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Question

Evaluate the definite integrals.
π40(2sec2x+x3+2)dx

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Solution

π40(2sec2x+x3+2)dx=2π40sec2xdx+π40x3dx+2π401dx=2[tanx]π40+14[x4]π40+2[x]π40=2tanπ4+14(π4)4+2(π4)=2+π2+π41024


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