If I1=∫1sin⁡θ x1+x2dx and I2=∫1cosec⁡θ 1xx2+1dx then the value of I1I12I2eI1+I2I22−11I12+I22−1, is

If I1=1sinθx1+x2dx and I2=1cosecθ1xx2+1dx then the value of I1I12I2eI1+I2I2211I12+I221, is

  1. A
  2. B
  3. C
  4. D

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    Solution:

    We have,

    I2=1cosecθ1xx2+1dx=1sinθt1+t2dt, where t=1x

     I2=I1I1I12I2eI1+I2I2211I12+I221=I1I12I1e0I12112I121

    =I1I12I11I1210I120                   [Applying R3R3R2]

    =I12I1+I1=0            [Expanding along R3]

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