Simplifying the integral $\int\frac{dx}{(3 + 2\sin x - \cos x)}$ by an easy approach

Question

$$I=\displaystyle\int\frac{dx}{(3 + 2\sin x - \cos x)}$$

If $$\tan\left(\frac{x}{2}\right)=u$$

or $$x=2\cdot\tan^{-1}(u)$$

Then,

$$\sin{x}=\dfrac{2u}{1+u^2}$$

$$\cos{x}=\dfrac{1-u^2}{1+u^2}$$

$$dx=\dfrac{2}{1+u^2}$$

Substitute $$\tan\left(\dfrac{x}{2}\right)=u$$

Let us simplify the integrand before integrating

$$\dfrac{1}{3+2\sin{x}-\cos{x}}$$

$$=\dfrac{1}{3+2\frac{2u}{1+u^2}-\frac{1-u^2}{1+u^2}}$$

$$=\dfrac{1}{3+\frac{4u-1+u^2}{1+u^2}}$$

$$=\dfrac{1}{\frac{4u-1+u^2+3+3u^2}{1+u^2}}$$

$$=\dfrac{1+u^2}{4u^2+4u+2}$$

$$=\dfrac{1+u^2}{(2u+1)^2+1}$$

$$I=\displaystyle\int\dfrac{1+u^2}{(2u+1)^2+1}\cdot\dfrac{2}{1+u^2}\ du$$

$$=\displaystyle\int\dfrac{1}{(2u+1)^2+1}\ 2\,du$$

Now,

Take :  $$v=2u+1$$

Therefore, $$dv=2\,du$$

$$I=\displaystyle\int\dfrac{1}{v^2+1}\ dv$$

$$I=\tan^{-1}(v)$$

Substitute everything back

$$I=\tan^{-1}(2u+1)$$

$$I=\tan^{-1}\left(2\tan\left(\frac{x}{2}\right)+1\right)$$

$$\boxed{\displaystyle\int\frac{dx}{(3 + 2\sin x - \cos x)} = \tan^{-1}\left(2\tan\left(\frac{x}{2}\right)+1\right)+C}$$

I know that my approach is also not so difficult but still I think there must be an relatively easy approach to this integral. I have tried many different things using trigonometric identities but nothing seems to bring to the solution easily. Kindly help me out.

Answer 1

Remember the following trigonometric equations, which are always very helpful to solve integrals: $$ \begin{matrix} 2\cos(u)^2 = 1 + \cos(2u) & & & \left(\cos(u) + \sin(u)\right)^2 = 1 + \sin(2u) \\ 2\sin(u)^2 = 1 - \cos(2u) & & & \left(\cos(u) - \sin(u)\right)^2 = 1 - \sin(2u) \\ \end{matrix} $$ Note that you can make a mnemonic to remember these equations.

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With them, you can solve the integral: $$ I = \int \dfrac{dx}{3+2\sin(x)-\cos(x)} $$ Using the following procedure:

1. Replace with $\;\;u=\frac{x}{2} \;\;\Rightarrow\;\; dx = 2 du$: $$ \begin{align} I &= 2 \int \dfrac{du}{3+2\sin(2u)-\cos(2u)} \end{align} $$

2. Rewrite the expression to a known trigonometric expression: $$ \begin{align} I&= 2 \int \dfrac{du}{2\left(1+\sin(2u)\right) + \left(1-\cos(2u)\right)} \end{align} $$

3. Replace with the corresponding trigonometric equations:

$$ \begin{align} I&= 2 \int \dfrac{du}{2\left(\cos(u) + \sin(u)\right)^2 + 2\sin(u)^2} \end{align} $$

4. Extract common factor by non-binomial term:

$$ \begin{align} I&= 2 \int \dfrac{1}{\left(\dfrac{\cos(u)}{\sin(u)} + 1\right)^2 + 1 } \cdot\dfrac{1}{2\sin(u)^2} du = \int \dfrac{1}{\left(\cot(u) + 1\right)^2 + 1 } \cdot \csc(u)^2 du \end{align} $$

5. Replace with $\;\;\omega=\cot(u) + 1 \;\;\Rightarrow\;\; d\omega = -\csc(u)^2 du$: $$ \begin{align} I&= \int \dfrac{-d\omega}{\omega^2 + 1 } = \mbox{arccot}(\omega) + C \end{align} $$

6. Returning to the variable x: $$ \omega =\cot(u) + 1 = \cot\left(\dfrac{x}{2}\right) + 1 \quad\Rightarrow\quad I = \mbox{arccot}\left(\cot\left(\dfrac{x}{2}\right) + 1\right) + C $$

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With a little more effort and using the same procedure, you could solve the following generic integral: $$ I = \int \dfrac{dx}{A+B\sin(x)+C\cos(x)} $$ The secret to this is:

1. In Step 2: choose the correct "known trigonometric expression" according to the signs of sine and cosine.

2. In Step 4: don't forget to remove the non-binomial term to reduce it to a simple expression.

Answer 2

Generally, consider the real function $$f(x)=\frac{1}{a+b\cos x+c\sin x}$$ With $a^2>b^2+c^2$ so that $f$ is defined on $\mathbb{R}$. It is not hard to check that the derivative of $$F(x)=\frac{x}{d}+\frac{2}{d}\arctan\left(\frac{c\cos x-b\sin x}{d+a+b\cos x+c\sin x}\right)$$ with $d=\sqrt{a^2-b^2-c^2}$, is $f(x)$. So $\int f(x)dx=F(x)+k$. The advantage of this expression of $F$ is that it is also defined on $\mathbb{R}$. In particular, $$\int \frac{1}{3-\cos x+2\sin x}=\frac{x}{2}+\arctan\left(\frac{2\cos x+\sin x}{5-\cos x+2\sin x}\right)$$

Answer 3

You could write $2\sin(x)-\cos(x) = \sqrt{5} (\frac{2}{\sqrt{5}} \sin x - \frac{1}{\sqrt{5}} \cos x) = \sqrt{5} \sin (x - \alpha)$, with $\tan \alpha = \frac{1}{2}$, so that the required integral becomes $\int \dfrac{dx}{3 + \sqrt{5} \sin(x-\alpha)}$ and then change the integration variable from $x$ to $y = x - \alpha$. Your integral is then of the form $\int \dfrac{dy}{A + B\sin y}$ which is a bit easier to solve using standard methods.

Answer 4

With $a=\tan^{-1}\frac1{\sqrt5}$ and $t=\frac12(x+a)$

\begin{align} \int\frac{dx}{3 + 2\sin x - \cos x} & = \int \frac{dx}{3-\sqrt5\cos(x+a)}\\ &= \int \frac{2d(\tan^2t)}{(3-\sqrt5)+(3+\sqrt5)\tan^2t}=\tan^{-1} \frac{2\tan t}{3-\sqrt5} +C \end{align}