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# Evaluate the integral

Question:

Evaluate the integral

$\int \frac{2 \tan x+3}{3 \tan x+4} d x$

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, $I=\int \frac{2 \tan x+3}{3 \tan x+4} d x$

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, $1=\int \frac{2 \tan x+3}{3 \tan x+4} d x=\int \frac{2 \frac{\sin x}{\cos x}+3}{3 \frac{\sin x}{\cos x}+4}=\int \frac{2 \sin x+3 \cos x}{4 \cos x+3 \sin x} d x$

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

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

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

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

Comparing both sides we have:

$C=0$

$3 B-4 A=2$

$4 B+3 A=3$

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

$A=1 / 25, B=18 / 25$ and $C=0$

Thus I can be expressed as:

$I=\int \frac{\frac{1}{25}(3 \cos x-4 \sin x)+\frac{19}{25}(4 \cos x+3 \sin x)}{4 \cos x+3 \sin x} d x$

$I=\int \frac{\frac{1}{25}(3 \cos x-4 \sin x)}{4 \cos x+3 \sin x} d x+\int \frac{\frac{18}{25}(4 \cos x+3 \sin x)}{4 \cos x+3 \sin x} d x$

$\therefore$ Let $I_{1}=\frac{1}{25} \int \frac{(3 \cos x-4 \sin x)}{4 \cos x+3 \sin x} d x$ and $I_{2}=\frac{18}{25} \int \frac{(4 \cos x+3 \sin x)}{4 \cos x+3 \sin x} d x$

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

$I_{1}=\frac{1}{25} \int \frac{(3 \cos x-4 \sin x)}{4 \cos x+3 \sin x} d x$

Let, $4 \cos x+3 \sin x=u$

$\Rightarrow(-4 \sin x+3 \cos x) d x=d u$

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

$I_{1}=\frac{1}{25} \int \frac{d u}{u}=\frac{1}{25} \log |u|+C_{1}$

$\therefore I_{1}=\frac{1}{25} \log |4 \cos x+3 \sin x|+C_{1} \ldots . .$ equation 2

As, $I_{2}=\frac{18}{25} \int \frac{(4 \cos x+3 \sin x)}{4 \cos x+3 \sin x} d x$

$\Rightarrow I_{2}=\frac{18}{25} \int d x=\frac{18 x}{25}+C_{2} \ldots . .$ equation 3

From equation 1,2 and 3 we have:

$I=\frac{1}{25} \log |4 \cos x+3 \sin x|+C_{1}+\frac{18 x}{25}+C_{2}$

$\therefore I=\frac{1}{25} \log |4 \cos x+3 \sin x|+\frac{18 x}{25}+C$