First slide

Introduction to integration

Question

$\int (\sin 2 x - \cos 2 x) dx = \frac{1}{\sqrt{2}} \sin (2 x - a) + b,$ then

Moderate

A

$a = \frac{5 π}{4}, b \in R$
B

$a = - \frac{5 π}{4}, b \in R$
C

$a = \frac{π}{4}, b \in R$
D

None of these

Solution

$\begin{aligned} \int (\sin 2 x - \cos 2 x) dx = \frac{1}{\sqrt{2}} \sin (2 x - a) + b \\ \Rightarrow \sqrt{2} \int (\frac{1}{\sqrt{2}} \sin 2 x - \frac{1}{\sqrt{2}} \cos 2 x) dx = \frac{1}{\sqrt{2}} \sin (2 x - a) + b \\ \Rightarrow - \sqrt{2} \int (\frac{1}{\sqrt{2}} \cos 2 x - \frac{1}{\sqrt{2}} \sin 2 x) dx = \frac{1}{\sqrt{2}} \sin (2 x - a) + b \\ \Rightarrow - \sqrt{2} \int \cos (2 x + \frac{π}{4}) dx = \frac{1}{\sqrt{2}} \sin (2 x - a) + b \\ \Rightarrow - \frac{\sqrt{2}}{2} \sin (2 x + \frac{π}{4}) + C = \frac{1}{\sqrt{2}} \sin (2 x - a) + b \\ \Rightarrow \frac{1}{\sqrt{2}} \sin (2 x + \frac{5 π}{4}) + C = \frac{1}{\sqrt{2}} \sin (2 x - a) + b \\ \Rightarrow a = - \frac{5 π}{4}, b \in R \end{aligned}$

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