如何直接看出0到pai/2定积分cost/(sint+cost)与sint/(sint+cost)相等?
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如何直接看出0到pai/2定积分cost/(sint+cost)与sint/(sint+cost)相等?
![如何直接看出0到pai/2定积分cost/(sint+cost)与sint/(sint+cost)相等?](/uploads/image/z/5982521-41-1.jpg?t=%E5%A6%82%E4%BD%95%E7%9B%B4%E6%8E%A5%E7%9C%8B%E5%87%BA0%E5%88%B0pai%2F2%E5%AE%9A%E7%A7%AF%E5%88%86cost%2F%28sint%2Bcost%29%E4%B8%8Esint%2F%28sint%2Bcost%29%E7%9B%B8%E7%AD%89%3F)
只需令x=pi/2-t,则当x=0,t=pi/2,当x=pi/2,t=0,dx=-dt,那么
∫(0,pi/2)cosx/(sinx+cosx)dx
=-∫(pi/2,0)sint/(sint+cost)dt
=∫(0,pi/2)sinx/(sinx+cosx)dx
所以
∫(0,pi/2)cosx/(sinx+cosx)dx=∫(0,pi/2)sinx/(sinx+cosx)dx
=(1/2)[∫(0,pi/2)cosx/(sinx+cosx)dx+∫(0,pi/2)sinx/(sinx+cosx)dx]
=(1/2)∫(0,pi/2)(sinx+cosx)/(sinx+cosx)dx
=(1/2)∫(0,pi/2)dx=pi/4
∫(0,pi/2)cosx/(sinx+cosx)dx
=-∫(pi/2,0)sint/(sint+cost)dt
=∫(0,pi/2)sinx/(sinx+cosx)dx
所以
∫(0,pi/2)cosx/(sinx+cosx)dx=∫(0,pi/2)sinx/(sinx+cosx)dx
=(1/2)[∫(0,pi/2)cosx/(sinx+cosx)dx+∫(0,pi/2)sinx/(sinx+cosx)dx]
=(1/2)∫(0,pi/2)(sinx+cosx)/(sinx+cosx)dx
=(1/2)∫(0,pi/2)dx=pi/4
如何直接看出0到pai/2定积分cost/(sint+cost)与sint/(sint+cost)相等?
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