x^2*(cosx)^2dx=多少？
x^2*(cosx)^2的积分为1/6 x³+1/4x² *sin2x+1/4cos2x-1/8sin2x+C
解： ∫（ x² cos²x）dx= ∫（ x² （cos2x+1）/2）dx
=1/2∫（ x² cos2x+x²）dx
=1/2∫x²dx+1/2∫ x² cos2xdx
=1/6 x³+1/2∫ x² cos2xdx
=1/6 x³+1/4∫ x² dsin2x
=1/6 x³+1/4x² *sin2x-1/4∫sin2x d x²
=1/6 x³+1/4x² *sin2x-1/2∫xsin2x d x
=1/6 x³+1/4x² *sin2x+1/4∫x d cos2x
=1/6 x³+1/4x² *sin2x+1/4cos2x-1/4∫cos2x d x
=1/6 x³+1/4x² *sin2x+1/4cos2x-1/8sin2x+C
扩展资料：
1、分部积分法是微积分学中的一类重要的、基本的计算积分的方法。它的主要原理是将不易直接求结果的积分形式，转化为等价的易求出结果的积分形式的。
2、分部积分法的公式为：∫μ（x）v'（x）dx=∫μ（x）dv（x）=μ（x）*v（x）-∫v（x）dμ（x）
3、分部积分计算例题：
（1）∫xcosxdx=∫xdsinx=xsinx-∫sinxdx=xsinx+cosx+C
（2）∫xarctanxdx=∫arctanxd（x²/2）
=x²/2*arctanx-1/2∫x²darctanx
=x²/2*arctanx-1/2∫x²/（x²+1）dx
=x²/2*arctanx-1/2∫dx+1/2∫1/（x²+1）dx
=x²/2*arctanx-1/2∫dx+1/2arctanx+C
参考资料来源：百度百科-积分
参考资料来源：百度百科-分部积分法
方法如下，
请作参考：
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绛旓細濡傛灉2鏄痗osx鐨勫钩鏂 鈭玿锛坈osx锛塣2dx=鈭玿(1+cos2x)/2dx =鈭玿/2dx+1/2鈭玞os2xdx =x^2/4+1/4鈭玞os2xd2x =x^2/4+1/4sin2x+C 濡傛灉2鏄痻鐨勫钩鏂 鈭玿cos(x^2)dx=1/2鈭玞os(x^2)dx^2=1/2sin(x^2)+C
绛旓細涓鑸繖閮介氳繃鍒嗛儴绉垎鎶婂畠鍙樹负閫掓帹鍏紡锛岀劧鍚庡綊绾冲嚭閫氶」鍏紡 棰樼洰1浠x=sint甯﹀叆鍚庯紝鍙互鍙樹负(cost)^(m+1) dt鐨勭Н鍒 瀹冨拰棰樼洰2璁＄畻绫讳技 浠2涓轰緥瀛(S琛ㄧず绉垎锛塉m = -S[x(sinx)^(m-1)dcosx] = xcosx(sinx)^(m-1) - (m-1)S[x(sinx)^(m-2)(cosx)^2dx]= xcosx(sinx)^(m-...
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绛旓細绛旓細鈭2xcosx^2dx鈮犫埆sinx^2dx^2 鍥犱负锛屸埆2xcosx^2dx = 鈭玞osx^2dx^2 = sinx^2+C 鑰 鈭玸inx^2dx^2 =-cosx^2+C 娉ㄦ剰锛 sinx^2涓巗in²x 鐨勫尯鍒 鎵浠ワ紝瀹冧滑涓嶇浉绛
绛旓細鏈涢噰绾
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