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Question

The integral π01+4sin2x24sinx2 dx equals to:
  1. 434
  2. π4
  3. 2π3443
  4. 434π3


Solution

The correct option is D 434π3
π01+4sin2x24sinx2 dx=π0(12sinx2)2 dx
=π012sinx2 dx
Put x2=t  dx=2dt and t=0,t=π2
π012sinx2 dx=2π20|12sint| dt
=2⎢ ⎢ ⎢π60(12sint) dt+π2π6(2sint1) dt⎥ ⎥ ⎥
=2([t+2cost]π/60+[2costt]π/2π/6)
=2(π6+2322π2+232+π6)
=2(4322π6)
=434π3

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