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Question: Integrate the expression \(\int{\left( \sqrt{\cot x}+\sqrt{\tan x} \right)dx}\)...

Integrate the expression ∫(cot⁡x+tan⁡x)dx\int{\left( \sqrt{\cot x}+\sqrt{\tan x} \right)dx}

Explanation

Solution

To solve this question, we will make use of transformations and use substitution method. We will first transform the given expression into sin x and cos x form. We know that tan x = sin⁡xcos⁡x\dfrac{\sin \text{x}}{\cos \text{x}} and cot x = cos⁡xsin⁡x\dfrac{\cos \text{x}}{\sin \text{x}}. Then we will perform cross multiplication to make the denominators common. Then we will make alterations in the denominator and use substitution to simplify the expression and reduce it to a known integral.

Complete step by step answer:
The integral given to us is ∫(cot⁡x+tan⁡x)dx\int{\left( \sqrt{\cot x}+\sqrt{\tan x} \right)dx}.
We will write cot x as cos⁡xsin⁡x\dfrac{\cos \text{x}}{\sin \text{x}} and tan x as sin⁡xcos⁡x\dfrac{\sin \text{x}}{\cos \text{x}}.
The integral will change as ∫(cos⁡xsin⁡x+sin⁡xcos⁡x)dx\int{\left( \sqrt{\dfrac{\cos x}{\sin x}}+\sqrt{\dfrac{\sin x}{\cos x}} \right)dx}
We will now perform cross multiplication to make the denominator same.
The integral will now become ∫(cos⁡x+sin⁡xsin⁡xcos⁡x)dx\int{\left( \dfrac{\cos x+\sin x}{\sqrt{\sin x\cos x}} \right)dx}
Now, we will multiple and divide the integral by 2\sqrt{2}. It will be multiplied in to the denominator and go under the square root.
Thus, the integral will become 2∫(cos⁡x+sin⁡x2sin⁡xcos⁡x)dx\sqrt{2}\int{\left( \dfrac{\cos x+\sin x}{\sqrt{2\sin x\cos x}} \right)dx}
Now, we will add 1 and subtract 1 in the denominator under the root.
The integral will become 2∫(cos⁡x+sin⁡x1−1+2sin⁡xcos⁡x)dx\sqrt{2}\int{\left( \dfrac{\cos x+\sin x}{\sqrt{1-1+2\sin x\cos x}} \right)dx}
We will take the negative sign as common in the denominator.
⇒2∫(cos⁡x+sin⁡x1−(1−2sin⁡xcos⁡x))dx\Rightarrow \sqrt{2}\int{\left( \dfrac{\cos x+\sin x}{\sqrt{1-\left( 1-2\sin x\cos x \right)}} \right)dx}
From the trigonometric identities, we know that sin⁡2x+cos⁡2x=1{{\sin }^{2}}x+{{\cos }^{2}}x=1. Thus, we will write 1 in the denominator as sin⁡2x+cos⁡2x{{\sin }^{2}}x+{{\cos }^{2}}x.
The integral will change as 2∫(cos⁡x+sin⁡x1−(sin⁡2x+cos⁡2x−2sin⁡xcos⁡x))dx\sqrt{2}\int{\left( \dfrac{\cos x+\sin x}{\sqrt{1-\left( {{\sin }^{2}}x+{{\cos }^{2}}x-2\sin x\cos x \right)}} \right)dx}
Now, we know that a2+b2−2ab=(a−b)2{{a}^{2}}+{{b}^{2}}-2ab={{\left( a-b \right)}^{2}}. We will apply this property in the denominator.
Thus, the integral now is 2∫(cos⁡x+sin⁡x1−(sin⁡x−cos⁡x)2)dx\sqrt{2}\int{\left( \dfrac{\cos x+\sin x}{\sqrt{1-{{\left( \sin x-\cos x \right)}^{2}}}} \right)dx}
Now we take sin⁡x−cos⁡x\sin x-\cos x as variable u.
⇒sin⁡x−cos⁡x=u\Rightarrow \sin x-\cos x=u
We know that the derivative of sin x is cos x and derivative of cos x is ─sin x.
⇒d(sin⁡x−cos⁡x)=du ⇒(cos⁡x+sin⁡x)dx=du \begin{aligned} & \Rightarrow d\left( \sin x-\cos x \right)=du \\\ & \Rightarrow \left( \cos x+\sin x \right)dx=du \\\ \end{aligned}
After substitution the integral changes as follow:
⇒2∫du1−u2\Rightarrow \sqrt{2}\int{\dfrac{du}{\sqrt{1-{{u}^{2}}}}}
We know that ∫11−x2dx=sin⁡−1x\int{\dfrac{1}{\sqrt{1-{{x}^{2}}}}dx=}{{\sin }^{-1}}x
Thus, now we carry out the integration.
⇒2sin⁡−1(u)+C\Rightarrow \sqrt{2}{{\sin }^{-1}}\left( u \right)+C
Where C is a constant.
But we know that sin⁡x−cos⁡x=u\sin x-\cos x=u.

Therefore, ∫(cot⁡x+tan⁡x)dx=2sin⁡−1(sin⁡x+cos⁡x)+C\int{\left( \sqrt{\cot x}+\sqrt{\tan x} \right)dx}=\sqrt{2}{{\sin }^{-1}}\left( \sin x+\cos x \right)+C

Note: It is always a good practice to simplify the trigonometric ratios of tan, cot, sec and cosec into sin and cos. It simplifies the operations which are needed. Moreover, students should always simplify the given expression before they perform the actual integration.