求∫sin^2xcos^2xdx的不定积分,详细过程 - —— ∫ 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
∫sin^2xcos^2xdx - —— ∫sin^2xcos^4xdx=∫(1-cos^2x)cos^4xdx=∫cos^4xdx-∫cos^6xdx 利用积分表得到.
数学积分问题,§sin^2xcos^2xdx §表示积分号 ^表示正弦的幂为2x 谢谢帮助了 - —— ∫sin^2x cos^2x dx=∫1/4*(sin2x)^2 dx=∫1/4*(1-cos4x)/2 dx=1/8*∫(1-cos4x)dx=1/8*(x-1/4*sin4x+C)=x/8-sin4x/32+C
积分符号1/sin^2xcos^2xdx - 作业帮 —— ∫sin^2xcos^2xdx=1/4 ∫sin^2 (2x)dx =1/4 ∫(1-cos4x)/2 dx =1/8 x-1/32 sin4x +C
跪求大神解答高数∫sin^2xcos^2*3xdx - —— ∫sin^版2xcos^权3xdx =∫(sinx)^2cosx(1-(sinx)^2)dx =∫(sinx)^2cosxdx-∫(sinx)^4cosxdx =∫(sinx)^2d(sinx)-∫(sinx)^4d(sinx) =(sinx)^3/3-(sinx)^5/5+C
∫ sin^2xcos^2dx 求思路 - —— ∫ sin²xcos²x dx= ∫ (sinxcosx)² dx= ∫ (1/2 * sin2x)² dx= (1/4)∫ sin²2x dx= (1/4)∫ (1 - cos4x)/2 dx= (1/8)[x - (1/4)sin4x] + C