Integration | mathgarage

Posts Tagged ‘Integration’

Integrating 1/(sin(x)+sec(x)) from 0 to pi/2

Posted: July 31, 2023 in Mathematics
Tags: Integration, math

$\displaystyle \int_0^{\pi/2} \frac{1}{\sin x + \sec x} \text{ d}x$

Solution

$\displaystyle I = \int_0^{\pi/2} \frac{\cos{x}}{\sin{x}\cos{x}+1} \text{ d}x = \int_0^{\pi/2} \frac{2\cos{x}}{2\sin{x}\cos{x}+2} \text{ d}x$

$\displaystyle = \int_0^{\pi/2} \frac{\cos x - \sin x}{2\sin x \cos x + 2} \text{ d}x + \int_0^{\pi/2} \frac{\cos x + \sin x}{2\sin x \cos x + 2} \text{ d}x$

$\displaystyle = \int_0^{\pi/2} \frac{\cos x - \sin x}{(\sin x + \cos x)^2 + 1} \text{ d}x - \int_0^{\pi/2} \frac{\cos x + \sin x}{(\sin x - \cos x)^2 - 3} \text{ d}x$

Let u = sin x + cos x and v = sin x – cos x,

$\displaystyle = \int_1^1 \frac{\text{d}u}{u^2+1} - \int_{-1}^1 \frac{\text{d}v}{v^2-3} = 2 \int_0^1 \frac{\text{d}v}{3-v^2}$

$\displaystyle = \frac{1}{\sqrt3} \int_0^1 \left(\frac{1}{\sqrt3+v} + \frac{1}{\sqrt3-v}\right) \text{d}v$

$\displaystyle = \frac{1}{\sqrt3}\left(\log(\sqrt3+v) - \log(\sqrt3-v)\right)\bigg|_0^1$

$\displaystyle = \frac{1}{\sqrt3} \left(\log(\sqrt3+1) - \log(\sqrt3-1)\right)$

$\displaystyle = \frac{1}{\sqrt3} \log\frac{\sqrt3+1}{\sqrt3-1}$

Done.

MIT Integration Bee 2023 Semi-final Problem 1

Posted: May 19, 2023 in Mathematics
Tags: Calculus, Integration, math, MIT, solution

I was browsing the net couple days ago and saw this integral. At first it looks quite difficult. I have tried numerous different substitutions and they don’t work. And later I tried using complex number and it seems to work. So, here is my solution.

Solution:

By Euler’s formula, cosine function is the real part of a complex number.

$e^{i\theta} = \cos \theta + i \sin \theta$

Hence,

$\displaystyle \int e^{\cos x} \cos(2x + \sin x) \text{ d}x = \Re \int e^{\cos x} e^{i(2x+\sin x)} \text{ d}x$

Re-arrange the exponents we’ll get

$\displaystyle = \Re \int e^{\cos x + i\sin x} e^{2ix} \text{ d}x = \Re \int e^{e^{ix}} e^{2ix}\text{ d}x$

Now, since

$\displaystyle \frac{\text{d}}{\text{d}x} e^{e^{ix}}=ie^{e^{ix}}e^{ix}$

We can use integration by parts,

$\displaystyle = \Re \; \frac{1}{i} \int e^{ix} \text{ d}e^{e^{ix}}$

$\displaystyle = \Re \; \frac{1}{i}\left(e^{ix}e^{e^{ix}} - \int e^{e^{ix}}\cdot i e^{ix} \text{ d}x \right)$

$\displaystyle = \Re \; \frac{1}{i}\left(e^{ix}e^{e^{ix}} - e^{e^{ix}}\right) + C$

$\displaystyle = \Re \; \frac{1}{i} e^{e^{ix}} \left(e^{ix}-1\right) + C$

$\displaystyle = \Re \; \frac{1}{i} e^{\cos x + i\sin x} (\cos x + i\sin x - 1) + C$

$\displaystyle = \Re \; \frac{1}{i} e^{\cos x + i\sin x} (\cos x - 1 + i\sin x) + C$

By using the double angle identity of sine and cosine,

$\displaystyle = \Re \; \frac{1}{i} e^{\cos x + i\sin x} \left(-2\sin^2 \frac{x}{2} + i2\sin \frac{x}{2} \cos\frac{x}{2}\right) + C$

$\displaystyle = \Re \; e^{\cos x + i\sin x} \left(2\sin\frac{x}{2}\cos\frac{x}{2} + i\left(2\sin^2 \frac{x}{2}\right)\right) + C$

$\displaystyle = \Re \; 2\sin\frac{x}{2} \cdot e^{\cos x + i\sin x} \left(\cos\frac{x}{2} + i\sin \frac{x}{2}\right) + C$

$\displaystyle = \Re \; 2\sin\frac{x}{2} \cdot e^{\cos x + i\sin x} e^{i\frac{x}{2}} + C$

$\displaystyle = \Re \; 2\sin\frac{x}{2} \cdot e^{\cos x} e^{i\left(\frac{x}{2}+\sin x\right)} + C$

$\displaystyle = 2 e^{\cos x} \sin\frac{x}{2} \cos\left(\frac{x}{2} + \sin x\right) + C$

Integrating the 2022nd power of cosine x from 0 to 2Pi

Posted: May 16, 2022 in Mathematics
Tags: Integration, math

So here is the definite integral:

$\displaystyle \int_0^{2\pi} \cos^{2022}{x} \text{ d}x$

And here is the solution:

The trick here is to use Euler’s formula and the rest is kinda straight forward.

In general, for any natural number n, we have the following,

$\displaystyle \int_0^{2\pi} \cos^n{x} \text{ d}x = \int_0^{2\pi} \left(\frac{e^{ix}+e^{-ix}}{2}\right)^n \text{d}x = \frac{1}{2^n}\int_0^{2\pi} \sum_{k=0}^n \binom{n}{k} e^{i(n-k)x}e^{-ikx} \text{ d}x$

$\displaystyle = \frac{1}{2^n} \sum_{k=0}^n \binom{n}{k} \int_0^{2\pi} e^{i(n-2k)x} \text{ d}x$

If $n-2k\neq0$ , then

$\displaystyle \int_0^{2\pi} e^{i(n-2k)x}\text{ d}x = \frac{e^{i(n-2k)x}}{i(n-2k)} \bigg|_0^{2\pi} = 0$

So, the integral vanishes unless $n-2k=0$ or $n=2k$ .

If n = 2k, then

$\displaystyle \frac{1}{2^n}\sum_{k=0}^n\binom{n}{k}\int_0^{2\pi}e^{i(n-2k)x}\text{ d}x = \frac{1}{2^n}\binom{n}{n/2}2\pi = \frac{\pi}{2^{n-1}}\binom{n}{n/2}$

Since the integral vanishes when n is odd, we can rewrite it as the following,

$\displaystyle \int_0^{2\pi} \cos^{2n}{x}\text{ d}x = \frac{\pi}{2^{2n-1}}\binom{2n}{n}, \quad\text{for }n\in\mathbb{N}$

For example, when n = 1011, then

$\displaystyle \int_0^{2\pi} \cos^{2022}{x}\text{ d}x = \frac{\pi}{2^{2021}}\binom{2022}{1011}$

Interesting Integral Problem from the JEE Advanced exam

Posted: April 6, 2021 in Mathematics
Tags: exa, Integration, JEE, math

Last week I was watching some youtube videos about some Korean undergraduates trying to do the JEE Advanced exam. Apparently this is a highschool exam, an entrance exam for some University or Institution or some sort in India. Some look quite difficult and some are interesting like math contest type questions. I may share them sometimes later. Here is one of the problems.

Divide the numerator and the denominator by $(\sqrt{\cos \theta})^5$ , the integral becomes

$\displaystyle I = 3 \int_0^{\pi/2} \frac{\sec^2 \theta}{(1+\sqrt{\tan \theta})^5} \text{ d}\theta$

Let

$w=1+\sqrt{\tan \theta}, \qquad (w-1)^2 = \tan \theta, \qquad 2(w-1)\text{ d}w = \sec^2 \theta \text{ d}\theta$

Then the integral becomes

$\displaystyle I = 3 \int_1^{\infty} \frac{2(w-1)\text{ d}w}{w^5} = 6 \int_1^\infty w^{-4}-w^{-5} \text{ d}w$

$\displaystyle = 6 \left(\frac{w^{-3}}{-3}+\frac{w^{-4}}{4}\right)\bigg|_1^\infty = 6\left(\frac13-\frac14\right) = \frac12$

Solving Integral Using Symmetry

Posted: December 9, 2020 in Mathematics
Tags: Integration, math

Saw an integral online this evening on Facebook. It is one of those that you won’t see as an example on regular textbooks but the teacher or prof will throw it on the exam or assignments. Or you will see them on integration bee contest. Here is the problem.

$\displaystyle \int_3^5 \frac{\log(x+2)}{\log(-x^2+8x+20)} \text{ d}x$

First of all, factorize the expression on the denominator.

$-x^2+8x+20 = 20+8x-x^2 = (10-x)(2+x)$

Then the integral becomes

$\displaystyle \int_3^5 \frac{\log(2+x)}{\log((10-x)(2+x))} \text{ d}x = \int_3^5 \frac{\log(2+x)}{\log(10-x)+\log(2+x)} \text{ d}x$

Next, we want to turn the integral into something like this,

$\displaystyle \int_{-a}^a \frac{f(x)}{f(x)+f(-x)} \text{ d}x$

So we let u = x – 4, and the integral becomes

$\displaystyle \int_{-1}^1 \frac{\log(6+u)}{\log(6-u)+\log(6+u)} \text{ d}u$

Then, let u = –v,

$\displaystyle I = \int_{-1}^1 \frac{\log(6+u)}{\log(6-u)+\log(6+u)} \text{ d}u = \int_{-1}^1 \frac{\log(6-v)}{\log(6+v)+\log(6-v)} \text{ d}v$

Final step, since u and v are dummy variables, we add the two integrals together.

$\displaystyle 2I = \int_{-1}^1 \text{ d}x = 2 \qquad \therefore I = 1$

Remark:

Whenever you see the symmetry in the integral that takes on the following form,

$\displaystyle I=\int_{-a}^a \frac{f(x)}{f(x)+f(-x)} \text{ d}x$

You will always have

$\displaystyle I=\int_{-a}^a \frac{f(x)}{f(x)+f(-x)} \text{ d}x = \int_{-a}^a \frac{f(-x)}{f(x)+f(-x)} \text{ d}x$

It can be easily shown algebraically or using a general graph. Then

$\displaystyle 2I = \int_{-a}^a \frac{f(x)+f(-x)}{f(x)+f(-x)} \text{ d}x = \int_{-a}^a \text{ d}x = 2a$

and hence

$I = a$

Integration using Leibniz Rule

Posted: September 12, 2020 in Mathematics
Tags: Integration, Leibniz Rule, math

Leibniz Integral Rule states that given a function f(x, t) is continuous and continuously differentiable such as the partial derivatives exist and continuous and both the limit a(x) and b(x) are continuous and continuously differentiable, then

$\displaystyle \frac{\text{d}}{\text{d}x} \int_{a(x)}^{b(x)} f(x,t) \text{ d}t = f(x,b(x))\cdot\frac{\text{d}}{\text{d}x}b(x) - f(x,a(x))\cdot\frac{\text{d}}{\text{d}x}a(x) + \int_{a(x)}^{b(a)} \frac{\partial}{\partial x} f(x,t) \text{ d}t$