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).