Question

Xsinx /sec x + tan x integration of limit 0 to pie

Answer 1

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∫ x sinx ln|secx + tanx| dx

Since you've already determine the derivative of ln|secx + tanx| is secx, we can skip that part

∫ x sinx ln|secx + tanx| dx

= ∫ ln|secx + tanx| * xsinx dx

integrate by parts:

u = ln|secx + tanx| ==> du = secx dx

dv = x sinx dx ==> v = -xcosx + ∫ cosx dx = -xcosx + sinx

∫ u dv

= uv - ∫ v du

= secx (-xcosx + sinx) - ∫ (-xcosx + sinx) (secx dx)

= -x + tanx + ∫ x dx - ∫ tanx dx

= -x + tanx + (1/2)x² - ln|secx| + C

with the limits 0 to π/4

= [ -0 + tan0 + (1/2)(0²) - ln|sec0| ] - [-π/4 + tan(π/4) + (1/2)(π/4)² - ln|sec(π/4)| ]

= [ 0 ] - [-π/4 + 1 + π/32 - (1/2)ln2 ]

= (1/2)ln2 - 1 - 7π/32

= -1.3406498

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