Determine the integral ∫π2π4cot3xdx
∫π2π4cot3xdx=∫π2π4cotxcot2xdxApply Trigonometric Identities csc2x=1+cot2x∫π2π4cot3xdx=∫π2π4cotx(csc2x−1)dx∫π2π4cot3xdx=∫π2π4(cotxcsc2x−cotx)dx∫π2π4cot3xdx=∫π2π4cotxcsc2xdx−∫π2π4cotxdx
We integrate the equation term by term
@ 1st term
∫π2π4cotxcsc2xdx
Let u=cotx, then du=−csc2xdx, so csc2xdx=−du. When x=π4,u=1 and when x=π2,u=0. Therefore,
∫π2π4cotxcsc2xdx=∫01u⋅−du∫π2π4cotxcsc2xdx=−∫01udu∫π2π4cotxcsc2xdx=−[u1+11+1]01∫π2π4cotxcsc2xdx=−[u22]01∫π2π4cotxcsc2xdx=−(0)22+(1)22∫π2π4cotxcsc2xdx=12
@ 2nd term
∫π2π4cotxdx=[ln(sinx)]π2π4∫π2π4cotxdx=ln(sinπ2)−ln(sinπ4)∫π2π4cotxdx=ln(1)=ln(√22)∫π2π4cotxdx=−ln(√22)
Combine the results of the integration term by term
∫π2π4cotx3dx=12−(−ln(√22))∫π2π4cotx3dx=12+ln(√22)
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