- #1

- 168

- 21

## Homework Statement:

- ##\int \frac{\cos x + \sqrt 3}{1 + 4\sin \left(x+ \pi/3\right) + 4\sin^2 \left(x+\pi/3\right)} dx##

## Homework Equations:

- ## 1 + 4\sin \left(x+ \pi/3\right) + 4\sin^2 \left(x+\pi/3\right) = \left(1+2\sin\left(x+\pi/3\right)\right)^2##

If we look at the denominator of this integral $$\int \frac{\cos x + \sqrt 3}{1 + 4\sin \left(x+ \pi/3\right) + 4\sin^2 \left(x+\pi/3\right)} dx$$ then we can see that ## 1 + 4\sin \left(x+ \pi/3\right) + 4\sin^2 \left(x+\pi/3\right) = \left(1+2\sin\left(x+\pi/3\right)\right)^2## and ## 2\sin\left(x+\pi/3\right)\ = \sin x + \sqrt 3 \cos x##. Now, our integral reduces to

$$\int \frac{\cos x + \sqrt 3}{ 1+\sin x + \sqrt 3 \cos x} dx $$. After this I tried so many things, but no substitution worked. Therefore, I finally looked for the solution in my book and there it was written

My question how does just dividing by ##\sin^2 x## made everything come out to be? I know that we did something like this $$ \frac{\cos x/ \sin^2 x + \sqrt 3/\sin^2 x}{\left(1/\sin x + \sin x/\sin x + \sqrt 3 \cos x/\sin x\right)^2} \\

\int \frac{\csc x \cot x + \sqrt 3 csc^2 x}{\left(\csc x + 1 + \sqrt 3 \cot x\right)^2} dx \\ \textrm{taking the denominator as u} \\

du = -\csc x \cot x - \sqrt 3 \csc^2 x dx \\

\int \frac{-du}{u^2} \\

\frac{1}{u} = \frac{1}{\csc x +1+ \sqrt 3 \cot x} +C $$ What I want to know is how that dividing the numerator and denominator by ##\sin^2 x## did the whole thing? It's like a magic and I'm no magician I'm a a mathematics student so how can I solve integrals like this in future? How to learn these magic tricks?

$$\int \frac{\cos x + \sqrt 3}{ 1+\sin x + \sqrt 3 \cos x} dx $$. After this I tried so many things, but no substitution worked. Therefore, I finally looked for the solution in my book and there it was written

.Dividing by ##\sin ^2 x## we get ## \int \frac{\csc x \cot x + \sqrt 3 csc^2 x}{\left(\csc x + 1 + \sqrt 3 \cot x\right)^2} dx = \frac{1}{\csc x +1+ \sqrt 3 \cot x} + C##

My question how does just dividing by ##\sin^2 x## made everything come out to be? I know that we did something like this $$ \frac{\cos x/ \sin^2 x + \sqrt 3/\sin^2 x}{\left(1/\sin x + \sin x/\sin x + \sqrt 3 \cos x/\sin x\right)^2} \\

\int \frac{\csc x \cot x + \sqrt 3 csc^2 x}{\left(\csc x + 1 + \sqrt 3 \cot x\right)^2} dx \\ \textrm{taking the denominator as u} \\

du = -\csc x \cot x - \sqrt 3 \csc^2 x dx \\

\int \frac{-du}{u^2} \\

\frac{1}{u} = \frac{1}{\csc x +1+ \sqrt 3 \cot x} +C $$ What I want to know is how that dividing the numerator and denominator by ##\sin^2 x## did the whole thing? It's like a magic and I'm no magician I'm a a mathematics student so how can I solve integrals like this in future? How to learn these magic tricks?