Question

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

Answer 1

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

Complete step by step solution:
Let us consider the given function $\dfrac{1}{\sin x\sqrt{\sin x\cos x}}.$ 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 $\sin x.$
Then, we will get $\dfrac{\sin x}{{{\sin }^{2}}x\sqrt{\sin x\cos x}}.$
We will use the identity $\sqrt{ab}=\sqrt{a}\sqrt{b}.$
Now, from the above identity, we will get $\sqrt{\sin x\cos x}=\sqrt{\sin x}\sqrt{\cos x}.$
We will get \[\dfrac{1}{\sin x\sqrt{\sin x\cos x}}=\dfrac{\sin x}{{{\sin }^{2}}x\sqrt{\sin x}\sqrt{\cos x}}.\]
Also, we know that $\dfrac{a}{\sqrt{a}}=\sqrt{a}.$ So, we will get $\dfrac{\sin x}{\sqrt{\sin x}}=\sqrt{\sin x}.$
Thus, we will get \[\dfrac{1}{\sin x\sqrt{\sin x\cos x}}=\dfrac{\sqrt{\sin x}}{{{\sin }^{2}}x\sqrt{\cos x}}.\]
Now, we need to further simplify the above expression using the identity $\dfrac{\sqrt{a}}{\sqrt{b}}=\sqrt{\dfrac{a}{b}.}$
We will get the expression as \[\dfrac{1}{\sin x\sqrt{\sin x\cos x}}=\dfrac{1}{{{\sin }^{2}}x}\sqrt{\dfrac{\sin x}{\cos x}}.\]
We know that $\dfrac{\sin x}{\cos x}=\tan x.$
So, we will get \[\dfrac{1}{\sin x\sqrt{\sin x\cos x}}=\dfrac{\sqrt{\tan x}}{{{\sin }^{2}}x}.\]
We know the trigonometric identity $\tan x=\dfrac{1}{\cot x}.$ From this, we will get $\sqrt{\tan x}=\dfrac{1}{\sqrt{\cot x}}.$
Similarly, since $\sin x=\dfrac{1}{\text{cosec}x},$ we will get $\cos e{{c}^{2}}x=\dfrac{1}{{{\sin }^{2}}x}.$
So, we will get the equation as \[\dfrac{1}{\sin x\sqrt{\sin x\cos x}}=\dfrac{\text{cosec}x}{\sqrt{\cot x}}.\]
Now that we have simplified the given expression, we can find the integral using integration by substitution.
We will get $\int{\dfrac{1}{\sin x\sqrt{\sin x\cos x}}dx=\int{\dfrac{\cos e{{c}^{2}}x}{\sqrt{\cot x}}dx.}}$
Let us put $\cot x=t.$ Then, we will get $-\cos e{{c}^{2}}xdx=dt.$ So, we will get $\cos e{{c}^{2}}xdx=-dt.$
So, now the integral will become \[\int{\dfrac{1}{\sin x\sqrt{\sin x\cos x}}dx=\int{\dfrac{-dt}{\sqrt{t}}.}}\]
We know that $\int{\dfrac{1}{\sqrt{x}}dx=2\sqrt{x}+C.}$
So, we can write this integral as $\int{\dfrac{1}{\sin x\sqrt{\sin x\cos x}}dx=-2\sqrt{t}+c.}$
Now, since $t=\cot x,$ when we substitute the value, we will get $\int{\dfrac{1}{\sin x\sqrt{\sin x\cos x}}dx=-2\sqrt{\cot x}+c.}$
Hence the integral is $\int{\dfrac{1}{\sin x\sqrt{\sin x\cos x}}dx=-2\sqrt{\cot x}+c}$

Note: We should always remember the trigonometric identities. Remember that $\cos x=\dfrac{1}{\sec x}.$ 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.