Question

If $\int \left(\sin 2x-\cos 2x\right)dx={\displaystyle \frac{1}{\sqrt{2}}}\sin \left(2x-a\right)+b$ then
A. $a={\displaystyle \frac{5\pi }{4}},b\in R$
B.$a={\displaystyle \frac{-5\pi }{4}},b\in R$
C. $a={\displaystyle \frac{\pi }{4}},b\in R$
D.None of these

Answer 1

Hint: We consider the left-hand side of the expression and try to solve it by integrating step by step. In order to achieve the final expression, we need to do some adjustment operations in the expression. Once the final form of the expression is obtained, we can easily determine the required values of a and b.

Formula Used:
In order to solve this expression some important formulas are used that is,
$\int \sin axdx={\displaystyle \frac{-\cos ax}{a}}+c$
$\int \cos axdx={\displaystyle \frac{\sin ax}{a}}+c$
Where, $c$ is the constant of integration.
$\left(\sin A\cos B+\cos A\sin B\right)=\sin \left(A+B\right)$


Complete step by step solution
Given:
The expression $\int \left(\sin 2x-\cos 2x\right)dx={\displaystyle \frac{1}{\sqrt{2}}}\sin \left(2x-a\right)+b$.
Now, we will be solving the left hand side of the above expression, we have,$$\begin{array}{c}\int \left(\sin 2x-\cos 2x\right)dx=\left({\displaystyle \frac{-\cos 2x}{2}}\right)-\left({\displaystyle \frac{\sin 2x}{2}}\right)+c\\ ={\displaystyle \frac{-1}{2}}\left(\cos 2x+\sin 2x\right)+c\end{array}$$
On rearranging the term by multiplying and dividing it with $\sqrt{2}$ we get,$$\begin{array}{c}\int \left(\sin 2x-\cos 2x\right)dx={\displaystyle \frac{1}{\sqrt{2}}}\left[\cos 2x\left({\displaystyle \frac{-1}{\sqrt{2}}}\right)+\sin 2x\left({\displaystyle \frac{-1}{\sqrt{2}}}\right)\right]+c\\ ={\displaystyle \frac{1}{\sqrt{2}}}\left(\sin 2x\cos a+\cos 2x\sin a\right)+c\end{array}$$
Where, we know that $\cos a={\displaystyle \frac{-1}{\sqrt{2}}}$ and $\sin a={\displaystyle \frac{-1}{\sqrt{2}}}$.
Now, we know that the term $$\left(\sin 2x\cos a+\cos 2x\sin a\right)$$ can be written as the formula,$$\left(\sin 2x\cos a+\cos 2x\sin a\right)=\sin \left(2x+a\right)$$
We know that the values for both $\cos$ and $\sin$ are negative in the third quadrant so, we get,$\begin{array}{c}\cos a=\sin a\\ ={\displaystyle \frac{-1}{\sqrt{2}}}\end{array}$
So, the value of a will be equal to,
$a=-{\displaystyle \frac{5\pi }{4}}$
Then finally, the integral reduces to,
$\int \left(\sin 2x-\cos 2x\right)dx={\displaystyle \frac{1}{\sqrt{2}}}\sin \left(2x+{\displaystyle \frac{5\pi }{4}}\right)+c$
Also, the above expression can also be written as:
$\int \left(\sin 2x-\cos 2x\right)dx={\displaystyle \frac{1}{\sqrt{2}}}\sin \left(2x-a\right)+b$
Where, $a$ and $b$ are arbitrary constants, whose values are $a={\displaystyle \frac{-5\pi }{4}}$ and $b\in R$ (b belongs to real number).
The correct option is (B) that is $a={\displaystyle \frac{-5\pi }{4}},b\in R$.


Note: We have to reduce the expression in simplest form as given in the question that is why we had to do some rearrangements in the expression by multiplying-dividing with a number and using trigonometric relations.