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.