Question

The $ \int {\sqrt {1 + \sin 2x} dx} $ equals
A $ \sin x + \cos x + c $
B $ \sin x - \cos x + c $
C $ \cos x - \sin x + c $
D none of these

Answer 1

Hint: In order to determine the answer of above indefinite integral use the method of Integration by substitution by substituting $ 9 - {x^2} $ with $ {t^2} $ . After getting the result, always remember to substitute the original variable.
Formula:
 $
  \int {\sin xdx = - \cos x + C} \\
  \int {\cos xdx = \sin x + C} \;
  $

Complete step-by-step answer:
We are given integral $ \int {\sqrt {1 + \sin 2x} dx} $
 $ I = \int {\sqrt {1 + \sin 2x} dx} $ -(1)
Rewriting the above integral equation using the double angle formula of sine as $ \sin 2x = 2\sin x\cos x $ and the trigonometric identity which says $ 1 = {\sin ^2}x + {\cos ^2}x $ .we have
 $ I = \int {\sqrt {{{\sin }^2}x + {{\cos }^2}x + 2\sin x\cos x} dx} $
Now applying the identity $ {A^2} + {B^2} + 2AB = {\left( {A + B} \right)^2} $ by considering A as $ \sin x $ and B as $ \cos x $ , we get
 $ I = \int {\sqrt {{{\left( {\sin x + \cos x} \right)}^2}} } dx $
Simplifying further ,we get
 $ I = \int {\left( {\sin x + \cos x} \right)} dx $
As we know the integration gets distributed into the terms, we have
 $
  I = \int {\sin xdx} + \int {\cos xdx} \\
  I = - \cos x + \sin x + C \;
  $
Rearranging the terms
 $ I = \sin x - \cos x + C $ , here C is the constant of integration
Therefore , the integral of function $ \int {\sqrt {1 + \sin 2x} dx} = \sin x - \cos x + C $ .So Correct option is (B)
So, the correct answer is “Option B”.

Note: 1.Use standard formula carefully while evaluating the integrals.
2. Indefinite integral=Let $ f(x) $ be a function .Then the family of all its primitives (or antiderivatives) is called the indefinite integral of $ f(x) $ and is denoted by $ \int {f(x)} dx $
3.The symbol $ \int {f(x)dx} $ is read as the indefinite integral of $ f(x) $ with respect to x.
4.C is known as the constant of integration.