一道微积分竞赛题,求大神!请注意根号里面是sin-sin²x.还有不带根号解答的就别白费劲了~.
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![一道微积分竞赛题,求大神!请注意根号里面是sin-sin²x.还有不带根号解答的就别白费劲了~.](/uploads/image/z/13240533-21-3.jpg?t=%E4%B8%80%E9%81%93%E5%BE%AE%E7%A7%AF%E5%88%86%E7%AB%9E%E8%B5%9B%E9%A2%98%2C%E6%B1%82%E5%A4%A7%E7%A5%9E%21%E8%AF%B7%E6%B3%A8%E6%84%8F%E6%A0%B9%E5%8F%B7%E9%87%8C%E9%9D%A2%E6%98%AFsin-sin%26%23178%3Bx.%E8%BF%98%E6%9C%89%E4%B8%8D%E5%B8%A6%E6%A0%B9%E5%8F%B7%E8%A7%A3%E7%AD%94%E7%9A%84%E5%B0%B1%E5%88%AB%E7%99%BD%E8%B4%B9%E5%8A%B2%E4%BA%86%7E.)
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一道微积分竞赛题,求大神!请注意根号里面是sin-sin²x.还有不带根号解答的就别白费劲了~.
一道微积分竞赛题,求大神!
请注意根号里面是sin-sin²x.还有不带根号解答的就别白费劲了~.
一道微积分竞赛题,求大神!请注意根号里面是sin-sin²x.还有不带根号解答的就别白费劲了~.
∫(π/2→π)√(sinx-sin^2(x))dx
=∫(π/2→0)√(sinx-sin^2(x))d(π-x)
=∫(0→π/2)√(sinx-sin^2(x))dx
所以原式=2∫(0→π/2)√(sinx-sin^2(x))dx
=2∫(0→1)√(t-t^2)/√(1-t^2)*dt (t=arcsinx)
=2∫(0→1)√(t/(1+t))dt
=2∫(0→1)t/√(t^2+t)*dt
=4∫(0→1)t/√((2t+1)^2-1)*dt
令2t+1=1/cosy
则原式=4∫(0→arccos(1/3))(1-cosy)/(2cosy)*1/tany*1/2*siny/cos^2(y)dy
=∫(0→arccos(1/3))(1-cosy)/cos^2(y)*dy
=∫(0→arccos(1/3))sec^(y)dy-∫(0→arccos(1/3))dy/cosy
=tany|(0→arccos(1/3))-ln|secy+tany||(0→arccos(1/3))
=2√2-ln(3+2√2)