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Evaluate the integral $\int{{{\tan }^{-1}}\left( \dfrac{\sin 2x}{1+\cos 2x} \right)dx}$
$\left( a \right)x+c$
$\left( b \right)\dfrac{{{x}^{2}}}{2}+c$
$\left( c \right){{x}^{2}}+c$
$\left( d \right)\dfrac{{{x}^{2}}}{4}+c$

Last updated date: 03rd Mar 2024
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Hint: We are asked to find the integral of the given term. Since this is a complex one we need to first solve the stuff inside the bracket and we need to simplify that. In this case we will use several identities of double angle involving cosine and sine and then after simplifying the matter inside the bracket we will integrate the whole term. We should be aware of all the trigonometric identities.

Complete step-by-step solution:
We have $\int{{{\tan }^{-1}}\left( \dfrac{\sin 2x}{1+\cos 2x} \right)dx}$.
The term inside the bracket is:
$\dfrac{\sin 2x}{1+\cos 2x}$
We can use the following identity in place of $sin2x$:
$\sin 2x=2\sin x\cos x$
And in place of $cos2x$, we can use the following formula:
$\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x$
Putting in the fraction and simplifying it, we have:
$\dfrac{\sin 2x}{1+\cos 2x}=\dfrac{2\sin x\cos x}{1+\left( {{\cos }^{2}}x-{{\sin }^{2}}x \right)}$
Now, we use the following important identity:
${{\cos }^{2}}x+{{\sin }^{2}}x=1$
Using the same we get:
${{\cos }^{2}}x=1-{{\sin }^{2}}x$
Putting this in the denominator:
$\dfrac{\sin 2x}{1+\cos 2x}=\dfrac{2\sin x\cos x}{2{{\cos }^{2}}x}$
So, we can say that:
$\dfrac{\sin 2x}{1+\cos 2x}=\tan x$
We obtained this using the formula:
$\dfrac{\sin x}{\cos x}=\tan x$
Putting this term inside the bracket we have:
$\int{{{\tan }^{-1}}\left( \tan x \right)dx=\int{xdx}}$
And we know that:
Hence, option $\left( b \right)\dfrac{{{x}^{2}}}{2}+c$ is correct.

Note: If the formulae related to the double angle of sine and cosine are not known then this becomes a very difficult integral to operate because the terms inside bracket cannot be solved so easily. Moreover, you should be aware about the integrals of some basic algebraic terms such as the polynomial terms.

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