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

Hint: 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.