∫3π2π2[2sinx]dx
Put x=π2+t⇒dx=dt
=∫3π2π2[2sin(π2+t)]dx
=∫3π2π2[2cost]dt
=∫π301dt+∫2π3π30dt+∫2π3π2(−1)dt+∫π2π3(−2)dt
=[t]π30+0−[t]2π3π2−2[t]π2π3
=π3−0+0−(2π3−π2)−2[π−2π3]
=π3−2π3+π2−2π+4π3
=π−2π+4π3+π2−2π
=3π3+π2−2π
=π+π2−2π
=π2−π
=π−2π2
=−π2
∴∫3π2π2[2sinx]dx=−π2