Question

The integration of \[\int {\sqrt {1 - \sin x} dx} \]
A. \[2\sqrt {1 + \sin x} + C\]
B. \[2\sqrt {1 - \sin x} + C\]
C. \[2\sqrt {1 - 2\sin x} + C\]
D. \[2\sqrt {1 - \sin 2x} + C\]

Answer 1

Hint: Applying the integration directly to the function may be complicated. So, to solve this question we are multiplying and dividing by the conjugate of this function and then we are going to simplify and then we use the substitution method to solve further and then we are applying the integration to the function.
 \[\int {{x^m}dx = \dfrac{{{x^{m + 1}}}}{{m + 1}}} \] where \[x\] is the function and \[m\] is the power of the function

Complete step-by-step answer:
 Consider the given function \[I = \int {\sqrt {1 - \sin x} dx} \] . Now we multiply and divide the given function with its conjugate. The conjugate for this function is \[\sqrt {1 + \sin x} \]
 \[
  I = \int {\sqrt {1 - \sin x} \cdot \dfrac{{\sqrt {1 + \sin x} }}{{\sqrt {1 + \sin x} }}} dx \\
   = \int {\dfrac{{\sqrt {(1 - \sin x)(1 + \sin x)} }}{{\sqrt {1 + \sin x} }}dx} \;
 \]
We use the standard algebraic formula and that is \[(a + b)(a - b) = {a^2} - {b^2}\] . So, we have
 \[ = \int {\dfrac{{\sqrt {({1^2} - {{\sin }^2}x)} }}{{\sqrt {1 + \sin x} }}} dx\] since \[{1^2} = 1\] we have
 \[ = \int {\dfrac{{\sqrt {(1 - {{\sin }^2}x} )}}{{\sqrt {1 + \sin x} }}} dx\]
We have the trigonometric identity \[{\sin ^2}x + {\cos ^2}x = 1\] \[ \Rightarrow 1 - {\sin ^2}x = {\cos ^2}x\]
Therefore, we have \[ = \int {\dfrac{{\sqrt {{{\cos }^2}x} }}{{\sqrt {1 + \sin x} }}} \]
 \[ = \int {\dfrac{{\cos x}}{{\sqrt {1 + \sin x} }}} dx\]
Let we substitute \[1 + \sin x = t\]
On differentiating the above function w.r.t, x we have
 \[
  \dfrac{d}{{dx}}(1 + \sin x) = \dfrac{{dt}}{{dx}} \\
  \cos x = \dfrac{{dt}}{{dx}} \;
 \]
Since the differentiation of constant is zero and \[\sin x\] is \[\cos x\]
 \[ \Rightarrow \cos xdx = dt\]
Therefore, the I can be rewritten as \[I = \int {\dfrac{1}{{\sqrt t }}dt} \]
Square root written in the form power we have \[I = \int {\dfrac{1}{{{t^{\dfrac{1}{2}}}}}} \cdot dt\]
 \[ = \int {{t^{\dfrac{{ - 1}}{2}}}dt} \]
[note: \[\int {{x^m}dx = \dfrac{{{x^{m + 1}}}}{{m + 1}}} + C\] , where C is integration constant]
On integration of I we have \[I = \dfrac{{{t^{\left( {\dfrac{{ - 1}}{2} + 1} \right)}}}}{{\left( {\dfrac{{ - 1}}{2} + 1} \right)}} + C\] ,where \[C\] is integration constant
By simplifying, \[I = \dfrac{{{t^{\dfrac{1}{2}}}}}{{\dfrac{1}{2}}} + C\] , where \[C\] is integration constant
 \[ \Rightarrow I = 2\sqrt t + C\] , where \[C\] is integration constant
By re-substituting the value of t we have
 \[I = 2\sqrt {1 + \sin x} + C\] , where \[C\] is integration constant
So, the correct answer is “Option A”.

Note: Here we are rationalizing the function to solve further. If we apply directly the integration to the function it may be complicated to solve or we may go wrong to solve the function. So we are rationalizing the function to solve the integration easier. And the second main thing is that here we use the substitution method and we have to look at the function carefully to substitute and by substitution we should not make the function much more complicated.