Question

Integrate the given function.

f(x) = cosec x . sec x

Answer 1

ANSWER:

To Integrate:

• f(x) = cosec x . sec x

Solution:

We are given that,

$\implies f(x)=\csc x\cdot \sec x$

First, we need to simplify the value of f(x),

$\implies f(x)=\dfrac{1}{\sin x}\cdot \sec x$

$\implies f(x)=\left(\dfrac{1}{\sin x}\times\dfrac{\cos x}{\cos x}\right)\cdot \sec x$

$\implies f(x)=\dfrac{\cos x}{\sin x} \cdot \dfrac{1}{\cos x}\cdot \sec x$

$\implies f(x)=\dfrac{1}{\tan x}\cdot\sec x\cdot \sec x$

$\implies f(x)=\dfrac{1}{\tan x}\cdot \sec^2x$

So,

$\implies f(x)=\dfrac{\sec^2x}{\tan x}$

We need to calculate the integral of,

$\displaystyle\implies\int f(x) dx$

So,

$\displaystyle\implies\int \dfrac{\sec^2x}{\tan x} dx$

Let us assume tan x = u

So,

$\implies u=\tan x$

Now, differentiating both sides w.r.t. x,

$\implies \dfrac{du}{dx}=\dfrac{d}{dx}\tan x$

$\implies \dfrac{du}{dx}=\sec^2x$

So,

$\implies du=\sec^2x\: dx$

We had,

$\displaystyle\implies\int \dfrac{\sec^2x}{\tan x} dx$

Substituting the values of tan x and sec^2x dx,

$\displaystyle\implies\int \dfrac{1}{u} du$

$\implies \ln(u) + C$

Substituting u for tan x,

$\implies \ln(\tan x) + C$

But, this function is valid only for positive value of tan x. So,

$\implies \ln(|\tan x|) + C$

Hence,

$\displaystyle\bf\implies\int \dfrac{sec^2x}{tan x} dx = ln(|tan x|) + C$