∫π/43π/4x1+sinxdx=

I=∫π/43π/4x1+sinxdx∴∫bax f(x)=a+b2∫abf(x)dxif (a+b−x)=f(x)here f(x)=11+sinxf(π4+3π4−x)=f(π−x) =11+sin(π−x)=11+sinx I=π4+3π42∫π/43π/411+sinxdx=π2∫π/43π/411+sinxdx=π2∫π/43π/4(1−sinx)cos2xdx=π2∫π/43π/4(sec2x−secxtanx)dx=π2[(tanx−secx)π/43π/4]=π2[(−1−(−2)−(1−2)]=π2(2(2−1))=π(2−1)

∫π/43π/4x1+sinxdx=

I=∫π/43π/4x1+sinxdx∴∫bax f(x)=a+b2∫abf(x)dxif (a+b−x)=f(x)here f(x)=11+sinxf(π4+3π4−x)=f(π−x) =11+sin(π−x)=11+sinx I=π4+3π42∫π/43π/411+sinxdx=π2∫π/43π/411+sinxdx=π2∫π/43π/4(1−sinx)cos2xdx=π2∫π/43π/4(sec2x−secxtanx)dx=π2[(tanx−secx)π/43π/4]=π2[(−1−(−2)−(1−2)]=π2(2(2−1))=π(2−1)

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$\int_{π / 4}^{3 π / 4} \frac{x}{1 + \sin x} d x =$

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$\sqrt{2} π$

$(\sqrt{2} - 1) π$

$π$

answer is B.

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Detailed Solution

$I = \int_{π / 4}^{3 π / 4} \frac{x}{1 + \sin x} d x$

$(∴ \begin{array}{l} \int_{b}^{a} x f (x) = \frac{a + b}{2} \int_{a}^{b} f (x) d x \\ \begin{array}{l} i f (a + b - x) = f (x) \\ h e r e f (x) = \frac{1}{1 + \sin x} \\ f (\frac{π}{4} + \frac{3 π}{4} - x) = f (π - x) \\ = \frac{1}{1 + \sin (π - x)} = \frac{1}{1 + \sin x} \end{array} \end{array}) \begin{array}{l} \end{array}$

$\begin{array}{l} I = \frac{\frac{π}{4} + \frac{3 π}{4}}{2} \int_{π / 4}^{3 π / 4} \frac{1}{1 + \sin x} d x \\ = \frac{π}{2} \int_{π / 4}^{3 π / 4} \frac{1}{1 + \sin x} d x \\ = \frac{π}{2} \int_{π / 4}^{3 π / 4} \frac{(1 - \sin x)}{\cos^{2} x} d x \\ = \frac{π}{2} \int_{π / 4}^{3 π / 4} (\sec^{2} x - \sec x \tan x) d x \\ = \frac{π}{2} [{(\tan x - \sec x)}_{π / 4}^{3 π / 4}] \\ = \frac{π}{2} [(- 1 - (- \sqrt{2}) - (1 - \sqrt{2})] \\ = \frac{π}{2} (2 (\sqrt{2} - 1)) \\ = π (\sqrt{2} - 1) \end{array}$