Question

What is the value of integral $\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sqrt {\sin x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}} dx$?
A. $\dfrac{\pi }{2}$
B. $\dfrac{\pi }{4}$
C. $0$
D. $1$

Answer 1

Hint: We will use the property of definite integrals of changing limits that integration area under $x = a$ to $x = b$ in $f(x)$ is same as integration under $x = a$ to $x = a + b - x$and then with some basic algebraic manipulation we will try to get the answer.

Formula Used:
$\int\limits_a^b {f(x)dx = \int\limits_a^{a + b - x} {f(a + b - x)dx} } $
$\int\limits_0^a {1dx = a} $
$\sin (\dfrac{\pi }{2} - x) = \cos x$

Complete step by step solution:
Given -$I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sqrt {\sin x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}} dx$
Adding and subtracting $\sqrt {\cos x} $in the numerator of $\dfrac{{\sqrt {\sin x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}$
$\dfrac{{\sqrt {\sin x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }} = \dfrac{{\sqrt {\sin x} + \sqrt {\cos x} - \sqrt {\cos x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}$
On simplifying the equation
$ \Rightarrow \dfrac{{\sqrt {\sin x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }} = \dfrac{{\sqrt {\sin x} + \sqrt {\cos x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }} - \dfrac{{\sqrt {\cos x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}$
$ \Rightarrow \dfrac{{\sqrt {\sin x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }} = 1 - \dfrac{{\sqrt {\cos x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}$
$ \Rightarrow \dfrac{{\sqrt {\sin x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }} + \dfrac{{\sqrt {\cos x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }} = 1$ —(1)
We know by the properties of integral that $\int\limits_a^b {f(x)dx = \int\limits_a^{a + b - x} {f(a + b - x)dx} } $
Here $a = 0$ and $b = \dfrac{\pi }{2}$
Hence $a + b - x = (\dfrac{\pi }{2} - x)$
Therefore $\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sqrt {\sin x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}} dx$=$\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sqrt {\cos x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}} dx$ But $\sin (\dfrac{\pi }{2} - x) = \cos x$
Hence $\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sqrt {\sin x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}} dx$=$\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sqrt {\cos x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}} dx$
So $I = \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sqrt {\cos x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}} dx$
Thus $\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sqrt {\cos x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}} dx + \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sqrt {\sin x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}} dx = 2I$
We know that $\int\limits_a^b {f(x)} + \int\limits_a^b {g(x)} = \int\limits_a^b {f(x) + g(x)} $
Using the above formula –
$\int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sqrt {\cos x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}} dx + \int\limits_0^{\dfrac{\pi }{2}} {\dfrac{{\sqrt {\sin x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}} dx = \int\limits_0^{\dfrac{\pi }{2}} {\,\dfrac{{\sqrt {\cos x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }} + } \dfrac{{\sqrt {\sin x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }}dx$
But by equation (1)
$\dfrac{{\sqrt {\sin x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }} + \dfrac{{\sqrt {\cos x} }}{{\sqrt {\sin x} + \sqrt {\cos x} }} = 1$
Thus $2I = \int\limits_0^{\dfrac{\pi }{2}} 1 dx$
$ \Rightarrow I = \dfrac{\pi }{2}$

Option ‘A’ is correct

Note: In such types of questions involving the complicated functions to be integrated, students may get wrong if they solve them directly without using the integration properties. Hence the questions having definite integration with complicated functions, first of all try to simplify the function as much as possible to a known standard form of integral using the integration properties and other trigonometric or algebraic formulas and once it is simplified to a known form then apply integral rules to integrate it.