Question

Evaluate the integral \[\int{\sqrt{\dfrac{a-x}{x}}dx}\]
(A) \[\sqrt{a-x}\sqrt{x}+a{{\tan }^{-1}}\left( \dfrac{\sqrt{x}}{\sqrt{a-x}} \right)+C\]
(B) \[\dfrac{\sqrt{a-x}}{\sqrt{x}}+a{{\tan }^{-1}}\left( \dfrac{\sqrt{x}}{\sqrt{a-x}} \right)+C\]
(C) \[\dfrac{\sqrt{a-x}}{\sqrt{x}}-a{{\tan }^{-1}}\left( \dfrac{\sqrt{x}}{\sqrt{a-x}} \right)+C\]
(D) \[\sqrt{a-x}\sqrt{x}-a{{\tan }^{-1}}\left( \dfrac{\sqrt{x}}{\sqrt{a-x}} \right)+C\]

Answer 1

Hint: We are given an expression and we are asked to integrate it. We cannot integrate the expression directly, so we will first let \[x=a{{\sin }^{2}}\theta \] and then rearrange the expression further. We will use the expression, \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] and compute further to get the expression in terms of cosine function as, \[a\int{2{{\cos }^{2}}\theta d\theta }\] . Then, we will make use of the trigonometric identity, which is, \[\cos 2\theta =2{{\cos }^{2}}\theta -1\]. We will then integrate the expression and we will have the expression in terms of inverse functions of sines and cosines which we will convert to tangent functions. Hence, we will have the required integration value.

Complete step by step solution:
According to the given question, we are given an expression which we have to integrate and find the most suitable answer from the options given.
The expression that we have is,
\[\int{\sqrt{\dfrac{a-x}{x}}dx}\]----(1)
We cannot solve it directly, so we will let \[x=a{{\sin }^{2}}\theta \] and substitute in the above expression.
Also,
\[x=a{{\sin }^{2}}\theta \]----(2)
\[\Rightarrow \dfrac{dx}{d\theta }=2a\sin \theta .\left( \cos \theta \right)\]----(3)
Now, substituting the equation (2) and (3) in equation (1), we get,
\[\Rightarrow \int{\sqrt{\dfrac{a-a{{\sin }^{2}}\theta }{a{{\sin }^{2}}\theta }}dx}\]
Taking ‘a’ in the numerator common, we get,
\[\Rightarrow \int{\sqrt{\dfrac{a(1-{{\sin }^{2}}\theta )}{a{{\sin }^{2}}\theta }}(2a\sin \theta \cos \theta )d\theta }\]
We know that, \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] and so we have,
\[\Rightarrow \int{\sqrt{\dfrac{{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta }}(2a\sin \theta \cos \theta )d\theta }\]
Simplifying the expression further, we get,
\[\Rightarrow 2a\int{\dfrac{\cos \theta }{\sin \theta }(\sin \theta \cos \theta )d\theta }\]
Cancelling out the sine functions, we get,
\[\Rightarrow 2a\int{\dfrac{\cos \theta }{1}(\cos \theta )d\theta }\]
So, we get the new expression in terms of cosine function and we have,
\[\Rightarrow a\int{2{{\cos }^{2}}\theta d\theta }\]
We know the trigonometric identity that is, \[\cos 2\theta =2{{\cos }^{2}}\theta -1\]. Applying this identity, we get the expression as,
\[\Rightarrow a\int{\left( \cos 2\theta +1 \right)d\theta }\]
We will now integrate the terms and we get,
\[\Rightarrow a\left[ \dfrac{\sin 2\theta }{2}+\theta \right]\]-----(4)
 From equation (2), we will obtain the value of theta and so we have,
\[x=a{{\sin }^{2}}\theta \]
\[\Rightarrow {{\sin }^{2}}\theta =\dfrac{x}{a}\]
\[\Rightarrow \sin \theta =\sqrt{\dfrac{x}{a}}\]----(5)
\[\theta ={{\sin }^{-1}}\sqrt{\dfrac{x}{a}}\]----(6)
We get the equation (4) as,
\[\Rightarrow a\left[ \dfrac{2\sin \theta \cos \theta }{2}+\theta \right]\]-----(7)
\[\Rightarrow a\left[ \sin \theta \cos \theta +\theta \right]\]-----(8)
We will substitute equation (6) in equation (8) and we get,
\[\Rightarrow a\left[ \sin \left( {{\sin }^{-1}}\sqrt{\dfrac{x}{a}} \right)\cos \theta +{{\sin }^{-1}}\sqrt{\dfrac{x}{a}} \right]\]
We have the expression as,
\[\Rightarrow a\left[ \sqrt{\dfrac{x}{a}}\cos \theta +{{\sin }^{-1}}\sqrt{\dfrac{x}{a}} \right]\]-----(9)
The value of \[\cos \theta \] can be found by using equation (5), we get,
\[\cos \theta =\sqrt{1-{{\sin }^{2}}\theta }\]
\[\Rightarrow \cos \theta =\sqrt{1-{{\left( \sqrt{\dfrac{x}{a}} \right)}^{2}}}\]
\[\Rightarrow \cos \theta =\sqrt{1-\dfrac{x}{a}}=\sqrt{\dfrac{a-x}{a}}\] -----(10)
We will now substitute the equation (10) in equation (9), we get the expression as,
\[\Rightarrow a\left[ \sqrt{\dfrac{x}{a}}\left( \sqrt{\dfrac{a-x}{a}} \right)+{{\sin }^{-1}}\sqrt{\dfrac{x}{a}} \right]\]
Opening up the brackets, we get,
\[\Rightarrow a\left( \sqrt{\dfrac{x}{a}}\sqrt{\dfrac{a-x}{a}} \right)+a\;\; {{\sin }^{-1}}\sqrt{\dfrac{x}{a}}\]
Cancelling up the common terms, we get,
\[\Rightarrow \sqrt{x}\sqrt{a-x}+a\; {{\sin }^{-1}}\sqrt{\dfrac{x}{a}}\]----(11)
We can write the sine inverse in the equation (11) in terms of tangent function.
From the equation (5) and equation (10), we get the value of the tangent function as,
\[\tan \theta =\dfrac{\sin \theta }{\cos \theta }=\dfrac{\sqrt{\dfrac{x}{a}}}{\sqrt{\dfrac{a-x}{a}}}=\sqrt{\dfrac{x}{a-x}}\]
That is, \[\theta ={{\tan }^{-1}}\sqrt{\dfrac{x}{a-x}}\]
\[\Rightarrow \sqrt{x}\sqrt{a-x}+a\; {{\tan }^{-1}}\sqrt{\dfrac{x}{a-x}}+C\]
Therefore, the correct option is (A) \[\sqrt{a-x}\sqrt{x}+a\;{{\tan }^{-1}}\left( \dfrac{\sqrt{x}}{\sqrt{a-x}} \right)+C\]

Note: The above solution is quite lengthy and tricky as well, so it is advised to carry out the integration clearly and step wise. The basic integration of the trigonometric functions must be known or else the solution can get very complicated. Also, since the options given to us had tangent function so we had to convert the obtained answer in sine function to the equivalent tangent function.