Integrate ∫sinxcosxdx.
Calculate the given expression:
Given, ∫sinxcosxdx
Let, I=∫sinxcosxdx
Let, sinx=t
Differentiating both sides with respect to x,
cosx=dtdx⇒cosxdx=dt
⇒I=∫tdt[∵sinx=t,cosxdx=dt]⇒I=t22+C[∵∫xn=xn+1n+1]⇒I=sin2x2+C
Therefore,∫sin(x)cos(x)dx=sin2(x)2+C.