Let
I=3π/4∫π/4x1+sinxdx ⋯(i)
Using property
b∫af(x)dx=b∫af(a+b−x)dx
We get,
⇒I=3π/4∫π/4π−x1+sinxdx ⋯(ii)
Adding (i) and (ii):
⇒2I=3π/4∫π/4π1+sinxdx=3π/4∫π/4π(1−sinx)cos2xdx =π3π/4∫π/4(sec2x−secxtanx)dx⇒I=π2[tanx−secx]3π/4π/4⇒I=π(√2−1)
⇒[(√2+1)I]=[π]=3