Question

Find the integral of $\dfrac{{\left( {\sin x + \cos x} \right)}}{{\left[ {3 + \sin 2x} \right]}}dx$

Answer 1

Hint: The process of integration is to obtain the antiderivative of a function. Also, it should be known that integration is the inverse process of differentiation and is known as the anti-differentiation. The integrals are of two types, definite integral and indefinite integral. A definite integral contains upper and lower limits whereas an indefinite integral does not contain upper and lower limits.
Here, we are given an indefinite integral and we are asked to calculate the integral of$\dfrac{{\left( {\sin x + \cos x} \right)}}{{\left[ {3 + \sin 2x} \right]}}dx$

Formula used:
   The formulae that we need to apply in the given problem are as follows.
$\sin 2x = 2\sin x\cos x$
${\sin ^2}x + {\cos ^2}x = 1$
${a^2} + {b^2} - 2ab = {\left( {a - b} \right)^2}$
$\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x$
$\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x$
$\int {\dfrac{{dx}}{{{a^2} - {x^2}}}} = \dfrac{1}{{2a}}\log \left( {\dfrac{{a + x}}{{a - x}}} \right) + c$, where $c$ is a constant.

Complete step by step answer:
We are asked to calculate the integral of$\dfrac{{\left( {\sin x + \cos x} \right)}}{{\left[ {3 + \sin 2x} \right]}}dx$
$\int {\dfrac{{\left( {\sin x + \cos x} \right)}}{{\left[ {3 + \sin 2x} \right]}}dx} $$ = \int {\dfrac{{\left( {\sin x + \cos x} \right)}}{{\left[ {4 - 1 + \sin 2x} \right]}}dx} $
                                    $ = \int {\dfrac{{\left( {\sin x + \cos x} \right)}}{{\left[ {4 - \left( {1 - \sin 2x} \right)} \right]}}dx} $
                                    $ = \int {\dfrac{{\left( {\sin x + \cos x} \right)}}{{\left[ {4 - \left( {1 - 2\sin x\cos x} \right)} \right]}}dx} $ (Here we applied the formula$\sin 2x = 2\sin x\cos x$)
                                    $ = \int {\dfrac{{\left( {\sin x + \cos x} \right)}}{{\left[ {4 - \left( {{{\sin }^2}x + {{\cos }^2}x - 2\sin x\cos x} \right)} \right]}}dx} $
                                    $ = \int {\dfrac{{\left( {\sin x + \cos x} \right)}}{{\left[ {4 - {{\left( {\sin x - \cos x} \right)}^2}} \right]}}dx} $ ………….$\left( 1 \right)$ (Here we used the formula${a^2} + {b^2} - 2ab = {\left( {a - b} \right)^2}$)
 Now, we consider$\sin x - \cos x = t$ …………$\left( 2 \right)$
We need to differentiate$\left( 2 \right)$on both sides.
$\left( {\cos x + \sin x} \right)dx = dt$ …….$\left( 3 \right)$ (Here we applied$\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x$and $\dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x$)
We shall substitute$\left( 2 \right)$and$\left( 3 \right)$in$\left( 1 \right)$.
$\int {\dfrac{{\left( {\sin x + \cos x} \right)}}{{\left[ {3 + \sin 2x} \right]}}dx} $$ = \int {\dfrac{{\left( {\sin x + \cos x} \right)}}{{\left[ {4 - {{\left( {\sin x - \cos x} \right)}^2}} \right]}}dx} $
     $ = \int {\dfrac{{dt}}{{\left[ {4 - {{\left( t \right)}^2}} \right]}}} $
     $ = \int {\dfrac{{dt}}{{\left[ {{2^2} - {t^2}} \right]}}} $
We need to apply the integral formula$\int {\dfrac{{dx}}{{{a^2} - {x^2}}}} = \dfrac{1}{{2a}}\log \left( {\dfrac{{a + x}}{{a - x}}} \right) + c$to the above equation.
$\int {\dfrac{{\left( {\sin x + \cos x} \right)}}{{\left[ {3 + \sin 2x} \right]}}dx} $$ = \int {\dfrac{{dt}}{{\left[ {{2^2} - {t^2}} \right]}}} $(Here$a = 2$and$x = t$)
     $ = \dfrac{1}{{2 \times 2}}\log \left( {\dfrac{{2 + t}}{{2 - t}}} \right) + C$
     $ = \dfrac{1}{4}\log \left( {\dfrac{{2 + t}}{{2 - t}}} \right) + C$
We have assumed$\sin x - \cos x = t$. So, we need to substitute it in the above equation.
$\int {\dfrac{{\left( {\sin x + \cos x} \right)}}{{\left[ {3 + \sin 2x} \right]}}dx} $$ = \dfrac{1}{4}\log \left( {\dfrac{{2 + \sin x - \cos x}}{{2 - \left( {\sin x - \cos x} \right)}}} \right) + C$
     $ = \dfrac{1}{4}\log \left( {\dfrac{{2 + \sin x - \cos x}}{{2 - \sin x + \cos x}}} \right) + C$
Therefore, $\int {\dfrac{{\left( {\sin x + \cos x} \right)}}{{\left[ {3 + \sin 2x} \right]}}dx} $$ = \dfrac{1}{4}\log \left( {\dfrac{{2 + \sin x - \cos x}}{{2 - \sin x + \cos x}}} \right) + C$ and $C$ is the constant of integration.

Note: We all know that differentiation is the process of finding the derivation of the functions whereas process integration is to calculate the antiderivative of a function. Hence, these two processes are said to be inverse to each other, (i.e.) the integration is the inverse process of differentiation and it is known as the anti-differentiation. Here we have applied trigonometric identities, algebraic identities, and integral formulas to obtain the required result. Hence, we get$\int {\dfrac{{\left( {\sin x + \cos x} \right)}}{{\left[ {3 + \sin 2x} \right]}}dx} $$ = \dfrac{1}{4}\log \left( {\dfrac{{2 + \sin x - \cos x}}{{2 - \sin x + \cos x}}} \right) + C$and$C$ is the constant of integration.