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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

 

 

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.

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, $tan(θ)=\frac{x}{\sqrt{4-x^2}}, cot(θ)=\frac{1}{tan(θ)} = \frac{\sqrt{4-x^2}}{x}, \int \frac{\sqrt{4-x^2}}{x^2}dx = -cot(θ) -θ +C = -\frac{\sqrt{4-x^2}}{x}-sin^{-1}(\frac{x}{2})+C.$

 

 

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

 

 

Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007].
  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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