Answer: [tex]\sin x+6[/tex]
[tex]-\cos x+6x+5-3\pi[/tex]
Step-by-step explanation:
Given
[tex]f''(x)=\cos x[/tex]
Integrating the equation
[tex]\Rightarrow f''(x)=\dfrac{d^2y}{dx^2}=\cos x\\\\\Rightarrow \int \frac{\mathrm{d^2} y}{\mathrm{d} x^2}=\int \cos x\\\\\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x}=\sin x+c[/tex]
Put [tex]x=\frac{\pi}{2}\ \text{and}\ \frac{dy}{dx}=7[/tex]
[tex]\Rightarrow 7=\sin \frac{\pi }{2}+c\\\Rightarrow c=7-1\\\Rightarrow c=6[/tex]
[tex]\Rightarrow \dfrac{dy}{dx}=\sin x+6[/tex]
again integrating both sides of the equation
[tex]\Rightarrow \int \frac{\mathrm{d} y}{\mathrm{d} x}=\int (\sin x+6)dx\\\\\Rightarrow y=-\cos x+6x+c_1[/tex]
Put [tex]x=\frac{\pi}{2}\ \text{and}\ y=5[/tex]
[tex]\Rightarrow 5=-\cos \frac{\pi}{2}+6(\frac{\pi}{2})+c_1\\\Rightarrow c_1=5-3\pi[/tex]
[tex]\therefore f(x)=y=-\cos x+6x+5-3\pi[/tex]
