Thursday, January 10, 2019

Single Variable Calculus, Chapter 8, 8.2, Section 8.2, Problem 38

Determine the integral π2π4cot3xdx


π2π4cot3xdx=π2π4cotxcot2xdxApply Trigonometric Identities csc2x=1+cot2xπ2π4cot3xdx=π2π4cotx(csc2x1)dxπ2π4cot3xdx=π2π4(cotxcsc2xcotx)dxπ2π4cot3xdx=π2π4cotxcsc2xdxπ2π4cotxdx


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=01uduπ2π4cotxcsc2xdx=01uduπ2π4cotxcsc2xdx=[u1+11+1]01π2π4cotxcsc2xdx=[u22]01π2π4cotxcsc2xdx=(0)22+(1)22π2π4cotxcsc2xdx=12


@ 2nd term


π2π4cotxdx=[ln(sinx)]π2π4π2π4cotxdx=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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