> 数学 >
不定积分 :∫ xcos^2xdx
人气:383 ℃ 时间:2020-06-28 01:44:02
解答
∫ xcos^2xdx
=∫ x(1+cos2x/2)dx
=1/2∫ xdx+1/2∫xcos2xdx
=x²/4+1/4∫xdsin2x
=x²/4+1/4*xsin2x-1/4∫sin2xdx
=x²/4+1/4*xsin2x-1/8∫sin2xd2x
=x²/4+1/4*xsin2x+1/8*cos2x+C
推荐
猜你喜欢
© 2026 79432.Com All Rights Reserved.
电脑版|手机版