# Evaluate the integral

Question:

Evaluate the integral

$\int \frac{1}{\mathrm{p}+\mathrm{q} \tan \mathrm{x}} \mathrm{dx}$

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, $I=\int \frac{1}{p+q \tan x} d x$

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, $1=\int \frac{1}{\mathrm{p}+\mathrm{q} \tan \mathrm{x}} \mathrm{dx}=\int \frac{1}{\mathrm{p}+\mathrm{q} \frac{\sin x}{\cos x}} \mathrm{dx}=\int \frac{\cos x}{\mathrm{p} \cos x+\mathrm{q} \sin x} \mathrm{dx}$

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

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

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

$\Rightarrow \cos x=-\sin x(B q+A p)+\cos x(B p+A q)+C$

Comparing both sides we have:

$C=0$

$B p+A q=1$

$B q+A p=0$

On solving above equations, we have:

$A=\frac{q}{p^{2}+q^{2}} B=\frac{p}{p^{2}+q^{2}}$ and $C=0$

Thus I can be expressed as:

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

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

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

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

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

Let, $u=p \cos x+q \sin x \Rightarrow d u=(-p \sin x+q \cos x) d x$

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

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

Let, $u=p \cos x+q \sin x \Rightarrow d u=(-p \sin x+q \cos x) d x$

So, $I_{1}$ reduces to:

$I_{1}=\frac{q}{p^{2}+q^{2}} \int \frac{d u}{u}=\frac{q}{p^{2}+q^{2}} \log |u|+C_{1}$

$\left.\therefore\right|_{1}=\frac{q}{p^{2}+q^{2}} \log |(p \cos x+q \sin x)|+C_{1} \ldots . .$ equation 2

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

$\therefore \mathrm{l}_{2}=\frac{\mathrm{px}}{\mathrm{p}^{2}+\mathrm{q}^{2}}+\mathrm{C}_{2} \ldots \ldots$ equation 3

From equation 1,2 and 3 we have:

$I=\frac{q}{p^{2}+q^{2}} \log |(p \cos x+q \sin x)|+C_{1}+\frac{p x}{p^{2}+q^{2}}+C_{2}$

$\therefore I=\frac{q}{p^{2}+q^{2}} \log |(p \cos x+q \sin x)|+\frac{p x}{p^{2}+q^{2}}+C$