lower limit is 0 upper limit is 2x. ∫ sin 2 θ ⁡ d θ =?

Question

lower limit is 0 upper limit is 2x.  ∫      sin          2      θ        ⁡  d  θ  =?

Porter Mata · Accepted Answer

∫          0              2      x            ﻿        sin          2        ⁡  t  d  t  =      ∫          0              2      x            sin          2        ⁡  t  d  t  =      ∫          0              2      x        (            1      −      cos      ⁡      2      t        2    )  d  t  =      ∫          0              2      x            1    2    d  t  −      ∫          0              2      x                  cos      ⁡      2      t        2    d  t  =  x  −            sin      ⁡      4      x        4        ∫          0              2      x        sin  ⁡  t  cos  ⁡  t  d  t  =      1    2        ∫          0              2      x        sin  ⁡  2  t  d  t  =  [  −            cos      ⁡      2      t        4        ]          0              2      x        =      1    2        sin          2        ⁡  2  x  =  2      cos          2        ⁡  x      sin          2        ⁡  x

vroos5p · Accepted Answer

a)      sin          2        ⁡  θ  =  (  1  −      cos          2        ⁡  θ  )      /    2so the intgeral becomes   1      /    2  ∫  1  −      cos          2        ⁡  θ from 0 to   2  π  =  1      /    2  [  θ  −  1      /    2      sin          2        ⁡  θ  ] from 0 to   2  π  =  1      /    2  [  (  2  π  −  0  )  −  (  0  −  0  )  ]  =  πb) let   u  =  sin  ⁡  θ  d  u  =  cos  ⁡  θ  d  θthe integral become ∫udu  =  1      /    2      u          2        =  1      /    2      sin          2        ⁡  θ from 0 to   2  π=0

lower limit is 0 upper limit is 2x. int sin^2 theta dtheta=?

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