(u^2-1)u^2du = \int (u^4-u^2)du = [u^5 – u^3] +c $$ $$=sec^5x – sec^3x +c$$

case 2

This is a case where n is an even positive integer. We split sec2x to form a differential of tan x along with dx. We then use the identity ‘sec2x = 1 + tan2x’ to convert the remaining even powers of x into powers of tan(x). This prepares the integrand for substitution in u = tan(x). consider:

$$\int tan^5xsec^4xdx = \tan^5xsec^2xsec^2xdx = \int tan^5x(1+tan^2x)sec^xdx $$

u = tan(x) and du = sec2xdx

$$\int u^5(1+u^2)du = \int (u^5 + u^7)du = \frac{1}{6}tan^6x + \frac{1}{8}tan^8x + c $$

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