Question
Question: Solve the following Integral. \[\int {\sqrt {1 - \sin 2x} \cdot dx} \]...
Solve the following Integral.
∫1−sin2x⋅dx
Solution
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-
sin2x+cos2x=1 sin2x=2sinxcosx
Complete step-by-step answer:
We have to find out the value of ∫1−sin2x⋅dx
Let us assume our function f(x)=1−sin2x
Now let us simplify the function using trigonometric identity.
sin2x+cos2x=1 sin2x=2sinxcosx
Using the identities in the function we get:
Now let us use algebraic identity to solve the term.
∵a2+b2−2ab=(a−b)2
Using the algebraic identity in the given function we get
∴f(x)=(sinx−cosx)2 ⇒f(x)=(sinx−cosx)
So, now we will find the integration
⇒∫f(x)dx=∫(sinx−cosx)dx ⇒∫f(x)dx=∫(sinx)dx−∫(cosx)dx
As we know the basic rules for integration of sine and cosine term are given by:
∫(sinx)dx=−cosx ∫(cosx)dx=sinx
Using the same in the integral we get:
⇒∫f(x)dx=∫(sinx)dx−∫(cosx)dx ∴∫f(x)dx=−cosx−sinx+c
Or ∫1−sin2x⋅dx=−cosx−sinx+c
Hence, the result of integration is ∫1−sin2x⋅dx=−cosx−sinx+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.