Question

# ∫2π−2πsin6x(sin6x+cos6x)(1+e−x)dx=

A
2π
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B
π
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C
π2
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D
4π
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Solution

## The correct option is B πThe given problem can be easily solved if we know the properties of definite integration. ∫a−af(x)dx=∫a0(f(x)+f(−x))dx Using this, we get - ∫2π0sin6x(sin6x+cos6x)(1+e−x)+sin6x(sin6x+cos6x)(1+ex)dx = ∫2π0sin6x(sin6x+cos6x)dx We also know another property which is - ∫2a0f(x)dx=2∫a0f(x)dx if f(2a-x) = f(x) Using this property we get - = 2∫π0sin6x(sin6x+cos6x)dx Again using this property again we can write - 4∫π20sin6x(sin6x+cos6x)dx We know an another property which is - ∫baf(x)dx=∫a0f(a+b−x)dx Let I = ∫π20sin6x(sin6x+cos6x)dx then I = ∫π20cos6x(cos6x+sin6x)dx Adding both we get - 2I = ∫π201.dx 2I = π2 I = π4 So, the value of the integral which is asked to find will be = 4.I = π

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