∫cost/(sint+cost)dt在0到π取積分

∫cost/(sint+cost)dt在0到π取積分

∫cost/(sint + cost)dt=(1/2)∫[(cost + sint)+(cost - sint)]/(sint + cost)dt=(1/2)∫[1 +(cost - sint)/(sint + cost)] dt= t/2 +(1/2)ln|sint + cost| + C設ƒ;(t)= cost/(sint + cost)∫(0→…