%%

The integral $$\int \, $$cos(loge x) dx is equal to : (where C is a constant of integration)

JEE · Math · previous-year question

  1. A.$${x \over 2}$$[sin(loge x) $$-$$ cos(loge x)] + C
  2. B.x[cos(loge x) + sin(loge x)] + C
  3. C.$${x \over 2}$$[cos(loge x) + sin(loge x)] + Ccorrect
  4. D.x[cos(loge x) $$-$$ sin(loge x)] + C

Answer

C. $${x \over 2}$$[cos(loge x) + sin(loge x)] + C

Explanation

$${\rm I} = \int {\cos \left( {\ell nx} \right)} dx$$ $${\rm I} = \cos (\ln x).x + \int {\sin \left( {\ell nx} \right)dx} $$ $${\rm I} = \cos \left( {\ell nx} \right)x + \left[ {\sin \left( {\ell nx} \right).x - \int {\cos \left( {\ell nx} \right)dx} } \right]$$ $${\rm I} = {x \over 2}\left[ {\sin \left( {\ell nx} \right) + \cos \left( {\ell nx} \right)} \right] + C$$

Practice more JEE questions

Answer thousands more real JEE previous-year questions free, get graded instantly, and climb the global ranked leaderboard.

Practice JEE free →

More JEE Math questions