# Evaluate the integral

Question:

Evaluate the integral

$\int \frac{1}{4+3 \tan x} d x$

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, $I=\int \frac{1}{4+3 \tan x} d x$

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

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

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

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

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

Comparing both sides we have:

$C=0$

$3 B-4 A=0$

$4 B+3 A=1$

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

$A=3 / 25, B=4 / 25$ and $C=0$

Thus I can be expressed as:

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

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

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

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

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

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

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

So, $I_{1}$ reduces to:

$I_{1}=\frac{3}{25} \int \frac{\mathrm{du}}{\mathrm{u}}=\frac{3}{25} \log |\mathrm{u}|+C_{1}$

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

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

$\Rightarrow I_{2}=\frac{4}{25} \int d x=\frac{3 x}{25}+C_{2} \ldots \ldots$ equation 3

From equation 1,2 and 3 we have:

$\mathrm{I}=\frac{3}{25} \log |4 \cos \mathrm{x}+3 \sin \mathrm{x}|+\mathrm{C}_{1}+\frac{4 \mathrm{x}}{25}+\mathrm{C}_{2}$

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