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Evaluate \[\int {\dfrac{{\cos x - \sin x}}{{\cos x - \sin x}} \cdot \left( {2 + 2\sin 2x} \right)dx} \]?

Last updated date: 13th Jul 2024
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Hint: Here in this question given an Indefinite integral, we have to find the integrated value of a given trigonometric function. This can be solved, by substituting some trigonometric function and algebraic identities and later integrated by using the standard trigonometric formula of integration. And by further simplification we get the required solution.

Complete step by step solution:
Integration is the inverse process of differentiation. An integral which is not having any upper and lower limit is known as an indefinite integral.
Consider the given function.
\[ \Rightarrow \int {\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}} \cdot \left( {2 + 2\sin 2x} \right)dx} \]--------(2)
Take 2 as common in the function, then
\[ \Rightarrow \int {\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}} \cdot 2\left( {1 + \sin 2x} \right)dx} \]
Where, 2 as constant take it out from the integration.
\[ \Rightarrow 2\int {\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}} \cdot \left( {1 + \sin 2x} \right)dx} \]
Now, by using a algebraic identity \[{\sin ^2}x + {\cos ^2}x = 1\] and the double angle formula \[\sin 2x = 2\sin x\cos x\] on substituting, we have
 \[ \Rightarrow 2\int {\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}} \cdot \left( {{{\sin }^2}x + {{\cos }^2}x + 2\sin x\cos x} \right)dx} \]---------(2)
The term \[\left( {{{\sin }^2}x + {{\cos }^2}x + 2\sin x\cos x} \right)\] similar like a algebraic identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\],
Here \[a = \sin x\] and \[b = \cos x\], then equation (2) becomes
\[ \Rightarrow 2\int {\dfrac{{\cos x - \sin x}}{{\cos x + \sin x}} \cdot {{\left( {\sin x + \cos x} \right)}^2}dx} \]
On cancelling like terms \['\cos x + \sin x'\] on both numerator and denominator, we have
\[ \Rightarrow 2\int {\left( {\cos x - \sin x} \right) \cdot \left( {\sin x + \cos x} \right)dx} \]---------(3)
The function \[\left( {\cos x - \sin x} \right) \cdot \left( {\sin x + \cos x} \right)\] similar like a algebraic identity \[{a^2} - {b^2} = \left( {a + b} \right)\left( {a - b} \right)\],
Here \[a = \cos x\] and \[b = \sin x\], then equation (3) becomes
\[ \Rightarrow 2\int {\cos {x^2} - {{\sin }^2}xdx} \]-------(4)
Again, by using a double angle formula of trigonometry i.e., \[\cos 2x = \cos {x^2} - {\sin ^2}x\], then equation (4) becomes.
\[ \Rightarrow 2\int {\cos 2xdx} \]
On integrating with respect to \[x\], using a standard formula \[\int {\cos dx} = \sin x + c\], then we have
\[ \Rightarrow 2\dfrac{{\sin 2x}}{2} + C\]
On simplification, we get
\[ \Rightarrow \sin 2x + C\]
Where, \[C\] is an integrating constant.
Hence, it’s a required solution.
So, the correct answer is “$\sin 2x + C$”.

Note: By simplifying the question using the substitution of different trigonometric formulas we can integrate the given function easily. If we apply integration directly it may be complicated to solve further. So, simplification is needed. We must know the differentiation and integration formulas. The standard integration formulas for the trigonometric ratios must know.