JEE MAIN - Mathematics (2019 - 12th January Morning Slot - No. 16)
The integral $$\int \, $$cos(loge x) dx is equal to : (where C is a constant of integration)
$${x \over 2}$$[sin(loge x) $$-$$ cos(loge x)] + C
x[cos(loge x) + sin(loge x)] + C
$${x \over 2}$$[cos(loge x) + sin(loge x)] + C
x[cos(loge x) $$-$$ sin(loge x)] + C
Explanation
$${\rm I} = \int {\cos \left( {\ell nx} \right)} dx$$
$${\rm I} = \cos (\ln x).x + \int {\sin \left( {\ell nx} \right)dx} $$
$${\rm I} = \cos \left( {\ell nx} \right)x + \left[ {\sin \left( {\ell nx} \right).x - \int {\cos \left( {\ell nx} \right)dx} } \right]$$
$${\rm I} = {x \over 2}\left[ {\sin \left( {\ell nx} \right) + \cos \left( {\ell nx} \right)} \right] + C$$
$${\rm I} = \cos (\ln x).x + \int {\sin \left( {\ell nx} \right)dx} $$
$${\rm I} = \cos \left( {\ell nx} \right)x + \left[ {\sin \left( {\ell nx} \right).x - \int {\cos \left( {\ell nx} \right)dx} } \right]$$
$${\rm I} = {x \over 2}\left[ {\sin \left( {\ell nx} \right) + \cos \left( {\ell nx} \right)} \right] + C$$
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