# Integration of Trigonometric Functions II

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# Recall

In calculus, an antiderivative or indefinite integral, G, of a function g, is the function that can be differentiated to obtain the original function, that is, G’ = f.

G(x) = $\int f(x)dx$

Trigonometry is a branch of mathematics that deals with the relationships between the sides and angles of triangles.

This is the second article in our two-part series about Integration of Trigonometric Functions and is a continuation of the first one, so if you haven’t taken a look at it yet, I recommend you read it first and come back.

# Solved examples

• $\int \frac{dx}{x^2\sqrt{x^2+4}}$ =[Trigonometric substitution, x = 2tan(θ), dx = 2sec2(θ)dθ] $\int \frac{2sec^2(θ)dθ}{4tan^2(θ)\sqrt{4tan^2(θ)+4}} = \int \frac{sec^2(θ)dθ}{4·tan^2(θ)\sqrt{tan^2(θ)+1}} =$[Recall that sec2(x)=1+tan2(x)] $\int \frac{sec^2(θ)dθ}{4·tan^2(θ)\sqrt{sec^2(θ)}} = \int \frac{sec(θ)dθ}{4·tan^2(θ)} = \frac{1}{4}\int \frac{1}{cos(θ)}\frac{cos^2(θ)}{sin^2(θ)}dθ = \frac{1}{4}\int \frac{cos(θ)}{sin^2(θ)}dθ$ =[Derivatives of Trigonometric Formulas, csc’(θ) = -csc(θ)cot(θ) = $-\frac{1}{sin(θ)}·\frac{1}{tan(θ)}=-\frac{1}{sin(θ)}\frac{cos(θ)}{sin(θ)} = \frac{-cos(θ)}{sin^2(θ)}$] = $-\frac{csc(θ)}{4}+C.$

• $\int \frac{\sqrt{x^2-4}}{x^4}dx$ [Trigonometric substitution, think about $x^2-4=x^2-2^2$, so it needs to be a right triangle’s leg, Figure 5, x = 2sec(θ), dx = 2sec(θ)tan(θ)dθ] = $\int \frac{\sqrt{4sec^2(θ)-4}}{(2sec(θ))^4}2sec(θ)tan(θ)dθ = \int \frac{\sqrt{4(sec^2(θ)-1)}}{16sec^4(θ)}2sec(θ)tan(θ)dθ = \frac{1}{4}\int \frac{\sqrt{sec^2(θ)-1}}{sec^4(θ)}sec(θ)tan(θ)dθ$ =[Recall sec2(x) = 1 + tan2(x)] $\frac{1}{4}\int \frac{\sqrt{tan^2(θ)}}{sec^4(θ)}sec(θ)tan(θ)dθ = \frac{1}{4}\int \frac{tan(θ)}{sec^4(θ)}sec(θ)tan(θ)dθ = \frac{1}{4}\int \frac{tan^2(θ)}{sec^3(θ)}dθ = \frac{1}{4}\int tan^2(θ)·cos^3(θ)dθ = \frac{1}{4}\int \frac{sin^2(θ)}{cos^2(θ)}·cos^3(θ)dθ = \frac{1}{4}\int sin^2(θ)·cos(θ)dθ$ =[Integration substitution u = sin(θ), du = cos(θ)dθ] $\frac{1}{4}\int u^2du = \frac{1}{4}\frac{1}{3}u^3 = \frac{1}{12}sin^3(θ)$ =[Backtrack our substitution, Figure 5, sin(θ) = $\frac{\sqrt{x^2-4}}{x}$] $\frac{1}{12}(\frac{(\sqrt{x^2-4})^3}{x^3}) = \frac{(x^2-4)^{\frac{3}{2}}}{12x^3}+C$

• Let’s compute the area of the section inside the circle with radius a and height b (Figure 1.a.).

Area = $\int_{0}^{b} \sqrt{a^2-y^2}dy$ =[Trigonometric substitution, y = asin(θ), $\sqrt{a^2-y^2} = \sqrt{a^2-a^2sin^2(θ)} = a\sqrt{1-sin^2(θ)} = a\sqrt{cos^2(θ)}$ = acos(θ) = x, dy = acos(θ)dθ] $\int_{0}^{b} acos(θ)(acos(θ)dθ) = a^2\int cos^2(θ)dθ =$[It was previously calculated] $a^2(\frac{θ}{2}+\frac{sin(2θ)}{4}) = a^2(\frac{θ}{2}+\frac{sin(θ)cos(θ)}{2})$ = [sin(θ) = ya ⇒ θ = arcsin(ya)]$\frac{a^2arcsin(y/a)}{2}+ \frac{y(\sqrt{a^2-y^2})}{2}\bigg|_{0}^{b} = \frac{a^2arcsin(b/a)}{2}+ \frac{b(\sqrt{a^2-b^2})}{2} = \frac{a^2θ_0}{2}+ \frac{b(\sqrt{a^2-b^2})}{2}$ where $θ_0 = arcsin(\frac{b}{a})$.

Geometric interpretation. If you have a circle with radius r and a central angle θ (measured in radians), the area of the sector, aka Area1, is given by $\frac{1}{2}r^2θ = \frac{1}{2}a^2θ_0$. Area2 = $\frac{b(\sqrt{a^2-b^2})}{2}$ is the area of the upper triangle.

• $\int \frac{\sqrt{4-x^2}}{x^2}dx$ =[Trigonometry substitution, x = 2sin(θ), dx = 2cos(θ)dθ] $\int \frac{\sqrt{4-4sin^2(θ)}}{4sin^2(θ)}2cos(θ)dθ = \int \frac{\sqrt{4(1-sin^2(θ))}}{4sin^2(θ)}2cos(θ)dθ = \int \frac{\sqrt{4(cos^2(θ))}}{4sin^2(θ)}2cos(θ)dθ = \int \frac{2cos(θ)}{4sin^2(θ)}2cos(θ)dθ = \int \frac{cos^2(θ)}{sin^2(θ)}dθ = \int \frac{1-sin^2(θ)}{sin^2(θ)}dθ = \int \frac{1}{sin^2(θ)}dθ - \int dθ = \int csc^2(θ)dθ - \int dθ$ =[Derivate of Trigonometric Formulas, cot’(x) = -csc2(x)] -cot(θ) -θ +C.

Now, we need to backtrack our substitution (Figure 1), $x = 2sin(θ) ⇒ sin(θ)=\frac{x}{2}.$ By the Pythagorean theorem, $c^2+x^2 = 2 ⇒c = \sqrt{4-x^2}$. Besides, $tan(θ)=\frac{x}{\sqrt{4-x^2}}, cot(θ)=\frac{1}{tan(θ)} = \frac{\sqrt{4-x^2}}{x}, \int \frac{\sqrt{4-x^2}}{x^2}dx = -cot(θ) -θ +C = -\frac{\sqrt{4-x^2}}{x}-sin^{-1}(\frac{x}{2})+C.$

• $\int sec^4(x)dx$ =[Recall sec2(x) = 1 + tan2(x)] $\int (1 + tan^2(x))sec^2(x)dx$ [Apply substitution, u = tan(x), du = sec2(x)dx] $\int (1 + u^2)du = u + \frac{u^3}{3} + C = tan(x) + \frac{tan^3(x)}{3} + C.$

• $\int \frac{dx}{x^2\sqrt{1+x^2}}$ [Recall that sec2(x) = 1 + tan2(x), let’s apply substitution x = tan(θ), 1 + x2 = sec2(θ), dx = sec2(θ)dθ] $\int \frac{sec^2(θ)dθ}{(tan(θ))^2sec(θ)} = \int \frac{sec(θ)dθ}{(tan(θ))^2} = \int \frac{cos^2(θ)dθ}{cos(θ)sin^2(θ)} = \int \frac{cos(θ)dθ}{sin^2(θ)}$ [u = sin(θ), du = cos(θ)dθ] $\int \frac{du}{u^2} = \frac{-1}{u} + C = \frac{-1}{sin(θ)} + C = -csc(θ) + C$ [Figure 1.b.] = $-\frac{\sqrt{1+x^2}}{x}+C$

• $\int \frac{x^2}{\sqrt{9-x^2}}dx$ [Trigonometric Substitution, 💡$\sqrt{3^2-3^2sin^2(θ)}$, x = 3·sin(θ), dx = 3cos(θ)dθ]= $\int \frac{(3·sin(θ))^2}{\sqrt{3^2-3^2sin^2(θ)}}3cos(θ)dθ = \int \frac{9·sin^2(θ)}{3\sqrt{1-sin^2(θ)}}3cos(θ)dθ = \int \frac{9·sin^2(θ)}{\sqrt{1-sin^2(θ)}}cos(θ)dθ = \int \frac{9·sin^2(θ)}{\sqrt{cos^2(θ)}}cos(θ)dθ = 9\int sin^2(θ)dθ$ =[Half-Angle Formulas] $9\int \frac{1}{2}(1-cos(2θ))dθ = \frac{9}{2}\int (1-cos(2θ))dθ = \frac{9}{2}(θ-\frac{1}{2}sin(2θ)) =$[Double Angle Formulas, sin(2θ) = 2sin(θ)cos(θ)] $\frac{9}{2}(θ-sin(θ)cos(θ))$

Now, we need to backtrack our substitution (Figure 4), x = 3sin(θ) ⇒ $sin(θ)=\frac{x}{3}, θ = sin^{-1}(\frac{x}{3}), cos(θ)=\frac{\sqrt{9-x^2}}{3} ⇒ \int \frac{x^2}{\sqrt{9-x^2}}dx = \frac{9}{2}(θ-sin(θ)cos(θ)) = \frac{9}{2}(sin^{-1}(\frac{x}{3})-\frac{x}{3}\frac{\sqrt{9-x^2}}{3}) = \frac{9}{2}(sin^{-1}(\frac{x}{3})-\frac{x\sqrt{9-x^2}}{9})+C.$

• $\int_{1}^{\sqrt{3}} \frac{1}{(1+x^2)^{\frac{3}{2}}}dx = \int_{1}^{\sqrt{3}} \frac{1}{(\sqrt{1+x^2})^3}dx$= [Trigonometric substitution, x = tan(θ), Figure 2, dx = sec2(θ)dθ, $x = \sqrt{3} ⇒ tan(θ) = \sqrt{3} ⇒ θ = tan^{-1}(\sqrt{3}) = \frac{π}{3}, x = 1 ⇒ tan(θ) = 1 ⇒ θ = tan^{-1}(1) = \frac{π}{4}$] $\int_{\frac{π}{4}}^{\frac{π}{3}} \frac{1}{(\sqrt{1+tan^2(θ)})^3}sec^2(θ)dθ$ =[Recall sec2(x) = 1 + tan2(x)] $\int_{\frac{π}{4}}^{\frac{π}{3}} \frac{1}{(\sqrt{sec^2(θ)})^3}sec^2(θ)dθ = \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{1}{sec^3(θ)}sec^2(θ)dθ = \int_{\frac{π}{4}}^{\frac{π}{3}} \frac{1}{sec(θ)}dθ = \int_{\frac{π}{4}}^{\frac{π}{3}} cos(θ)dθ = sin(θ)\bigg|_{\frac{π}{4}}^{\frac{π}{3}} = sin(\frac{π}{3})-sin(\frac{π}{4}) = \frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2} = \frac{\sqrt{3}-\sqrt{2}}{2}$

# Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn, and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
8. MIT OpenCourseWare 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007.
9. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.
10. Professor Leonard.
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