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Question Number 82244 by M±th+et£s last updated on 19/Feb/20

$$findthefunctionoffwhenthis$$$$functioncontinueatinterval[-\mathrm{\infty },0]$$$${\int }_{-x^2}^{0}f\left(t\right)dt=\frac{d}{dx}[x(1-sin\left(\pi x\right)]$$

Commented by mr W last updated on 19/Feb/20

$${\int }_{-x^2}^{0}f\left(t\right)dt=\frac{d}{dx}[x(1-sin\left(\pi x\right)]$$$${\int }_{-x^2}^{0}f\left(t\right)dt=1-sin\left(\pi x\right)-\pi x\cos \left(\pi x\right)$$$$2xf(-x^2)=-2\pi \cos \left(\pi x\right)+{\pi }^{2}x\sin \left(\pi x\right)$$$$f(-x^2)=-\frac{\pi }{x}\cos \left(\pi x\right)+\frac{{\pi }^2}{2}\sin \left(\pi x\right)$$$$t=-x^2\Rightarrow x=\pm \sqrt{-t}$$$$f\left(t\right)=\mp \frac{\pi }{\sqrt{-t}}\cos \left(\pi \sqrt{-t}\right)\pm \frac{{\pi }^2}{2}\sin \left(\pi \sqrt{-t}\right)$$$$\Rightarrow f\left(x\right)=\mp \frac{\pi }{\sqrt{-x}}\cos \left(\pi \sqrt{-x}\right)\pm \frac{{\pi }^2}{2}\sin \left(\pi \sqrt{-x}\right)$$

Commented by M±th+et£s last updated on 19/Feb/20

$$godblessyousir$$

Answered by mind is power last updated on 19/Feb/20

$$\Rightarrow 2xf(-x^2)=\frac{d^2}{{dx}^2}[x-xsin\left(\pi x\right)]$$$$\Rightarrow f(-x^2)=\frac{1}{2x}\frac{d}{dx}[-sin\left(\pi x\right)-\pi xcos\left(\pi x\right)]$$$$\Rightarrow f(-x^2)=\frac{1}{2x}.[-2\pi cos\left(\pi x\right)+{\pi }^{2}xsin\left(\pi x\right)]$$$$f(-x^2)=-\frac{\pi cos\left(\pi x\right)}{x}+\frac{{\pi }^2}{2}sin\left(\pi x\right)$$$$-x^2=y\Rightarrow x=\underset{-}{+}\sqrt{-y}$$$$wegot2solution$$$$f\left(y\right)=-\frac{\pi cos\left(\pi \sqrt{-y}\right)}{\sqrt{-y}}+\frac{{\pi }^{2}sin\left(\pi \sqrt{-y}\right)}{2}y\ne 0$$$$f\left(0\right)=0$$$$$$$$and-f$$

Commented by M±th+et£s last updated on 19/Feb/20

$$godblessyousir$$

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