∫(1+x)sin ⁡xx2+2xcos2⁡ x−(1+x)sin ⁡2xdx

(1+x)sin xx2+2xcos2 x(1+x)sin 2xdx

  1. A

    12logesinx(x+1)cosx1sinx(x+1)cosx+1+C

  2. B

    12tan1{sinx(x+1)cosx}+C

  3. C

    12sin1{sinx(x+1)cosx}+C

  4. D

    12sin1(cosx+sinx)+C

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    Solution:

    I=(1+x)sin xdx(x+1)2cos2xcos2 x2(1+x)sin xcos xI=(1+x)sinxdx(x+1)2cos2x+sin2 x2(1+x)sin xcos x1

    I=(1+x)sinx(sinx(x+1)cosx)21dx

     Put  sinx(x+1)cosx=tsin x(x+1)dx=dt

     I=dtt21=12log t1t+1+C=12loge sin x(x+1)cos x1sin x(x+1)cos x+1+C

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