Question

The integral $\int{\sqrt{1+2\cot x(\csc x+\cot x)}}dx\left( 0<x<\dfrac{\pi }{2} \right)$ is equal to (Where $c$ is the constant of integration.).

A. $2\log \cos \dfrac{x}{2}+c$

B. $4\log \sin \dfrac{x}{2}+c$

C. $4\log \cos \dfrac{x}{2}+c$

D. $2\log \sin \dfrac{x}{2}+c$

Answer 1

Hint: First simplify the root by applying trigonometric identities. Then convert it into sine and cosine form.

Then make a substitution and apply integration and again substitute. You will get the answer.

Complete step-by-step answer:
Integration, in mathematics, a technique of finding a function $g(x)$ the derivative of which,$D(g(x))$, is equal to a given function $f(x)$. This is indicated by the integral sign $\int{{}}$ as in $\int{f(x)}$, usually called the indefinite integral of the function. The symbol$dx$represents an infinitesimal displacement along$x$. Thus$\int{f(x)dx}$ is the summation of the product of $f(x)$and $dx$. The definite integral, written as $\int\limits_{a}^{b}{f(x)dx}$

Some antiderivatives can be calculated by merely recalling which function has a given derivative, but the techniques of integration mostly involve classifying the functions according to which types of manipulations will change the function into a form the antiderivative of which can be more easily recognized.

Integration is the reverse of differentiation.

However,

If $y=2x+3,\dfrac{dy}{dx}=2$

If $y=2x+5,\dfrac{dy}{dx}=2$

If $y=2x,\dfrac{dy}{dx}=2$

So the integral of 2 can be $2x+3,2x+5,2x$ etc.

For this reason, when we integrate, we have to add a constant. So the integral of 2 is$2x+c$, where $c$ is a constant.

An "S" shaped symbol is used to mean the integral of, and$dx$is written at the end of the terms to be integrated, meaning "with respect to $x$". This is the same "$dx$" that appears in$\dfrac{dy}{dx}$.

Now taking $\int{\sqrt{1+2\cot x(\csc x+\cot x)}}dx$.

So now let’s try solving the integration $\int{\sqrt{1+2\cot x(\csc x+\cot x)}}dx$.

We know the identity that ${{\csc }^{2}}x-{{\cot }^{2}}x=1$.

So, in place of 1 in the integral above we get,

$\int{\sqrt{1+2\cot x(\csc x+\cot x)}}dx=\int{\sqrt{{{\csc }^{2}}x-{{\cot }^{2}}x+2\cot x(\csc x+\cot x)}}dx$

$\int{\sqrt{1+2\cot x(\csc x+\cot x)}}dx=\int{\sqrt{{{\csc }^{2}}x-{{\cot }^{2}}x+2\cot x\csc x+2{{\cot }^{2}}x}}dx$

So simplifying further we get,

$=\int{\sqrt{{{\csc }^{2}}x+{{\cot }^{2}}x+2\cot x\csc x}}dx$

So, we know the identity ${{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$.

So, using the identity, we get the integral

$\begin{align}

  & =\int{\sqrt{{{(\csc x+\cot x)}^{2}}}}dx \\

 & =\int{(\csc x+\cot x)}dx \\

\end{align}$

So now simplifying, that is converting above into $\sin x$ and $\cos x$, we get, the integral
$=\int{(\dfrac{1}{\sin x}+\dfrac{\cos x}{\sin x})}dx$

$=\int{\left( \dfrac{1+\cos x}{\sin x} \right)}dx$

So we know the identity $1+\cos x=2{{\cos }^{2}}\dfrac{x}{2}$ and $\sin x=2\sin \dfrac{x}{2}\cos \dfrac{x}{2}$.

$=\int{\left( \dfrac{2{{\cos }^{2}}\dfrac{x}{2}}{2\sin \dfrac{x}{2}\cos \dfrac{x}{2}} \right)}dx$
$=\int{\left( \dfrac{\cos \dfrac{x}{2}}{\sin \dfrac{x}{2}} \right)}dx$

So simplifying we get,

$=\int{\cot \dfrac{x}{2}}dx$

Now substituting $\dfrac{x}{2}=u$.

So differentiating we get $dx=2du$.

Now integration becomes,

$=2\int{\cot u}du$

$=2\int{\dfrac{\cos u}{\sin u}}du$

Now substituting$\sin u=v$.

Now differentiating we get $\cos udu=dv$.

$=2\int{\dfrac{1}{v}}dv$

So we get,

$=2\log v+c$

So now substituting we get,

$=2\log \sin u+c$

Again substituting we get,

$=2\log \sin \dfrac{x}{2}+c$

So we get $\int{\sqrt{1+2\cot x(\csc x+\cot x)}}dx=2\log \sin \dfrac{x}{2}+c$.

So the correct answer is option(D).


Note: Read the question carefully. You should know the basic things of integration. So here substitution is important. It depends on you what you are substituting. Also, you must know the rules of integration. Avoid silly mistakes because silly mistakes change the whole problem.