Given, I=∫cos(lnx)dx
Let, lnx=t⇒x=et⇒dx=etdt
∴I=∫etcostdt
∫etcostdt=21∫et((cost+sint)+(cost−sint))dt
∫etcostdt=21et(cost+sint)+C, where C is the constant of integration.
=2x(cos(lnx)+sin(lnx))+C.
JEE Main 2019 — Mathematics Calculus
The integral ∫cos(lnx)dx, is equal to
Held on 12 Jan 2019 · Verified 6 Jul 2026.
2x(cos(lnx)−sin(lnx))+C
x(cos(lnx)−sin(lnx))+C
x(cos(lnx)+sin(lnx))+C
2x(cos(lnx)+sin(lnx))+C
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