Solving Trigonometric Integrals

$\bullet$ Techniques used in this part: u-Substitution and Integration by Parts

$\bullet$ u-Sub: $\int f(g(x)) g’(x) dx = \int f(u) du$ where $u = g(u)$

$\bullet$ Integration by Parts: $\int f(x) g’(x) dx = f(x)g(x) - \int f’(x)g(x)dx$

$\bullet$ Trigonometric Properties:

Derivatives and Integrals:

Half-Angle / Double Angle / Square and other properties:

Integral of type 1 : $\int \sin^m(x)\cos^n(x) dx$

Example 1 : Evaluate the integral $\int \cos^3(x) dx.$

We we may do $\cos^3(x) = \cos^2(x) \cos(x) = (1-\sin^2(x))\cos(x)$. Then the integral we are evaluating will become $\int (1 - \sin^2(x))\cos(x)dx$. Now By u-Substitution, we assume that $\sin(x) = u$, then we also know that $du = \cos(x)dx$, thus our integral can be written as $\int (1 - u^2)du$, which we can evaluate it since it’s a polynomial. Hence we have $\int (1 + u^2)du = u - \frac{1}{3}u^3 + C$, lastly we replace $u$ with $\sin(x)$, and we get $\int (1 - u^2)du = \sin(x) - \frac{1}{3}\sin^3(x) + C$

Exercise 1 : Evaluate the integral $\int \sin^5(x)\cos^2(x) dx.$

General strategy for integrals like $\int \sin^m(x) \cos^n(x)dx$:

(1) If we have $n = 2k + 1, k \in \mathbb{Z}$, rewrite as $\int \sin^m(x)(1 - \sin^2(x))^{k}\cos(x) dx $

(2) If we have $m = 2k + 1, k \in \mathbb{Z}$, rewrite as $-\int(1 - \cos^2(x))^{k}\cos^n(x)\sin(x) dx$

(3) Both $m,n$ are odd, then we can do for both $m,n$, by the same strategy.

(4) Both $m,n$ are even, then we use the formula (1)(2)(4)(5) to reduce the power

Exercise 2: Evaluate the integral $\int_0^{\pi}\sin^2(x)dx$

Integral of type 2 : $\int \sec^m(x)\tan^n(x) dx$

We may do the same, since we have $\sec^2(x) = 1 + \tan^2(x)$, and keep in mind, that

Example 2 : Evaluate the integral $\int \tan^6(x)\sec^4(x)dx$

In this problem, we rewrite the original integral as $\int \tan^6(x) (1 + \tan^2(x))\sec^2 (x)dx$, and we let $u = \tan(x)$, thus we know $du = \sec^2(x)dx$, then it becomes $\int u^6 (1+u^2)du$, which can be calculated as $\int u^6 (1+u^2)du = \frac{1}{7}\tan^7(x)+\frac{1}{9}\tan^9(x)+C$

Exercise 3 : Evaluate the integral $\int \tan^5(x)\sec^7(x) dx$.

Exercise 4 : Evaluate the integral $\int \sec(x) dx.$

Exercise 5 : Evaluate the integral $\int \sec^3(x) dx.$

MORE EXERCISES ON INTEGRATION

$\bigstar \bigstar $ Evaluate the integral $\int \frac{\sin(2x)}{1+\cos^2(x)}dx$

$\bigstar \bigstar \bigstar$ Evaluate the integral $\int \frac{1}{1-\cos(x)}dx$

$\bigstar \bigstar \bigstar \bigstar \bigstar$ Evaluate the integral $\int \sqrt{(\tan(x))}dx$

Trigonometric Substitution

Example 3 : Evaluate the integral $\int \frac {1}{x^2\sqrt{x^2 + 4}} dx.$

In forms like this, we will perform a trig-sub to simplify the terms inside the square root. Usually we do $\sec(x)$ sub if inside the square root there is $ax^2 + b$, and we do $\sin(x)$ or $\cos(x)$ sub if inside the square root there is $ax^2 - b$. In this problem, we will do $\sec(x)$ sub. If we let $x = 2\tan(y)$, then we have $x^2 + 4 = 4\tan^2(y) + 4 = 4\sec^2(y)$, and we have $dx = 2\sec^2(y)dy$, so the integral now becomes $\int \frac{2\sec^2(y)}{4\tan^2(y)(2\sec(y))}dy$, which is $\frac{1}{4}\int \frac{\sec(y)}{\tan^2(y)}dy$, if we let $\tan(y) = \sin(y) / \cos(y)$ and $\sec(y) = 1 / \cos(y)$, we will get $\frac{1}{4}\int \frac{\cos(y)}{\sin^2(y)}dy$, and we can achieve this by another u-sub, let $u = \sin(y)$ thus we get

$\frac{1}{4}\int \frac{\cos(y)}{\sin^2(y)}dy = \frac{1}{4} \int \frac{1}{u^2}du$, which we get $ \frac{1}{4} \int \frac{1}{u^2}du = -\frac{1}{4}\csc(y) + C$, then we perform the last step, by the equation $x = 2\tan(y)$, we solve $\csc(y)$ in terms of $x$, which we get $-\frac{\sqrt{x^2 + 4}}{4x} + C$

Exercise 6 : Evaluate the integral $\int \frac{x}{\sqrt{x^2 + 4}}dx.$

Exercise 7 : Evaluate the integral $\int_0^a \frac{1}{(a^2 + x^2)^{3/2}} dx.$

Exercise 8 : Evaluate the integral $\int \frac{x}{\sqrt{3-2x-x^2}} dx.$

Recommended citation: Jiajun Zhang, (2024). Solving Trigonometric Integrals