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Question: If the integral of \(\left( \sin 2x-\cos 2x \right)dx=\sqrt{\dfrac{1}{2}}.\sin \left( 2x-a \right)+b...

If the integral of (sin⁡2x−cos⁡2x)dx=12.sin⁡(2x−a)+b\left( \sin 2x-\cos 2x \right)dx=\sqrt{\dfrac{1}{2}}.\sin \left( 2x-a \right)+b, then find aa and bb.

Explanation

Solution

For solving this question you should know about the integration of trigonometric functions. The integration of trigonometric functions can be solved by a general process of integrating. But in this problem both the solutions are given and asked for the values of variables aa and bb, so we will solve this and then by comparing both the answers we will get the values of variables aa and bb.

Complete step by step answer:
According to our question it is asked to us to find the values of aa and bbif the integral of (sin⁡2x−cos⁡2x)dx=12.sin⁡(2x−a)+b\left( \sin 2x-\cos 2x \right)dx=\sqrt{\dfrac{1}{2}}.\sin \left( 2x-a \right)+b. So, if we solve to the integration of sin⁡2x−cos⁡2x\sin 2x-\cos 2x, then we find that:
⇒∫(sin⁡2x−cos⁡2x)dx=12.sin⁡(2x−a)+b ⇒∫(sin⁡2x)dx−∫cos⁡2x.dx=12.sin⁡(2x−a)+b \begin{aligned} & \Rightarrow \int{\left( \sin 2x-\cos 2x \right)}dx=\dfrac{1}{\sqrt{2}}.\sin \left( 2x-a \right)+b \\\ & \Rightarrow \int{\left( \sin 2x \right)dx-\int{\cos 2x.dx=\dfrac{1}{\sqrt{2}}.\sin \left( 2x-a \right)+b}} \\\ \end{aligned}
So, by using the further basic formulas of integration, we will get,
⇒(−12cos⁡2x−12sin⁡2x)+c=12sin⁡(2x−a)+b ⇒sin⁡2x.cos⁡a2−cos⁡2x.sin⁡a2=(−12cos⁡2x−12sin⁡2x) ⇒cos⁡a=−12,sin⁡a=12 \begin{aligned} & \Rightarrow \left( -\dfrac{1}{2}{{\cos }{2}}x-\dfrac{1}{2}{{\sin }{2}}x \right)+c=\dfrac{1}{\sqrt{2}}\sin \left( 2x-a \right)+b \\\ & \Rightarrow \dfrac{\sin 2x.\cos a}{\sqrt{2}}-\dfrac{\cos 2x.\sin a}{\sqrt{2}}=\left( -\dfrac{1}{2}{{\cos }{2}}x-\dfrac{1}{2}{{\sin }{2}}x \right) \\\ & \Rightarrow \cos a=-\dfrac{1}{\sqrt{2}}, \sin a= \dfrac{1}{\sqrt{2}} \\\ \end{aligned}
So, if we take the inverse on both sides then we can get the value of a=π4a=\dfrac{\pi }{4}. And by taking the LHS as 12sin⁡(2x+5π4)+c\dfrac{1}{\sqrt{2}}\sin \left( 2x+\dfrac{5\pi }{4} \right)+c , we will get as follows,
=12sin⁡(2x−a)+b=\dfrac{1}{\sqrt{2}}\sin \left( 2x-a \right)+b
So, the values of aa and bbare −5π4-\dfrac{5\pi }{4} and constant respectively.

Note: In a right angled triangle the hypotenuse, the base (adjacent) and the perpendicular (opposite), that is the three sides of a right-angled triangle are from where the trigonometric ratios are derived. There are three primary trigonometric ratios in maths which are also known as trigonometric identities. And here these are all used to solve these questions. The formulas are as follows: cos⁡θ=adjacenthypotenuse\cos \theta =\dfrac{\text{adjacent}}{\text{hypotenuse}} and sin⁡θ=oppositehypotenuse\sin \theta =\dfrac{\text{opposite}}{\text{hypotenuse}}.