Evaluate the integral

Solution:

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, $I=\int \frac{5 \cos x+6}{2 \cos x+\sin x+3} d x$

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, $I=\int \frac{5 \cos x+6}{2 \cos x+\sin x+3} d x$

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

$\therefore 5 \cos x+6=A \frac{d}{d x}(2 \cos x+\sin x+3)+B(2 \cos x+\sin x+3)+C$

$\Rightarrow 5 \cos x+6=A(-2 \sin x+\cos x)+B(2 \cos x+\sin x+3)+C\left\{\because \frac{d}{d x} \cos x=-\sin x\right\}$

$\Rightarrow 5 \cos x+6=\sin x(B-2 A)+\cos x(2 B+A)+3 B+C$

Comparing both sides we have:

$3 B+C=6$

$2 B+A=5$

$B-2 A=0$

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

$A=1, B=2$ and $C=0$

Thus I can be expressed as:

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

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

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

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

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

Let, $2 \cos x+\sin x+3=u$

$\Rightarrow(-2 \sin x+\cos x) d x=d u$

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

$\mathrm{I}_{1}=\int \frac{\mathrm{du}}{\mathrm{u}}=\log |\mathrm{u}|+\mathrm{C}_{1}$

$\therefore \mathrm{I}_{1}=\log |2 \cos \mathrm{x}+\sin \mathrm{x}+3|+\mathrm{C}_{1} \ldots . .$ equation 2

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

$\Rightarrow I_{2}=2 \int d x=2 x+C_{2} \ldots . .$ equation 3

From equation 1,2 and 3 we have:

$I=\log |2 \cos x+\sin x+3|+C_{1}+2 x+C_{2}$

$\therefore|=\log | 2 \cos x+\sin x+3 \mid+2 x+C$