Question

What is the antiderivative of $\dfrac{1+\sin x}{1-\sin x}$?

Answer 1

In order to find the antiderivative of $\dfrac{1+\sin x}{1-\sin x}$, we can reverse the process of derivation. In this case, we will be rationalizing the denominator first, and then applying the general formulas and then integrating them accordingly, will give the antiderivative.

Complete step-by-step solution:
Now, let us learn more about anti derivatives. An anti derivative is an inverse derivative, primitive function, primitive integral or indefinite integral of a function f is a differentiable function $F$ whose derivative is equal to the original function $f$. This can be stated symbolically as $F'=f$.
Now let us start finding the derivative for the given function $\dfrac{1+\sin x}{1-\sin x}$.
Firstly let us equate it to $I$. We get,
$I=\int \dfrac{1+\sin x}{1-\sin x}dx$
Upon rationalizing the denominator, we get
$\int \dfrac{\left( 1+\sin x \right)\left( 1+\sin x \right)}{\left( 1-\sin x \right)\left( 1+\sin x \right)}dx$
On applying the general formulas $\left( a+b \right)\left( a+b \right)={{\left( a+b \right)}^{2}}$,$\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$.
We can simplify them as,
$\int \dfrac{{{\left( 1+\sin x \right)}^{2}}}{{{1}^{2}}-{{\sin }^{2}}x}dx$
As we know that, $1-{{\sin }^{2}}x={{\cos }^{2}}x$
So let us replace it and rewrite the equation.
$I=\int \dfrac{1+{{\sin }^{2}}x+2\sin x}{{{\cos }^{2}}x}dx$
Now, let us separate the terms with their denominators separately.
$\int \dfrac{1}{{{\cos }^{2}}x}dx+\int \dfrac{{{\sin }^{2}}x}{{{\cos }^{2}}x}dx+2\int \dfrac{\sin x}{{{\cos }^{2}}x}dx$
Now let us replace the terms with known trigonometric ratios.
$\int {{\sec }^{2}}xdx+\int {{\tan }^{2}}xdx+2\int \dfrac{\sin x}{{{\cos }^{2}}x}dx$
Now let us consider $u=\cos x$
Upon differentiating,
$du=-\sin xdx$
Now replacing in the equation, we get
$\int {{\sec }^{2}}xdx+\int {{\tan }^{2}}xdx-2\int \dfrac{1}{{{u}^{2}}}dx$
We also know that, ${{\tan }^{2}}x={{\sec }^{2}}x-1$
On substituting and replacing,
$I=2\int {{\sec }^{2}}xdx-\int 1dx+2\int \dfrac{1}{{{u}^{2}}}dx$
Upon simplifying, we get
$2\tan x+\dfrac{2}{u}-x$
Since our $u=\cos x$, we will be substituting
$2\tan x+\dfrac{2}{\cos x}-x+C$
$\therefore$ The anti derivative of $\dfrac{1+\sin x}{1-\sin x}$ is $2\tan x+\dfrac{2}{\cos x}-x+C$, where $C\in R$.

Note: An integral usually has a defined limit where as an anti derivative is usually a general case and will most always have a +C, the constant of integration, at the end of it. This is the only difference between the two other than that they are completely the same. One of the most common errors would be choosing the proper integration method so it is very important to choose properly.