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The value of integral   \int_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\frac{x}{1+\sin x}dx is

  • Option 1)

    \pi \sqrt{2}

  • Option 2)

    \pi\left ( \sqrt{2}-1 \right )

  • Option 3)

    \frac{\pi}{2} \left (\sqrt{2}+1 \right )

  • Option 4)

    2\pi \left (\sqrt{2}-1 \right )

 

Answers (2)

best_answer

As we learned,

 

lower and upper limit -

\int_{a}^{b}f\left ( x \right )dx= \left ( F\left ( x \right ) \right )_{a}^{b}

                = F\left ( b \right )-F\left ( a \right )

 

- wherein

Where a is lower and b is upper limit.

 

 

 

I= \int_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\frac{x}{1+\sin x}dx

=\int_{\frac{\pi }{4}}^{\frac{3\pi }{4}}\left ( x\sec x^{2}-x\sec x\tan x \right )dx

=\left [ x\tan x-\ln \left ( secx \right ) \right ]_{\frac{\pi }{4}}^{\frac{3\pi }{4}}-\left [ x\sec x-\ln \left ( \sec x+\tan x \right ) \right ]_{\frac{\pi }{4}}^{\frac{3\pi }{4}}

=\frac{\pi }{2}\left ( \sqrt{2}+1 \right )

 

 

 


Option 1)

\pi \sqrt{2}

Option 2)

\pi\left ( \sqrt{2}-1 \right )

Option 3)

\frac{\pi}{2} \left (\sqrt{2}+1 \right )

Option 4)

2\pi \left (\sqrt{2}-1 \right )

Posted by

Himanshu

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