Question

Find the value of the given integral $\int {\sqrt {1{\text{ - sinx}}} {\text{dx = }}}$
${\text{A}}{\text{. 2}}\sqrt {{\text{1 + sinx}}} + {\text{C}} \\\ {\text{B}}{\text{. 2}}\sqrt {{\text{1 - sinx}}} + {\text{C}} \\\ {\text{C}}{\text{. 2}}\sqrt {{\text{1 - 2sinx}}} + {\text{C}} \\\ {\text{D}}{\text{. 2}}\sqrt {{\text{1 - sin2x}}} + {\text{C}} \\\$

Answer 1

Hint:To compute the value of the integral, we use the trigonometric identity ${\text{si}}{{\text{n}}^2}\theta + {\text{co}}{{\text{s}}^2}\theta = 1$, we rearrange the terms inside the square root by using algebraic formula. And compute the integral values of sin and cos functions and add them.

Complete step-by-step answer:
Given Data, $\int {\sqrt {1{\text{ - sinx}}} {\text{dx = }}}$
We know, ${\text{si}}{{\text{n}}^2}\theta + {\text{co}}{{\text{s}}^2}\theta = 1$, ${\text{sin2}}\theta {\text{ = 2sin}}\dfrac{\theta }{2}{\text{cos}}\dfrac{\theta }{2}$.
Also we know${\left( {{\text{a + b}}} \right)^2} = {{\text{a}}^2} + {{\text{b}}^2}{\text{ + 2ab}}$.
Using these identities we transform the equation as
$\int {\sqrt {1{\text{ - sinx}}} {\text{dx = }}}$ $\int {\sqrt {{\text{si}}{{\text{n}}^2}\dfrac{{\text{x}}}{2}{\text{ + co}}{{\text{s}}^2}\dfrac{{\text{x}}}{2}{\text{ - sinx}}} {\text{dx}}}$
⟹$\int {\sqrt {1{\text{ - sinx}}} {\text{dx = }}}$ $\int {\sqrt {{\text{si}}{{\text{n}}^2}\dfrac{{\text{x}}}{2}{\text{ + co}}{{\text{s}}^2}\dfrac{{\text{x}}}{2}{\text{ - 2sin}}\dfrac{{\text{x}}}{2}{\text{cos}}\dfrac{{\text{x}}}{2}} {\text{dx}}}$
= $\int {\sqrt {{{\left( {{\text{sin}}\dfrac{{\text{x}}}{2}{\text{ + cos}}\dfrac{{\text{x}}}{2}} \right)}^2}} } {\text{dx}}$
= $\int {\left( {{\text{sin}}\dfrac{{\text{x}}}{2}{\text{ + cos}}\dfrac{{\text{x}}}{2}} \right)} {\text{dx}}$
= $\int {{\text{sin}}\dfrac{{\text{x}}}{2}{\text{dx + }}\int {{\text{cos}}\dfrac{{\text{x}}}{2}{\text{dx}}} }$ -- (1)
We know$\int {{\text{sinx}}} = - {\text{cosx, hence }}\int {{\text{sin}}\dfrac{{\text{x}}}{2}} = \dfrac{{ - {\text{cos}}\dfrac{{\text{x}}}{2}}}{{\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {\dfrac{{\text{x}}}{2}} \right)}} = \dfrac{{ - {\text{cos}}\dfrac{{\text{x}}}{2}}}{{\dfrac{1}{2}}}$.
Similarly we know, $\int {{\text{cosx}}} = {\text{sinx, hence }}\int {{\text{cos}}\dfrac{{\text{x}}}{2}} = \dfrac{{{\text{sin}}\dfrac{{\text{x}}}{2}}}{{\dfrac{{\text{d}}}{{{\text{dx}}}}\left( {\dfrac{{\text{x}}}{2}} \right)}} = \dfrac{{{\text{sin}}\dfrac{{\text{x}}}{2}}}{{\dfrac{1}{2}}}$
Hence, equation (1) becomes,
⟹$\dfrac{{ - {\text{cos}}\dfrac{{\text{x}}}{2}}}{{\dfrac{1}{2}}}$+$\dfrac{{{\text{sin}}\dfrac{{\text{x}}}{2}}}{{\dfrac{1}{2}}}$+C
⟹$- 2{\text{cos}}\dfrac{{\text{x}}}{2} + 2{\text{sin}}\dfrac{{\text{x}}}{2} + {\text{C}}$ (where C is the integration constant).
$\Rightarrow 2\left[ {{\text{sin}}\dfrac{{\text{x}}}{2} - {\text{cos}}\dfrac{{\text{x}}}{2}} \right] + {\text{C}} \\\ {\text{We can write this as,}} \\\ \Rightarrow {\text{2}}\sqrt {{{\left( {{\text{sin}}\dfrac{{\text{x}}}{2} - {\text{cos}}\dfrac{{\text{x}}}{2}} \right)}^2}} + {\text{C}} \\\ \Rightarrow {\text{2}}\sqrt {\left( {{\text{si}}{{\text{n}}^2}\dfrac{{\text{x}}}{2} + {\text{co}}{{\text{s}}^2}\dfrac{{\text{x}}}{2} - 2{\text{sin}}\dfrac{{\text{x}}}{2}{\text{cos}}\dfrac{{\text{x}}}{2}} \right)} + {\text{C}} \\\$ -- We used ${\left( {{\text{a - b}}} \right)^2} = {{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{ - 2ab}}$
We know, ${\text{si}}{{\text{n}}^2}\theta + {\text{co}}{{\text{s}}^2}\theta = 1$, ${\text{sin2}}\theta {\text{ = 2sin}}\dfrac{\theta }{2}{\text{cos}}\dfrac{\theta }{2}$. Hence the equation becomes,
$\Rightarrow 2\sqrt {1 - {\text{sinx}}} + {\text{C}}$
Therefore$\int {\sqrt {1{\text{ - sinx}}} {\text{dx = }}}$ $2\sqrt {1 - {\text{sinx}}} + {\text{C}}$. Hence Option B is the correct answer.

Note – In order to solve this kind of problems the key is to represent the given integral as a sum of the results of two or more integral values as we know the integral values of sin and cos functions, also as the angle inside the functions is$\dfrac{{\text{x}}}{2}$, we should be careful while applying the integration formula. Having adequate knowledge in the trigonometric identities of sin and cos functions like${\text{si}}{{\text{n}}^2}\theta + {\text{co}}{{\text{s}}^2}\theta = 1$, ${\text{sin2}}\theta {\text{ = 2sin}}\dfrac{\theta }{2}{\text{cos}}\dfrac{\theta }{2}$is required. Basic algebraic formulae like ${\left( {{\text{a + b}}} \right)^2} = {{\text{a}}^2} + {{\text{b}}^2}{\text{ + 2ab}}$ and ${\left( {{\text{a - b}}} \right)^2} = {{\text{a}}^2}{\text{ + }}{{\text{b}}^2}{\text{ - 2ab}}$ are also used.