First slide

Methods of integration

Question

$If \int \frac{\sin (\frac{π}{4} - x)}{2 + \sin 2 x} d x = A \tan^{- 1} (f (x)) + B, f (0) = 1, where \underset{}{\underset{}{A}, B} are constants, then the range of A . f (x) is$

Moderate

A

$[- 1, 1]$
B

$[- \sqrt{2}, \sqrt{2}]$
C

$[0, 1]$
D

$[- \sqrt{2}, 0]$

Solution

$\begin{array}{l} Consider \int \frac{\sin (\frac{π}{4} - x)}{2 + \sin 2 x} d x \\ = \int \frac{\frac{1}{\sqrt{2}} (\cos x - \sin x)}{2 + \sin 2 x} d x \\ Put \sin x + \cos x = t \end{array}$
$\begin{array}{l} \Rightarrow (\cos x - \sin x) d x = d t \\ And \sin 2 x = t^{2} - 1 \\ = \frac{1}{\sqrt{2}} \int \frac{d t}{t^{2} + 1} \\ = \frac{1}{\sqrt{2}} \tan^{- 1} t + c \\ = \frac{1}{\sqrt{2}} \tan^{- 1} (\sin x + \cos x) + c \\ = A \tan^{- 1} (f (x)) + B (given) \\ ∴ A = \frac{1}{\sqrt{2}}, f (x) = \sin x + \cos x \\ Now, A f (x) = \frac{1}{\sqrt{2}} (\sin x + \cos x) \\ = \sin (x + \frac{π}{4}) \\ ∴ range of A f (x) is [- 1, 1] \end{array}$

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