Tutorials on Advanced Math and Computer Science Concepts

# Trigonometric Integrals

Trigonometric functions appear in the natural world frequently, so understanding how to integrate them is essential to solving real problems. There are a number of situations we can have involving trigonometric function, and corresponding rules to solve them.

Example: Evaluate $\int cos^3(x)dx$

If we tried using the substitution rule, we would have $u = cos(x)$ and $du = -sin(x)dx$, which doesn't really help. However, we can manipulate this function to be easier to work with. If we factor our a $cos^2(x)$ term, we would have $\int (cos^2(x)cos(x))dx$

From here, we can use the trig identity, $sin^2(x) + cos^2(x) = 1$ to change the $cos^2(x)$ into $1-sin^2(x)$. This gives us the following integral.

$\int(1-sin^2(x))(cos(x))dx$

From here, we can now apply the substitution rule to get a solution. If we let u = sin(x), then du = cos(x)dx, giving us the integral $\int (1-u^2)du = u - \frac{1}{3}u^3 + C = sin(x) - \frac{1}{3}sin^3(x) + C$

When we have functions in the form $sin^m(x)cos^n(x)$, there is a set of rules we can follow to evaluate them correctly.

Strategy for evaluating $\int sin^m(x)cos^n(x)$

1. If the power of cosine, n, is odd (n = 2k+1), save one factor of cosine and use $cos^2(x) = 1-sin^2(x)$ to express the remaining factors in terms of sine: $\int sin^mcos^{2k+1}dx = \int sin^m(x)(cos^2(x))^kcos(x) dx = \int sin^m(x)(1-sin^2(x))^kcos(x) dx$, then substitute u = sin(x)
2. If the power of sine is odd (m = 2k+1), save one factor of sine and use $sin^2(x) = 1 - cos^2(x)$ to express the remaining factors in terms of cosine: $\int sin^{2k+1}(x)cos^n dx = \int (sin^2(x))^kcos^n(x)sin(x)dx = \int (1-cos^2(x))^k cos^n(x) sin(x) dx$, then substitute u = cos(x)
3. If both the power of sine is odd and the power of cosine is odd, then either 1 or 2 can be used.
4. If the powers of both sine and cosine are even, use the half angle identities $sin^2(x) = \frac{1}{2}(1-cos(2x))$ or $cos^2(x) = \frac{1}{2}(1+cos(2x)$

We can also evaluate functions that include tan(x) and sec(x)

Strategy for evaluating $\int tan^m(x)sec^n(x) dx$

1. If the power of secant is even (n = 2k), save a factor of $sec^2(x)$, and use $sec^2(x) = 1+tan^2(x)$ to express the remaining factors in terms of tan(x): $\int tan^m sec^{2k} dx = \int tan^m(sec^2(x))^{k-1}sec^2(x) dx = \int tan^m(x) (1+tan^2(x))^{k-1} sec^2(x) dx$, the substitute u = tan(x)
2. If the power of tangent is odd (m = 2k+1), save a factor of sec(x)tan(x) and use $tan^2(x) = sec^2(x) - 1$ to express the remaining factors in terms of sec(x): $\int tan^{2k+1}(x)sec^n(x) dx$
$= \int (tan^2(x))sec^{n-1}(x) sec(x)tan(x) dx$
$= \int (sec^2(x) - 1)^k sec^{n-1}(x)tan(x) dx$, then substitute u = sec(x).