Question

Integrate $\int{{{\left( \tan x \right)}^{\dfrac{1}{2}}}+{{\left( \cot x \right)}^{\dfrac{1}{2}}}dx}$.

Answer 1

Hint: In this question we have been given with a trigonometric expression to integrate. We will solve this question by first writing the expression in the form of square root. We will then simplify the expression to get the expression in the form of a substitution. We then use substitution to integrate the expression and resubstitute the terms to get the required solution.

Complete step by step answer:
We have the expression given to us as:
$\Rightarrow \int{\left( {{\left( \tan x \right)}^{\dfrac{1}{2}}}+{{\left( \cot x \right)}^{\dfrac{1}{2}}} \right)dx}$
Now we know the property of exponent that $\sqrt{a}={{a}^{\dfrac{1}{2}}}$therefore, we can write:
$\Rightarrow \int{\left( \sqrt{\tan x}+\sqrt{\cot x} \right)dx}$
Now we know that $\tan x=\dfrac{\sin x}{\cos x}$ and $\cot x=\dfrac{\cos x}{\sin x}$ therefore, we get:
$\Rightarrow \int{\left( \sqrt{\dfrac{\sin x}{\cos x}}+\sqrt{\dfrac{\cos x}{\sin x}} \right)dx}$
On splitting the root, we get:
$\Rightarrow \int{\left( \dfrac{\sqrt{\sin x}}{\sqrt{\cos x}}+\dfrac{\sqrt{\cos x}}{\sqrt{\sin x}} \right)dx}$
On taking the lowest common multiple, we get:
$\Rightarrow \int{\left( \dfrac{\sin x+\cos x}{\sqrt{\sin x\cos x}} \right)dx}$
On multiplying and dividing the expression by $\sqrt{2}$, we get:
$\Rightarrow \sqrt{2}\int{\left( \dfrac{\sin x+\cos x}{\sqrt{2\sin x\cos x}} \right)dx}$
Now on adding and subtracting $1$ in the square root of the denominator, we get:
$\Rightarrow \sqrt{2}\int{\left( \dfrac{\sin x+\cos x}{\sqrt{2\sin x\cos x+1-1}} \right)dx}$
Now we know that ${{\left( \sin x-\cos x \right)}^{2}}={{\sin }^{2}}x-2\sin x\cos x+{{\cos }^{2}}x$ and since ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, we get ${{\left( \sin x-\cos x \right)}^{2}}=1-2\sin x\cos x$. We get $-{{\left( \sin x-\cos x \right)}^{2}}=2\sin x\cos x-1$.
On substituting, we get:
$\Rightarrow \sqrt{2}\int{\left( \dfrac{\sin x+\cos x}{\sqrt{1-{{\left( \sin x-\cos x \right)}^{2}}}} \right)dx}$
Now let $\sin x-\cos x=t$
On differentiating, we get $\left( \sin x+\cos x \right)dx=dt$
On substituting, we get:
\[\Rightarrow \sqrt{2}\int{\dfrac{dt}{\sqrt{1-{{t}^{2}}}}}\]
Now we know that $\int{\dfrac{dx}{\sqrt{1-{{x}^{2}}}}={{\sin }^{-1}}t+c}$ therefore, we get:
\[\Rightarrow \sqrt{2}{{\sin }^{-1}}t+c\]
On substituting the value of $t$, we get:
\[\Rightarrow \sqrt{2}{{\sin }^{-1}}\left( \sin x-\cos x \right)+c\], which is the required answer.

Note: It is to be remembered that integration and differentiation are inverse of each other. If the integration of $a$ is $b$ then the derivative of $b$ is $a$. In this question we have used the substitution method to solve the integral. The various types of integration should be remembered. It is to be noted that multiplying and dividing an expression with a term or adding or subtracting the same term does not change the expression's value.