∫0π x1+sin⁡xdx is equal to

Let,I=∫0π x1+sin⁡xdx----iI=∫0π π−x1+sin⁡(π−x)dxthen,  I=∫0π π−x1+sin⁡xdx ∵∫0a f(x)dx=∫0a f(a−x)dx [∵sin⁡(π−x)=sin⁡x]… (ii) On adding Eqs. (i) and (ii), we get2I=∫0π π(1+sin⁡x)dx=π∫0π 1(1+sin⁡x)dx=π∫0π 1−sin⁡x(1+sin⁡x)(1−sin⁡x)dx[multiply numerator and denominator by (1 - sin x)] ⇒ 2I=π∫0π 1−sin⁡x1−sin2⁡xdx =π∫0π 1cos2⁡xdx−π∫0π sin⁡xcos2⁡xdx⇒ 2I=π∫0π sec2⁡xdx−π∫0π sec⁡x⋅tan⁡xdx⇒ 2I=π[tan⁡x−sec⁡x]0π⇒ 2I=π[tan⁡π−sec⁡π−(tan⁡0−sec⁡0)]⇒ 2I=π[0+1−0+1]⇒2I=2π⇒I=π