Question

If $I={\int }_0^{\pi /2}\ell n\left(\sin x\right)dx$ then ${\int }_{-\pi /4}^{\pi /4}\ell n(\sin x+\cos x)dx=$

• A. $\frac{l}{2}$
• B. $\frac{l}{4}$
• C. $\frac{l}{\sqrt{2}}$
• D. $I$

Answer 1

Answer: (A)

$\frac{l}{2}$

$A\left(say\right)={\int }_{\frac{-\pi }{4}}^{\frac{\pi }{4}}\ell n(\sin x+\cos x)dx\dots \left(1\right)$

Using $(a+b-x)$ property

$A={\int }_{\frac{-\pi }{4}}^{\frac{\pi }{4}}\ell n(-\sin x+\cos x)dx\dots ..\left(2\right)$

$\left(1\right)+\left(2\right)$

$2A={\int }_{\frac{-\pi }{4}}^{\frac{\pi }{4}}\ell n(-{\sin }^{2}x+co{s}^{2}x)dx$

$2A={\int }_{\frac{-\pi }{4}}^{\frac{\pi }{4}}\ell n\left(cos2x\right)dx$

$2x=t\to 2dx=dt$

$2A=\frac{1}{2}{\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\ell n\left(\cos t\right)dt$

$2A=\frac{1}{2}\times 2{\int }_0^{\frac{\pi }{2}}\ell n\left(\cos t\right)dt$

$2A={\int }_0^{\frac{\pi }{2}}\ell n\left(\sin t\right)dt=I$

$A=\frac{I}{2}$