Evaluate the integral

Ideas required to solve the problems:

* Integration by substitution: A change in the variable of integration often reduces an integral to one of the fundamental integration. If derivative of a function is present in an integration or if chances of its presence after few modification is possible then we apply integration by substitution method.

* Knowledge of integration of fundamental functions like sin, cos, polynomial, log etc and formula for some special functions.

Let, $I=\int \frac{3+2 \cos x+4 \sin x}{2 \sin x+\cos x+3} d x$

To solve such integrals involving trigonometric terms in numerator and denominators. We use the basic substitution method and to apply this simply we follow the undermentioned procedure-

If I has the form $\int \frac{a \sin x+b \cos x+c}{d \sin x+e \cos x+f} d x$

Then substitute numerator as -

$a \sin x+b \cos x+c=A \frac{d}{d x}(d \sin x+e \cos x+f)+B(d \sin x+e \cos x+c)+C$

Where $A, B$ and $C$ are constants

We have, $I=\int \frac{3+2 \cos x+4 \sin x}{2 \sin x+\cos x+3} d x$

As I matches with the form described above, So we will take the steps as described.

$\therefore 3+2 \cos x+4 \sin x=A \frac{d}{d x}(2 \sin x+\cos x+3)+B(2 \sin x+\cos x+3)+C$

$\Rightarrow 3+2 \cos x+4 \sin x=A(2 \cos x-\sin x)+B(2 \sin x+\cos x+3)+C\left\{\because \frac{d}{d x} \cos x=-\sin x\right\}$

$\Rightarrow 3+2 \cos x+4 \sin x=\sin x(2 B-A)+\cos x(B+2 A)+3 B+C$

Comparing both sides we have:

$3 B+C=3$

$B+2 A=2$

$2 B-A=4$

On solving for $A, B$ and $C$ we have:

$A=0, B=2$ and $C=-3$

Thus I can be expressed as:

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

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

$\therefore$ Let $I_{1}=2 \int \frac{(2 \sin x+\cos x+3)}{2 \sin x+\cos x+3} d x$ and $I_{2}=-3 \int \frac{1}{2 \sin x+\cos x+3} d x$

$\Rightarrow I=I_{1}+I_{2} \ldots .$ equation 1

$I_{1}=2 \int \frac{(2 \sin x+\cos x+3)}{2 \sin x+\cos x+3} d x$

So, I $_{1}$ reduces to:

$\mathrm{I}_{1}=2 \int \mathrm{dx}=2 \mathrm{x}+\mathrm{C}_{1} \ldots . .$ equation 2

As, $I_{2}=-3 \int \frac{1}{2 \sin x+\cos x+3} d x$

To solve the integrals of the form $\int \frac{1}{\operatorname{asin} x+b \cos x+c} d x$

To apply substitution method we take following procedure.

We substitute:

$\sin x=\frac{2 \tan \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}$ and $\cos x=\frac{1-\tan ^{2} \frac{x}{2}}{1+\tan ^{2} \frac{x}{2}}$

$\therefore I_{2}=-3 \int \frac{1}{2 \sin x+\cos x+3} d x$

$\Rightarrow I_{2}=-3 \int \frac{1}{2\left(\frac{2 \tan \frac{x}{2}}{1+\tan \frac{2 x}{2}}\right)+3\left(\frac{1-\tan \frac{2 x}{2}}{1+\tan \frac{2 x}{2}}\right)+3} d x$

$\Rightarrow I_{2}=-3 \int \frac{1+\tan ^{2} \frac{x}{2}}{4 \tan \frac{x}{2}+1-\tan \frac{2}{2}+3\left(1+\tan \frac{2}{2}\right)} d x$

$\Rightarrow I_{2}=-3 \int \frac{\sec ^{2} \frac{x}{2}}{2\left(2 \tan \frac{x}{2}+2+1 \tan ^{2} \frac{x}{2}\right)} d x$

Let, $\mathrm{t}=\tan \frac{\mathrm{x}}{2} \Rightarrow \mathrm{dt}=\frac{1}{2} \sec ^{2} \frac{\mathrm{x}}{2} \mathrm{dx}$

$\therefore I_{2}=-3 \int \frac{1}{\left(2 t+2+t^{2}\right)} d t$

As, the denominator is polynomial without any square root term. So one of the special integral will be used to solve $I_{2}$.

$I_{2}=-3 \int \frac{1}{\left(2 t+2+t^{2}\right)} d t$

$\Rightarrow I_{2}=-3 \int \frac{1}{\left(t^{2}+2(1) t+1\right)+1} d t$

$\therefore I_{2}=-3 \int \frac{1}{(t+1)^{2}+1}$ dt $\left\{\because a^{2}+2 a b+b^{2}=(a+b)^{2}\right\}$

As, $I_{2}$ matches with the special integral form

$\int \frac{1}{x^{2}+a^{2}} d x=\frac{1}{a} \tan ^{-1} \frac{x}{a}+C$

$\mathrm{I}_{2}=-3 \tan ^{-1}(\mathrm{t}+1)$

Putting value of twe have:

$\therefore I_{2}=-3 \tan ^{-1}\left(\tan \frac{x}{2}+1\right)+C_{2} \ldots \ldots$ equation 3

From equation 1,2 and 3 :

$I=2 x+C_{1}-3 \tan ^{-1}\left(\tan \frac{x}{2}+1\right)+C_{2}$

$\therefore \mathrm{I}=2 \mathrm{x}-3 \tan ^{-1}\left(\tan \frac{\mathrm{x}}{2}+1\right)+\mathrm{C} \ldots . .$ ans