> 数学 >
求不定积分∫[1/sin^2cos^2 (x)]dx
人气:466 ℃ 时间:2020-04-16 09:13:13
解答
∫[1/sin²xcos²x]dx
=∫[4/(2sinxcosx)²]dx
=∫[4/sin²2x]dx
=∫2csc²2xd2x
=-2∫(-csc²2x)d2x
=-2cot2x+C
推荐
猜你喜欢
© 2024 79432.Com All Rights Reserved.
电脑版|手机版