Question

Find the value of $\int {\dfrac{{\cos 4x + 1}}{{\cot x - \tan x}}dx = }$

Answer 1

The given function is indefinite since there is no limit given. The indefinite integral of a function is a differentiable function F whose derivative is equal to the original function f. The first fundamental theorem of calculus allows definite integrals to be computed in terms of indefinite integrals.
In the given function, we will try to bring all the identity in the same form, and then we will use the trigonometric identity $\cos 2\theta = 2{\cos ^2}\theta - 1$we further reduce the function and then it is integrated.

Complete step by step solution:
Let the given integral be $I$ such that:
$I = \int {\dfrac{{\cos 4x + 1}}{{\cot x - \tan x}}dx}$
We know the quotient identity of
$\cot x = \dfrac{{\cos x}}{{\sin x}}$
$\tan x = \dfrac{{\sin x}}{{\cos x}}$
This function can be written as

$$I = \int {\dfrac{{\cos 4x + 1}}{{\dfrac{{\cos x}}{{\sin x}} - \dfrac{{\sin x}}{{\cos x}}}}dx} \\\ = \int {\dfrac{{\cos 4x + 1}}{{\dfrac{{{{\cos }^2}x - {{\sin }^2}x}}{{\sin x\cos x}}}}} dx \\\$$

Now we know the trigonometric identity $\cos 2\theta = 2{\cos ^2}\theta - 1$
Hence by using this trigonometric identity, we can write the above function as
$I = \int {\dfrac{{2{{\cos }^2}\left( {2x} \right) - 1 + 1}}{{\dfrac{{\cos 2x}}{{\sin x\cos x}}}}} dx$ [Since$\cos 2\theta = 2{\cos ^2}\theta - 1$]
This can be further written as
$I = \int {\dfrac{{2{{\cos }^2}\left( {2x} \right)}}{{\cos 2x}}} \sin x\cos xdx$
As we know the trigonometric identity $2\sin \theta \cos \theta = \sin 2\theta$, hence we can write above identity as

$$I = \int {\cos \left( {2x} \right)} \times 2\sin x\cos xdx \\\ = \int {\cos 2x\sin 2xdx} \\\$$

Now by again using trigonometric identity $2\sin \theta \cos \theta = \sin 2\theta$, we can write
$I = \dfrac{1}{2}\int {\sin 4xdx}$
Hence by integrating this, we get

$$I = \dfrac{1}{2}\left[ { - \dfrac{{\cos 4x}}{4}} \right] \\\ = - \dfrac{1}{8}\cos 4x \\\$$

Therefore$\int {\dfrac{{\cos 4x + 1}}{{\cot x - \tan x}}dx = } - \dfrac{1}{8}\cos 4x$

Important equations used:
1. $\cos 2\theta = 2{\cos ^2}\theta - 1$
2. $2\sin \theta \cos \theta = \sin 2\theta$
3. $\int {\sin \left( {ax} \right)dx = - \dfrac{1}{a}\cos \left( {ax} \right)}$
4. $\int {dx = x + c}$

Note:
While substituting the real parameter of the question with the auxiliary parameter, one should be sure that it will not make the problem more complex. However, selecting an auxiliary parameter completely depends on the individual point of view.