Question

Find the integral of $\dfrac{{\sin 2x}}{{1 + {{\cos }^2}x}}dx$?

Answer 1

First assume ${\cos ^2}x$ to be equal to t and then find its derivative, then we will get the numerator and dx related to dt as well and thus we have an easy integral in t.

Complete step by step answer:
We are given that we are required to find the integral of $\dfrac{{\sin 2x}}{{1 + {{\cos }^2}x}}dx$.
Let us just assume that ${\cos ^2}x = t$ …………………..(1)
Differentiating both the sides of the above equation, we will then obtain the following questions:-
$\Rightarrow \dfrac{d}{{dx}}\left( {{{\cos }^2}x} \right) = \dfrac{d}{{dx}}\left( t \right)$
Simplifying the above equation, we will then obtain the following equation:-
$\Rightarrow 2\cos x\dfrac{d}{{dx}}\left( {\cos x} \right) = \dfrac{{dt}}{{dx}}$
Now, we know that: $\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x$
Simplifying the above equation further, we will then obtain the following equation:-
$\Rightarrow - 2\cos x\sin x = \dfrac{{dt}}{{dx}}$ ………………..(2)
Now, we also know that: $2\sin x\cos x = \sin 2x$
Putting this in equation number 2, we will then obtain the following equation:-
$\Rightarrow \dfrac{{dt}}{{dx}} = - \sin 2x$
We can write this as following equation:-
$\Rightarrow \sin 2xdx = - dt$
Putting these things in the given integral, we will then obtain the following equation:-
$\Rightarrow \int {\dfrac{{\sin 2x}}{{1 + {{\cos }^2}x}}dx} = \int {\dfrac{{ - dt}}{{1 + {t^2}}}}$
We can take out the negative sign out of the integral and then we will obtain the following equation:-
$\Rightarrow \int {\dfrac{{\sin 2x}}{{1 + {{\cos }^2}x}}dx} = - \int {\dfrac{{dt}}{{1 + {t^2}}}}$
Since, we know that: $\int {\dfrac{{dx}}{{1 + {x^2}}} = {{\tan }^{ - 1}}x + C}$
Therefore, we will get the following equation:-
$\Rightarrow \int {\dfrac{{\sin 2x}}{{1 + {{\cos }^2}x}}dx} = - {\tan ^{ - 1}}t + C$
Now, replacing y back by using equation number 1, we will then obtain the following equation:-
$\Rightarrow \int {\dfrac{{\sin 2x}}{{1 + {{\cos }^2}x}}dx} = - {\tan ^{ - 1}}\left( {{{\cos }^2}x} \right) + C$

Note: The students must notice that while calculating $\dfrac{d}{{dx}}\left( {{{\cos }^2}x} \right)$, we have used the chain rule as follows:
Let us assume that $f(x) = {x^2}$ and $g(x) = \cos x$.
Therefore, we have: $f(g(x)) = f(\cos x) = {\cos ^2}x$
Now, chain rule states that we have following expression:-
$\Rightarrow \dfrac{d}{{dx}}\left[ {f\left( {g(x)} \right)} \right] = f'(g(x))g'(x)$
Therefore, we have:-
$\Rightarrow \dfrac{d}{{dx}}\left( {{{\cos }^2}x} \right) = \dfrac{d}{{dt}}\left( {{t^2}} \right) \times \dfrac{d}{{dt}}\left( {\cos x} \right)$, where $t = \cos x$
Simplifying it, we will get:-
$\Rightarrow \dfrac{d}{{dx}}\left( {{{\cos }^2}x} \right) = 2t( - \sin x)$, where $t = \cos x$
Simplifying it further, we will get:-
$\Rightarrow \dfrac{d}{{dx}}\left( {{{\cos }^2}x} \right) = - 2\sin x\cos x$
We can also write this as:-
$\Rightarrow \dfrac{d}{{dx}}\left( {{{\cos }^2}x} \right) = - \sin 2x$
Remember the following formulas:-
• $\dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x$
• $2\sin x\cos x = \sin 2x$
• $\int {\dfrac{{dx}}{{1 + {x^2}}} = {{\tan }^{ - 1}}x + C}$