If I1=∫03πf(cos2x)dx and I2=∫0πf(cos2x)dx, then
I1=I2
3I1=I2
I1=3I2
I1=5I2
Explanation for correct option:
Find the relation:
Given that,
I1=∫03πf(cos2x)dx and
I2=∫0πf(cos2x)dx
Now,
I1=∫03πf(cos²x)dx=3∫0πf(cos²x)dx[∵∫0nTf(x)dx=n∫0Tf(x)dx,f(x)isperiodicwithT]=3I2
Hence, the correct option is C.