Question

The value of integral $\int {\dfrac{{\sin x + \cos x}}{{\sqrt {1 + \sin 2x} }}dx} $is
1). $\sqrt {\left( {1 + \cos 2x} \right)} + c$
2). $\sqrt {x - c} $
3). $x + c$
4). $\sqrt {\left( {x + 2x} \right)} + c$

Answer 1

Hint: We have to simplify this function using the appropriate trigonometric properties and algebraic formulas. First, we have to simplify the function before integrating and then, after simplifying we have to integrate the function with respect to x and after that we will get the required answer.

Complete step-by-step solution:
$ = \int {\dfrac{{\sin x + \cos x}}{{\sqrt {1 + \sin 2x} }}dx} $
Using trigonometric properties sin2 x + cos2 x = 1 and sin 2x = 2 sin x cos x, we will rewrite the denominator of this function.
$ = \int {\dfrac{{\sin x + \cos x}}{{\sqrt {{{\sin }^2}x + {{\cos }^2}x + 2\sin x\cos x} }}dx} $
The terms in the denominator make a formula which is a2 + b2 + 2ab = (a + b)2.
So, we will again rewrite the denominator of the equation by using this formula.
$ = \int {\dfrac{{\sin x + \cos x}}{{\sqrt {{{\left( {\sin x + \cos x} \right)}^2}} }}dx} $
$ = \int {\dfrac{{\sin x + \cos x}}{{\sin x + \cos x}}dx} $
Now, there are the same terms in numerator and denominator, so by solving them we will get 1.
$ = \int {dx} $
Integration of dx with respect to x is x. so,
$ = x + c$
The value after integrating the function is $x + c$.
So, option (3) is the correct answer.

Note: Integration is the calculation of integrals. It is a reverse of differentiation, where we reduce the function into parts. In this case above we are solving an indefinite integral (integral which has no upper or lower limit).
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 trigonometric properties and algebraic formulas. Some identities are used in this problem but students should remember all of them.