$\int\limits_0^1ln(\sin\frac{πx}{2})d

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Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Definite Integration

Question:

$\int\limits_0^1ln(\sin\frac{πx}{2})dx$ is equal to:

Options:

$ln\, 2$

$-ln\, 2$

$\pi\,ln\, 2$

$-\pi\,ln\, 2$

Correct Answer:

$-ln\, 2$

Explanation:

$\int\limits_0^1ln(\sin\frac{πx}{2})dx$ ⇒ Let $\frac{πx}{2}=x⇒dx=\frac{2}{π}dx$

$⇒I=\frac{2}{π}\int\limits_0^{π/2}ln\,\sin x\,dx=\frac{2}{π}(-\frac{π}{2}ln\,2)=-ln\, 2$