Answer
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Hint- In order to solve the given integral, first we will simplify the given trigonometric term with the help of trigonometric identity to bring it in simpler form of sine and cosine before integrating it. We will use these two identities-
$
{\sin ^2}x + {\cos ^2}x = 1 \\
\sin 2x = 2\sin x\cos x \\
$
Complete step-by-step answer:
We have to find out the value of \[\int {\sqrt {1 - \sin 2x} \cdot dx} \]
Let us assume our function $f\left( x \right) = \sqrt {1 - \sin 2x} $
Now let us simplify the function using trigonometric identity.
$
{\sin ^2}x + {\cos ^2}x = 1 \\
\sin 2x = 2\sin x\cos x \\
$
Using the identities in the function we get:
\[
f\left( x \right) = \sqrt {\left( 1 \right) - \left( {\sin 2x} \right)} \\
f\left( x \right) = \sqrt {\left( {{{\sin }^2}x + {{\cos }^2}x} \right) - \left( {2\sin x\cos x} \right)} \\
\]
Now let us use algebraic identity to solve the term.
$\because {a^2} + {b^2} - 2ab = {\left( {a - b} \right)^2}$
Using the algebraic identity in the given function we get
$
\therefore f\left( x \right) = \sqrt {{{\left( {\sin x - \cos x} \right)}^2}} \\
\Rightarrow f\left( x \right) = \left( {\sin x - \cos x} \right) \\
$
So, now we will find the integration
$
\Rightarrow \int {f\left( x \right)dx} = \int {\left( {\sin x - \cos x} \right)dx} \\
\Rightarrow \int {f\left( x \right)dx} = \int {\left( {\sin x} \right)dx} - \int {\left( {\cos x} \right)dx} \\
$
As we know the basic rules for integration of sine and cosine term are given by:
$
\int {\left( {\sin x} \right)dx} = - \cos x \\
\int {\left( {\cos x} \right)dx} = \sin x \\
$
Using the same in the integral we get:
$
\Rightarrow \int {f\left( x \right)dx} = \int {\left( {\sin x} \right)dx} - \int {\left( {\cos x} \right)dx} \\
\therefore \int {f\left( x \right)dx} = - \cos x - \sin x + c \\
$
Or \[\int {\sqrt {1 - \sin 2x} \cdot dx} = - \cos x - \sin x + c\]
Hence, the result of integration is \[\int {\sqrt {1 - \sin 2x} \cdot dx} = - \cos x - \sin x + c\]
Note- These types of problems cannot be solved directly, the basic idea for solving such problems is to simplify the term to be integrated before integration by the use of algebraic as well as trigonometric identities. Students must remember such identities, some of them are mentioned above.
$
{\sin ^2}x + {\cos ^2}x = 1 \\
\sin 2x = 2\sin x\cos x \\
$
Complete step-by-step answer:
We have to find out the value of \[\int {\sqrt {1 - \sin 2x} \cdot dx} \]
Let us assume our function $f\left( x \right) = \sqrt {1 - \sin 2x} $
Now let us simplify the function using trigonometric identity.
$
{\sin ^2}x + {\cos ^2}x = 1 \\
\sin 2x = 2\sin x\cos x \\
$
Using the identities in the function we get:
\[
f\left( x \right) = \sqrt {\left( 1 \right) - \left( {\sin 2x} \right)} \\
f\left( x \right) = \sqrt {\left( {{{\sin }^2}x + {{\cos }^2}x} \right) - \left( {2\sin x\cos x} \right)} \\
\]
Now let us use algebraic identity to solve the term.
$\because {a^2} + {b^2} - 2ab = {\left( {a - b} \right)^2}$
Using the algebraic identity in the given function we get
$
\therefore f\left( x \right) = \sqrt {{{\left( {\sin x - \cos x} \right)}^2}} \\
\Rightarrow f\left( x \right) = \left( {\sin x - \cos x} \right) \\
$
So, now we will find the integration
$
\Rightarrow \int {f\left( x \right)dx} = \int {\left( {\sin x - \cos x} \right)dx} \\
\Rightarrow \int {f\left( x \right)dx} = \int {\left( {\sin x} \right)dx} - \int {\left( {\cos x} \right)dx} \\
$
As we know the basic rules for integration of sine and cosine term are given by:
$
\int {\left( {\sin x} \right)dx} = - \cos x \\
\int {\left( {\cos x} \right)dx} = \sin x \\
$
Using the same in the integral we get:
$
\Rightarrow \int {f\left( x \right)dx} = \int {\left( {\sin x} \right)dx} - \int {\left( {\cos x} \right)dx} \\
\therefore \int {f\left( x \right)dx} = - \cos x - \sin x + c \\
$
Or \[\int {\sqrt {1 - \sin 2x} \cdot dx} = - \cos x - \sin x + c\]
Hence, the result of integration is \[\int {\sqrt {1 - \sin 2x} \cdot dx} = - \cos x - \sin x + c\]
Note- These types of problems cannot be solved directly, the basic idea for solving such problems is to simplify the term to be integrated before integration by the use of algebraic as well as trigonometric identities. Students must remember such identities, some of them are mentioned above.
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