Question

Evaluate the definite integral: $∫_0^\frac{π}{4}(2sec^2x+x^3+2)dx$

Answer 1

Let I=$∫_0^\frac{π}{4}(2sec^2x+x^3+2)dx$

$∫(2sec^2x+x^3+2)dx=tanx+\frac{x^4}{4}+2x=F(x)$

By second fundamental theorem of calculus,we obtain

$I=F(π/4)-F(0)$

$={(2tan\frac{π}{4}+\frac{1}{4}(π/4)4+2(\frac{π}{4}))-(2tan0+0+0)}$

$=2tan\frac{π}{4}+\frac{π^4}{4}5+\frac{π}{2}$

$=2+\frac{π}{2}+\frac{{π^4}}{{1024}}$