∫[π^2,0]cos√xdx
换元法
设t=√x,则x=t^2
=∫(0→π)cost dt²
=2∫(0→π)tcost dt
=2∫(0→π)tdsint
=2tsint - ∫(0→π)sintdt
=2tsint + cost |(0→π)
=-2
∫[π^2,0]cos√xdx=?
∫[π^2,0]cos√xdx
换元法
设t=√x,则x=t^2
=∫(0→π)cost dt²
=2∫(0→π)tcost dt
=2∫(0→π)tdsint
=2tsint - ∫(0→π)sintdt
=2tsint + cost |(0→π)
=-2