The value of ∫-π2π2log2-sinθ2+sinθdθ is
0
1
2
none of these
Explanation for the correct option:
Compute the required value:
Given: ∫-π2π2log2-sinθ2+sinθdx
⇒f(θ)=log2-sinθ2+sinθ
Put θ=-θ
f(-θ)=log2-sin(-θ)2+sin(-θ)=log2+sin(θ)2-sin(θ)=log2-sin(θ)2+sin(θ)-1=-log2-sin(θ)2+sin(θ)f(-θ)=-f(θ)
So, f(θ)=log2-sin(θ)2+sin(θ) is an odd function
∫-π2π2log2-sinθ2+sinθdθ is equal to 0.
Hence, option A is the correct answer.
The value of ∫x + 5(x − 2)2dx is :