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Question: Evaluate the following \(\int{\dfrac{1}{\sin x\sqrt{\sin x\cos x}}}dx=\) \(\left( a \right) \sqrt...

Evaluate the following ∫1sin⁡xsin⁡xcos⁡xdx=\int{\dfrac{1}{\sin x\sqrt{\sin x\cos x}}}dx=
(a)tan⁡x+c\left( a \right) \sqrt{\tan x}+c
(b)−2tan⁡x+c\left( b \right) -2\sqrt{\tan x}+c
(c)−2cot⁡x+c\left( c \right) -2\sqrt{\cot x}+c
(d)2cot⁡x+c\left( d \right) 2\sqrt{\cot x}+c

Explanation

Solution

We will multiply and divide the given function with sin⁡x.\sin x. Then, we will use the trigonometric identity cosecx=1sin⁡x.\text{cosecx}=\dfrac{1}{\sin x}. Also, we will use the trigonometric identity sin⁡xcos⁡x=tan⁡x\dfrac{\sin x}{\cos x}=\tan x and tan⁡x=1cot⁡x.\tan x=\dfrac{1}{\cot x}. And then we will do integration by substitution.

Complete step by step solution:
Let us consider the given function 1sin⁡xsin⁡xcos⁡x.\dfrac{1}{\sin x\sqrt{\sin x\cos x}}. We need to find the integral of the given function.
We need to make some necessary rearrangements for it.
We will multiply and divide the function with sin⁡x.\sin x.
Then, we will get sin⁡xsin⁡2xsin⁡xcos⁡x.\dfrac{\sin x}{{{\sin }^{2}}x\sqrt{\sin x\cos x}}.
We will use the identity ab=ab.\sqrt{ab}=\sqrt{a}\sqrt{b}.
Now, from the above identity, we will get sin⁡xcos⁡x=sin⁡xcos⁡x.\sqrt{\sin x\cos x}=\sqrt{\sin x}\sqrt{\cos x}.
We will get 1sin⁡xsin⁡xcos⁡x=sin⁡xsin⁡2xsin⁡xcos⁡x.\dfrac{1}{\sin x\sqrt{\sin x\cos x}}=\dfrac{\sin x}{{{\sin }^{2}}x\sqrt{\sin x}\sqrt{\cos x}}.
Also, we know that aa=a.\dfrac{a}{\sqrt{a}}=\sqrt{a}. So, we will get sin⁡xsin⁡x=sin⁡x.\dfrac{\sin x}{\sqrt{\sin x}}=\sqrt{\sin x}.
Thus, we will get 1sin⁡xsin⁡xcos⁡x=sin⁡xsin⁡2xcos⁡x.\dfrac{1}{\sin x\sqrt{\sin x\cos x}}=\dfrac{\sqrt{\sin x}}{{{\sin }^{2}}x\sqrt{\cos x}}.
Now, we need to further simplify the above expression using the identity ab=ab.\dfrac{\sqrt{a}}{\sqrt{b}}=\sqrt{\dfrac{a}{b}.}
We will get the expression as 1sin⁡xsin⁡xcos⁡x=1sin⁡2xsin⁡xcos⁡x.\dfrac{1}{\sin x\sqrt{\sin x\cos x}}=\dfrac{1}{{{\sin }^{2}}x}\sqrt{\dfrac{\sin x}{\cos x}}.
We know that sin⁡xcos⁡x=tan⁡x.\dfrac{\sin x}{\cos x}=\tan x.
So, we will get 1sin⁡xsin⁡xcos⁡x=tan⁡xsin⁡2x.\dfrac{1}{\sin x\sqrt{\sin x\cos x}}=\dfrac{\sqrt{\tan x}}{{{\sin }^{2}}x}.
We know the trigonometric identity tan⁡x=1cot⁡x.\tan x=\dfrac{1}{\cot x}. From this, we will get tan⁡x=1cot⁡x.\sqrt{\tan x}=\dfrac{1}{\sqrt{\cot x}}.
Similarly, since sin⁡x=1cosecx,\sin x=\dfrac{1}{\text{cosec}x}, we will get cos⁡ec2x=1sin⁡2x.\cos e{{c}^{2}}x=\dfrac{1}{{{\sin }^{2}}x}.
So, we will get the equation as 1sin⁡xsin⁡xcos⁡x=cosecxcot⁡x.\dfrac{1}{\sin x\sqrt{\sin x\cos x}}=\dfrac{\text{cosec}x}{\sqrt{\cot x}}.
Now that we have simplified the given expression, we can find the integral using integration by substitution.
We will get ∫1sin⁡xsin⁡xcos⁡xdx=∫cos⁡ec2xcot⁡xdx.\int{\dfrac{1}{\sin x\sqrt{\sin x\cos x}}dx=\int{\dfrac{\cos e{{c}^{2}}x}{\sqrt{\cot x}}dx.}}
Let us put cot⁡x=t.\cot x=t. Then, we will get −cos⁡ec2xdx=dt.-\cos e{{c}^{2}}xdx=dt. So, we will get cos⁡ec2xdx=−dt.\cos e{{c}^{2}}xdx=-dt.
So, now the integral will become ∫1sin⁡xsin⁡xcos⁡xdx=∫−dtt.\int{\dfrac{1}{\sin x\sqrt{\sin x\cos x}}dx=\int{\dfrac{-dt}{\sqrt{t}}.}}
We know that ∫1xdx=2x+C.\int{\dfrac{1}{\sqrt{x}}dx=2\sqrt{x}+C.}
So, we can write this integral as ∫1sin⁡xsin⁡xcos⁡xdx=−2t+c.\int{\dfrac{1}{\sin x\sqrt{\sin x\cos x}}dx=-2\sqrt{t}+c.}
Now, since t=cot⁡x,t=\cot x, when we substitute the value, we will get ∫1sin⁡xsin⁡xcos⁡xdx=−2cot⁡x+c.\int{\dfrac{1}{\sin x\sqrt{\sin x\cos x}}dx=-2\sqrt{\cot x}+c.}
Hence the integral is ∫1sin⁡xsin⁡xcos⁡xdx=−2cot⁡x+c\int{\dfrac{1}{\sin x\sqrt{\sin x\cos x}}dx=-2\sqrt{\cot x}+c}

Note: We should always remember the trigonometric identities. Remember that cos⁡x=1sec⁡x.\cos x=\dfrac{1}{\sec x}. Even though we have not used this in the above problem, we may need to use this more often while doing problems using trigonometric functions.