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Sagot :
ANSWER:
0.5647
STEP-BY-STEP EXPLANATION:
We have the following function:
[tex]f(x)=\cos 3x\cdot\cos 2x[/tex]The primitive function is the definite integral of the function, evaluated from 0 to 2.
Therefore, we calculate the integral and evaluate:
[tex]\begin{gathered} \int ^{\sqrt{2}}_0\mleft(\cos 3x\cdot\cos 2x\mright)dx \\ \text{ Let's rewrite the function as follows:} \\ \cos \: \: 3x\cdot\cos \: \: 2x=\frac{\cos(3x+2x)+\cos(3x-2x)}{2} \\ \cos \: \: 3x\cdot\cos \: \: 2x=\frac{1}{2}\cdot(\cos 5x+\cos x) \\ \int ^{\sqrt[]{2}}_0(\frac{1}{2}\cdot(\cos 5x+\cos x))dx=\frac{1}{2}\int ^{\sqrt[]{2}}_0\cos 5xdx+\frac{1}{2}\int ^{\sqrt[]{2}}_0\cos xdx \\ \text{ we integrate:} \\ \int ^{\sqrt[]{2}}_0\cos 5xdx=\frac{1}{5}\cdot\sin (5\sqrt[]{2})-\frac{1}{5}\cdot\sin (5\cdot0)=0.1417-0=0.1417 \\ \int ^{\sqrt[]{2}}_0\cos xdx=\sin (\sqrt[]{2})-\sin (0)=0.98776-0=0.98776 \\ \int ^{\sqrt[]{2}}_0(\cos 3x\cdot\cos 2x)dx=\frac{1}{2}(0.1417+0.98776) \\ \int ^{\sqrt[]{2}}_0(\cos 3x\cdot\cos 2x)dx=0.5647 \end{gathered}[/tex]The result is 0.5647
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