Question

Integral of $\sqrt {1 + 2\cot x(\cot x + \cos ecx}$ w.r.t x is?

Answer 1

Integration is the one of the two main operations of calculus; its inverse operation differentiation is the other.
Integration of trigonometric functions generally if the function is $\operatorname{Sin} (x)\,or\,\cos (x)$ is reverse of their respective derivatives.
We have $\int {a\cos xxdx = \dfrac{a}{n}\operatorname{Sin} nx + C}$
In all formulas the constant ‘a’ is supposed to be non-new, while c denotes the constant of integration.

Complete step-by-step solution
Given $\sqrt {1 + 2\cot x(\cot x + \cos ecx} .....(1)$
To simplify the question let us convert all trigonometric ratio in $'COS'$ or $'\operatorname{Sin} '$
We know $\cot x = \dfrac{cos}{{\sin x}}\ and\,\cos ex = \dfrac{1}{{\sin x}}$
Using these value in (2) we have
$\Rightarrow \sqrt {1 + 2{{\cot }^2}x + 2\cot x\operatorname{Cos} ex} ..................(2)$
$\Rightarrow \sqrt {1 + 2{{(\dfrac{{\cos x}}{{\sin x}})}^2} + 2(\dfrac{{\cos x}}{{\sin x}})(\dfrac{1}{{\sin x}})}$
$\Rightarrow \sqrt {1 + 2{{(\dfrac{{{{\cos }^2}x}}{{{{\sin }^2}x}})}^2} + 2(\dfrac{{\cos x}}{{{{\sin }^2}x}})}$
Taking the LCM
$\Rightarrow \sqrt {\dfrac{{{{\sin }^2}x + 2{{\cos }^2}x + 2\cos x}}{{{{\sin }^2}x}}} = \sqrt {\dfrac{{{{\sin }^2}x + 2{{\cos }^2}x + 2\cos x}}{{\sin x}}} ..................(3)$
We known ${\cos ^2}x + {\sin ^2}x = 1 \Rightarrow {\operatorname{Sin} ^2}x = 1 - c{a^2}x$
Using the value in (3) we get

$\Rightarrow \sqrt {\dfrac{{(1 - {{\cos }^2}x) + 2{{\cos }^2}x + 2\cos x}}{{\sin x}})}$
$\Rightarrow \sqrt {\dfrac{{1 - {{\cos }^2}x + 2{{\cos }^2}x + 2\cos x}}{{\sin x}})}$
$\Rightarrow \sqrt {\dfrac{{{{\cos }^2}x + 2\cos x + 1}}{{\sin x}}}$
$\Rightarrow \sqrt {\dfrac{{\cos x + 1}}{{\sin x}}}$ $\left[ {\because {{\cos }^2}x + 2\cos x + 1 = {{(1 + \cos x)}^2}} \right]$
$\Rightarrow \dfrac{{\cos x + 1}}{{\sin x}}$

Writing the $\sin x$, and the $\cos x$ in submultiples angle we have,
$\Rightarrow \dfrac{{(2{{\cos }^2}\dfrac{x}{2} - 1 + 1)}}{{\operatorname{Sin} x}}\left[ {\because \cos x = 2\cos^ 2\dfrac{x}{2} - 1} \right]$
$\Rightarrow \dfrac{{(2{{\cos }^2}\dfrac{x}{2} - 1 + 1)}}{{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}}\left[ {\because \sin x = 2\sin \dfrac{x}{2}\cos \dfrac{x}{2}} \right]$
$\Rightarrow \dfrac{{2{{\cos }^2}\dfrac{x}{2}}}{{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}}} \Rightarrow \dfrac{{2\cos \dfrac{x}{2}}}{{2\sin \dfrac{x}{2}}}$ $= \cot \dfrac{x}{2}$

Now, $\cot xdx = \log \left| {\sin x} \right| + c$
So, $\int {\cot \dfrac{x}{2}dx = \dfrac{{\log \left| {\sin \dfrac{x}{2}} \right|}}{{\dfrac{d}{{dx}}(\dfrac{x}{2})}}}$
$= \dfrac{{\log \left| {\sin \dfrac{x}{2}} \right|}}{{(\dfrac{1}{2})}} + c$ $\left[ {\because \dfrac{d}{{dx}}(\dfrac{x}{n}) = \dfrac{1}{n}} \right]$
$= 2\log \left| {\sin \dfrac{x}{2} + c} \right|$ where c is content of integration

Note
In integration of trigonometric function if x is replaced by any other function then derivative of that function is divided by the integration only
For Ex. $\int {\cos nxdx = \dfrac{1}{n}\sin nx + c}$
$\int {\cos nxdx = \dfrac{{\sin nxx}}{{\dfrac{d}{{dx}}(nx)}} + c}$
$\int {\cos nxdx = \dfrac{1}{n}\sin nx + c}$

These functions are used to relate the angles of a triangle with the sides of that triangle. Trigonometric functions are important when studying triangles and modelling periodic phenomena such as waves, sound, and light. To define these functions for the angle theta, begin with a right triangle. Each function relates the angle to two sides of a right triangle