Evaluate the integral

Evaluate the integral

$\int \frac{8 \cot x+1}{3 \cot x+2} d x$


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, $I=\int \frac{8 \cot x+1}{3 \cot x+2} d x$

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, $I=\int \frac{8 \cot x+1}{3 \cot x+2} d x=\int \frac{8 \frac{\cos x}{\sin x}+1}{3 \frac{\sin x}{\sin x}+2}=\int \frac{8 \cos x+\sin x}{3 \cos x+2 \sin x} d x$

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

$\therefore \sin x+8 \cos x=A \frac{d}{d x}(3 \cos x+2 \sin x)+B(3 \cos x+2 \sin x)+C$

$\Rightarrow \sin x+8 \cos x=A(-3 \sin x+2 \cos x)+B(3 \cos x+2 \sin x)+C\left\{\because \frac{d}{d x} \cos x=-\sin x\right\}$

$\Rightarrow \sin x+8 \cos x=\sin x(2 B-3 A)+\cos x(2 A+3 B)+C$

Comparing both sides we have:


$2 B-3 A=1$

$3 B+2 A=8$

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

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

Thus I can be expressed as:

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

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

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

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

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

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

$\Rightarrow(-3 \sin x+2 \cos x) d x=d u$

So, $I_{1}$ reduces to:

$I_{1}=\int \frac{d u}{u}=\log |u|+C_{1}$

$\therefore I_{1}=\log |3 \cos x+2 \sin x|+C_{1} \ldots . .$ equation 2

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

$\Rightarrow I_{2}=2 \int \mathrm{dx}=2 \mathrm{x}+\mathrm{C}_{2} \ldots . .$ equation 3

From equation 1,2 and 3 we have:

$I=\frac{1}{25} \log |3 \cos x+2 \sin x|+C_{1}+2 x+C_{2}$

$\therefore I=\frac{1}{25} \log |3 \cos x+2 \sin x|+2 x+C$


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