Question

Integrate $\dfrac{1}{{(1 + \sin x)}}?$

Answer 1

Here in this question, to solve the given integration we need to apply conjugate form and we should be aware of derivatives and integration of $\sin x$, $\cos x$, $\tan x$ because integration is a reverse process of differentiation is given by fundamental theorem of calculus and differentiation .

Complete step-by-step answer:
consider the given question $\dfrac{1}{{(1 + \sin x)}}$
[To integrate, we should be aware of Trigonometric function so that we can integrate easily]
Now we are integrating by taking in the form of $\int {\left( {\dfrac{1}{{1 + \sin x}}} \right)dx}$
\Rightarrow $$$$\int {\left( {\dfrac{1}{{1 + \sin x}} \times \dfrac{{1 - \sin x}}{{1 - \sin x}}} \right)dx} [here we have considered the conjugate form $\dfrac{1}{{(1 + \sin x)}}$,
The conjugate form of (1+sinx) is given as (1-sinx) and also we are multiplying and dividing conjugate form or else we can also consider $(a + b).(a - b) = {a^2} - {b^2}$]
\Rightarrow $$$$\int {\left( {\dfrac{{1 - \sin x}}{{1 - {{\sin }^2}x}}} \right)} dx [Here (1 -${\sin ^2}x$) is obtained by multiplying the denominator terms such as (1 + \sin x)$$$$ \times $$$$(1 - \sin x)]
\Rightarrow $$$$\int {\left( {\dfrac{{1 - \sin x}}{{{{\cos }^2}x}}} \right)} dx [Here we know that ${\sin ^2}x + {\cos ^2}x = 1$ so, $1 - {\sin ^2}x = {\cos ^2}x$]
Now integrating each term we get,
\Rightarrow $$$$\int {\left( {\dfrac{1}{{{{\cos }^2}x}} - \dfrac{{\sin x}}{{{{\cos }^2}x}}} \right)} dx
$\Rightarrow$ $\int {\dfrac{1}{{{{\cos }^2}x}}} dx$- $\int {\dfrac{{\sin x}}{{{{\cos }^2}x}}} dx$
\Rightarrow $$$$\int {({{\sec }^2}x - \tan x.\sec x)dx} $\because$ [$\left( {\dfrac{{\sin x}}{{\cos x}} = \tan x} \right)$ and $\dfrac{1}{{\cos x}} = \sec x$]
\Rightarrow $$$$\int {({{\sec }^2}x)dx} - $\int {(\tan x.\sec x)dx}$
\Rightarrow $$$$\tan x - \sec x + c [where c is the integrating constant]

Note: An integral assigns numbers to functions in a way that describes displacement, area, volume and other concepts that arise by combining infinitesimal data.
The process of finding integrals is called integration.
${\sin ^2}x + {\cos ^2}x = 1$

Integrals of sin(x) and cos(x) is given by

$\int {\sin (x)dx}$=$- \cos (x) + c$
$\int {\cos (x)dx}$=$\sin (x) + c$
Where as derivatives of sin(x) and cos(x) is given by
$\dfrac{d}{{dx}}(\sin x)$= cos(x) and $\dfrac{d}{{dx}}$(cos x) = -sin(x)
The derivative of a function at a chosen input value describes the rate of change of the function near that input value.
The process of finding a derivative is called differentiation.
Derivative is defined as the varying rate of a function with respect to an independent variable.