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Question: If \(I_{1} = \int_{0}^{3\pi}{f(\cos^{2}x)dx}\) and \(I_{2} = \int_{0}^{\pi}{f(\cos^{2}x)dx}\) then...

If I1=∫03πf(cos⁡2x)dxI_{1} = \int_{0}^{3\pi}{f(\cos^{2}x)dx} and I2=∫0πf(cos⁡2x)dxI_{2} = \int_{0}^{\pi}{f(\cos^{2}x)dx} then

A

I1=I2I_{1} = I_{2}

B

I1=2I2I_{1} = 2I_{2}

C

I1=3I2I_{1} = 3I_{2}

D

I1=4I2I_{1} = 4I_{2}

Answer

I1=3I2I_{1} = 3I_{2}

Explanation

Solution

f(cos⁡2x)=f(cos⁡2(3π−x))f(\cos^{2}x) = f(\cos^{2}(3\pi - x))

∴ I1=3∫0πf(cos⁡2x)dxI_{1} = 3\int_{0}^{\pi}{f(\cos^{2}x)dx} ⇒ I1=3I2I_{1} = 3I_{2}

(7) ∫abf(x)dx=∫abf(a+b−x)dx\int_{a}^{b}{f(x)}dx = \int_{a}^{b}{f(a + b - x)dx}

Note : ∫abf(x)dxf(x)+f(a+b−x)=12(b−a)\int_{a}^{b}{}\frac{f(x)dx}{f(x) + f(a + b - x)} = \frac{1}{2}(b - a)