∫0π x1+sin⁡xdx is equal to

0πx1+sinxdx is equal to

  1. A

    π

  2. B

    π2

  3. C

    2π

  4. D

    π8

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    Solution:

    Let,

    I=0πx1+sinxdx----iI=0ππx1+sin(πx)dxthen,  I=0ππx1+sinxdx 0af(x)dx=0af(ax)dx [sin(πx)=sinx] (ii) 

    On adding Eqs. (i) and (ii), we get

    2I=0ππ(1+sinx)dx=π0π1(1+sinx)dx=π0π1sinx(1+sinx)(1sinx)dx

    [multiply numerator and denominator by (1 - sin x)] 

     2I=π0π1sinx1sin2xdx =π0π1cos2xdxπ0πsinxcos2xdx 2I=π0πsec2xdxπ0πsecxtanxdx 2I=π[tanxsecx]0π 2I=π[tanπsecπ(tan0sec0)] 2I=π[0+10+1]2I=2πI=π

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