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Question

The value of -π/2π/2x3+x cos x+tan5 x+1 dx, is

(a) 0
(b) 2
(c) π
(d) 1

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Solution

(c) π

-π2π2x3+xcosx+tan5x+1dx=x44-π2π2+x sinx-π2π2--π2π2sinx dx+-π2π2tan3x sec2x-1dx+x -π2π2=π464-π464+π2-π2--cosx-π2π2+-π2π2tan3x sec2x dx--π2π2tan3x dx+π2+π2=π+0+tan4x4-π2π2--π2π2tanx sec2x dx--π2π2tanx dx=π-tan2x2-π2π2--logcosx-π2π2=π

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