Integrals of Trig Functions

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Product of Secants and Tangents

Products of secants and tangents have strategies, in the same way that the products of sines and cosines have strategies. Based on the even and odd powers, we evaluate $$\int \sec^n(x) \tan^m(x)\, dx.$$

If $n=2$, we can use the substitution $u=\tan(x)$, $du=\sec^2(x)\, dx$ to get $\int u^m\, du$.
If $n$ is even, we can use the identity $\sec^2(x)=1+\tan^2(x)$ to convert all but two powers of secant into tangents. (If $n=0$ and $m$ is large, things get tedious, but we can convert two powers of tangent into secants instead, thereby reducing the power of tangent. Then repeat on the reduce power of tangent.)
If $m=1$ (and $n>1$), we can substitute $u=\sec(x)$, $du=\sec(x)\tan(x)\, dx$ to get $\int u^{n-1}\, du$. (If $m$ and $n$ are both one, we know the antiderivative.)
If $m$ is odd, we can use the identity $\tan^2(x) = \sec^2(x)-1$ to convert all but one power of tangent into secants.
Other cases can be much harder. We look at more challenging integrals later in this module.

Alternatively, as we did with sines and cosines, we can observe the combinations of secant and tangent functions in the integrand, and look for a derivative to strip off.

Possible derivatives to strip off: $\sec^2x\,dx$ with the remaining factors of the integrand containing only tangent functions (which may involve substituting $\sec^2 x=\tan^2 x+1$); and $\sec x\tan x\,dx$, with the remaining factors of the integrand containing only secant functions (which may involve substituting $\tan^2 x=\sec^2 x-1$). For some integrands, this cannot be done, but we will not ask you to compute such integrals.

Some examples:

$\int\sec^6x\tan^5x\,dx=\int\sec^4x\tan^5x\sec^2x\,dx$. Here, we stripped off $\sec^2x\,dx$ since we know the remaining even power of $\sec x$ can be rewritten in terms of $\tan x$. Now we make our substitution to get $\int(\tan^2 x+1)^2\tan^5x\sec^2\,dx$ and let $u=\tan x$. This will work for any power of tangent.

$\int\sec^5x\tan^7x\,dx$. Here, we don't want to strip off $\sec^2x\,dx$, since the remaining power of secant cannot be turned into tangents. So try stripping off $\sec x\tan x\,dx$, getting $\int\sec^4x\tan^6x\sec x\tan x\,dx$. Now we can set $\tan^6x=(\sec^2x-1)^3$ and let $u=\sec x$.

The Fundamental Theorem of Calculus

The Indefinite Integral and the Net Change

Substitution

Area Between Curves

Volumes

Integration by Parts

Integrals of Trig Functions

Trig Substitutions

Partial Fractions

Strategies of Integration

Improper Integrals

Differential Equations

Models of Growth

Infinite Sequences

Infinite Series

Integral Test

Comparison Tests

Convergence of Series with Negative Terms

The Ratio and Root Tests

Strategies for testing Series

Power Series

Representing Functions as Power Series

Taylor and Maclaurin Series

Applications of Taylor Polynomials

Partial Derivatives

Multiple Integrals

Iterated Integrals over Rectangles

Double Integrals over General Regions

Product of Secants and Tangents