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Question

Mathematics Question on Integrals of Some Particular Functions

If I1=03πf(cos2x)dxI_{1} = \int\limits^{3\pi}_{0}f\left(\cos^{2}\,x\right)dx and I2=0πf(cos2x)dxI_{2} = \int\limits^{\pi}_{0} f\left(\cos^{2}\,x\right)dx, then

A

I1=I2I_1 = I_2

B

3I1=I23I_1 = I_2

C

I1=3I2I_1 = 3I_2

D

I1=5I2I_1 = 5I_2

Answer

I1=3I2I_1 = 3I_2

Explanation

Solution

I1=30πf(cos2x)dx=3I2I_{1}=3 \int\limits_{0}^{\pi} f\left(\cos ^{2} x\right) d x=3 I_{2}
[[\because period is π]\pi]