If ∫$$sin$$⁡π$$4$$−$$x2+$$$$sin$$⁡$$2xdx=Atan$$−$$1$$⁡$$fx+B$$,$$f0=1$$, where A ,B are constants, then the range of A.fx is

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If ∫$$sin$$⁡π$$4$$−$$x2+$$$$sin$$⁡$$2xdx=Atan$$−$$1$$⁡$$fx+B$$,$$f0=1$$, where A ,B  are constants, then the range  of A.fx is

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Consider ∫$$sin$$⁡π$$4$$−$$x2+$$$$sin$$⁡$$2xdx=$$∫$$12($$$$cos$$⁡x−$$sin$$⁡$$x)2+$$$$sin$$⁡$$2xdx$$ Put $$sin$$⁡$$x+$$$$cos$$⁡$$x=t$$⇒$$($$$$cos$$⁡x−$$sin$$⁡$$x)dx=dt$$ And $$sin$$⁡$$2x=t2$$−$$1=12$$∫$$dtt2+1=12tan$$−$$1$$⁡$$t+c=12tan$$−$$1$$⁡$$($$$$sin$$⁡$$x+$$$$cos$$⁡$$x)+c=Atan$$−$$1$$⁡(f(x))+B$$ $$($$ given $$)∴$$A=12$$,$$f(x)=$$$$sin$$⁡$$x+$$$$cos$$⁡x Now, A $$f(x)=12($$$$sin$$⁡$$x+$$$$cos$$⁡$$x)=$$$$sin$$⁡$$x+$$π$$4$$∴ range of $$Af(x)$$ is [−$$1$$,$$1$$]

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

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If ∫sin⁡π4−x2+sin⁡2xdx=Atan−1⁡fx+B,f0=1, where A ,B are constants, then the range of A.fx is

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