这个是积分计算题，算到最后实在搞不好了，请教会的人怎么写。尤其是cos四次方的积分。 考研高数 sin的平方x乘cos的四次方x 的 积分 怎么算
\u4e0d\u5b9a\u79ef\u5206\u6c42\u89e3\uff0c\u9898\u76ee\u5728\u7167\u7247\u4e0a\uff0c\u8981\u8be6\u89e3\uff0c\u6700\u540e\u597d\u662f\u5199\u597d\u4e86\u53d1\u4e00\u4e0b\u7167\u7247\uff0c\u7528\u6587\u5b57\u6253\u8d77\u6765\u770b\u7684\u4e0d\u6e05\u695a\u7b2c\u4e00\u5f20\u56fe\u7247
5. \u6ca1\u6709\u601d\u8003\u7684\u4f59\u5730\uff0c\u76f4\u63a5\u6362\u5143
u=arctan x\uff0c\u90a3\u4e48x=tan u\uff0cdx=du/cos² u\uff0c1+x²=1/cos² u,
\u56e0\u6b64\u6362\u5143\u4e4b\u540e\u7684\u88ab\u79ef\u51fd\u6570=e^u*|cos u|³/cos² u=e^u*|cos u|\uff0c
\u56e0\u4e3a\u628au\u9650\u5236\u5728-\u03c0/20\uff0c\u56e0\u6b64\u88ab\u79ef\u51fd\u6570=e^u*cos u\u3002
\u4ece\u5f62\u5f0f\u4e0a\u53ef\u4ee5\u770b\u51fa\u63a5\u4e0b\u6765\u8be5\u91c7\u53d6\u5206\u90e8\u79ef\u5206\uff1a
I=\u222be^u*cos u du=\u222bcos u de^u=e^u*cos u+\u222be^u*sin u du
=e^u*cos u+\u222bsin u de^u=e^u*cos u+e^u*sin u-\u222be^u*cos u du
=e^u*cos u+e^u*sin u-I
\u56e0\u6b64I=(e^u*cos u+e^u*sin u)/2+C
\u53c8\u56e0\u4e3au=arctan x\uff0c-\u03c0/2<u<\u03c0/2\uff0c\u6240\u4ee5cos u=1/sqrt(1+x²)\uff0csin u=x/sqrt(1+x²)\uff0c\u4ee3\u5165\u4e0a\u5f0f\u5e76\u5316\u7b80\u5f97\u5230\uff1a
I=exp(arctan x)*(x+1)/[2*sqrt(1+x²)]\u3002\u89e3\u6bd5\u3002
7.\u7b2c\u4e00\u79cd\u65b9\u6cd5\uff1a\u628a\u5206\u5b50\u5904\u76841\u6362\u6210 sin² x+cos² x\uff0c\u56e0\u4e3a\u5206\u6bcd\u662f\u56db\u6b21\u65b9\uff0c\u56e0\u6b64\u5206\u5b50\u5904\u7684\u201c1\u201d\u5e94\u8be5\u5e73\u65b9\uff0c\u4ece\u800c\u5f97\u5230\u9f50\u6b21\u7684\u88ab\u79ef\u51fd\u6570\u3002\u81ea\u5df1\u53ef\u4ee5\u53bb\u5c1d\u8bd5\u3002\u8fd9\u91cc\u5229\u7528\u53e6\u5916\u4e00\u79cd\u65b9\u6cd5\u3002\u770b\u5230\u5206\u6bcd\u542b\u6709(1/cos² x)^n\u8fd9\u79cd\u5f62\u5f0f\uff0c\u56de\u60f3\u4e00\u4e0btan x\u7684\u5bfc\u6570\u521a\u597d\u662f1/cos ² x\uff0c\u56e0\u6b64\u8003\u8651\u539f\u51fd\u6570\u662f\u5185\u51fd\u6570\u4e3atan x\u7684\u7b26\u5408\u51fd\u6570\uff0c\u8bbe\u4e4b\u4e3af(tan x)\uff0c\u90a3\u4e48
df(tan x)/dx=df/du \u00b7 du/dx=f'(tan x)/cos² x=(1/cos x)^4\uff0c
\u56e0\u6b64f'(tan x)=1/cos² x=(sin² x+cos² x)/cos² x=tan² x+1\uff0c
\u5373f'(u)=u²+1\uff0c\u4ece\u800cf(u)=u³/3+u+C\uff0c
\u56e0\u6b64\u539f\u51fd\u6570\u4e3af(tan x)=(tan x)³/3+tan x+C\u3002
\u7b2c\u4e8c\u5f20\u56fe\u7247\uff1a
4.\u5206\u5b50\u5206\u6bcd\u540c\u65f6\u4e58\u4ee5sin x\uff0c\u7136\u540e\u5206\u5b50\u5904\u7684sin d dx\u5c31\u53ef\u4ee5\u5199\u6210-dcos x\uff0c\u5206\u6bcd\u5199\u6210cos x\u7684\u5f62\u5f0f\uff0c\u7136\u540e\u4ee4u=cos x\u8fdb\u884c\u6362\u5143\u5373\u53ef\u3002
6.\u5206\u6bcd\u5904\u6839\u53f7\u5185\u7684\u90e8\u5206\u91c7\u7528\u914d\u65b9\u6cd5\uff1ax(4-x)=4x-x²-4+4=4-(x-2)²\uff0c\u7136\u540e\u4ee4x-2=2*cos u\u8fdb\u884c\u6362\u5143\u79ef\u5206\uff0c\u5927\u529f\u544a\u6210\u3002
8.\u4e5f\u6ca1\u6709\u601d\u8003\u7684\u4f59\u5730\uff0c\u4ee4 e^2x=tan u\uff0c\u6362\u5143\u79ef\u5206\u3002
\u5177\u4f53\u5982\u56fe\uff1a
\u5bf9\u4e8e\u4efb\u610f\u4e00\u4e2a\u5b9e\u6570x\u90fd\u5bf9\u5e94\u7740\u552f\u4e00\u7684\u89d2\uff08\u5f27\u5ea6\u5236\u4e2d\u7b49\u4e8e\u8fd9\u4e2a\u5b9e\u6570\uff09\uff0c\u800c\u8fd9\u4e2a\u89d2\u53c8\u5bf9\u5e94\u7740\u552f\u4e00\u786e\u5b9a\u7684\u6b63\u5f26\u503csinx\uff0c\u8fd9\u6837\uff0c\u5bf9\u4e8e\u4efb\u610f\u4e00\u4e2a\u5b9e\u6570x\u90fd\u6709\u552f\u4e00\u786e\u5b9a\u7684\u503csinx\u4e0e\u5b83\u5bf9\u5e94\uff0c\u6309\u7167\u8fd9\u4e2a\u5bf9\u5e94\u6cd5\u5219\u6240\u5efa\u7acb\u7684\u51fd\u6570\uff0c\u8868\u793a\u4e3ay=sinx\u3002
\u8fde\u7eed\u51fd\u6570\uff1a
\u4e00\u5b9a\u5b58\u5728\u5b9a\u79ef\u5206\u548c\u4e0d\u5b9a\u79ef\u5206\uff1b\u82e5\u5728\u6709\u9650\u533a\u95f4[a,b]\u4e0a\u53ea\u6709\u6709\u9650\u4e2a\u95f4\u65ad\u70b9\u4e14\u51fd\u6570\u6709\u754c\uff0c\u5219\u5b9a\u79ef\u5206\u5b58\u5728\uff1b\u82e5\u6709\u8df3\u8dc3\u3001\u53ef\u53bb\u3001\u65e0\u7a77\u95f4\u65ad\u70b9\uff0c\u5219\u539f\u51fd\u6570\u4e00\u5b9a\u4e0d\u5b58\u5728\uff0c\u5373\u4e0d\u5b9a\u79ef\u5206\u4e00\u5b9a\u4e0d\u5b58\u5728\u3002
\u5c06\u6240\u6c42\u79ef\u5206\u5316\u4e3a\u4e24\u4e2a\u79ef\u5206\u4e4b\u5dee\uff0c\u79ef\u5206\u5bb9\u6613\u8005\u5148\u79ef\u5206\u3002\u5b9e\u9645\u4e0a\u662f\u4e24\u6b21\u79ef\u5206\u3002
\u6709\u7406\u51fd\u6570\u5206\u4e3a\u6574\u5f0f\uff08\u5373\u591a\u9879\u5f0f\uff09\u548c\u5206\u5f0f\uff08\u5373\u4e24\u4e2a\u591a\u9879\u5f0f\u7684\u5546\uff09\uff0c\u5206\u5f0f\u5206\u4e3a\u771f\u5206\u5f0f\u548c\u5047\u5206\u5f0f\uff0c\u800c\u5047\u5206\u5f0f\u7ecf\u8fc7\u591a\u9879\u5f0f\u9664\u6cd5\u53ef\u4ee5\u8f6c\u5316\u6210\u4e00\u4e2a\u6574\u5f0f\u548c\u4e00\u4e2a\u771f\u5206\u5f0f\u7684\u548c\uff0e\u53ef\u89c1\u95ee\u9898\u8f6c\u5316\u4e3a\u8ba1\u7b97\u771f\u5206\u5f0f\u7684\u79ef\u5206\u3002
I = I1-I2
I1 = ∫<0,π>dt∫<0,2>r(cost+sint)rdr
= (8/3)∫<0,π>(cost+sint)dt
= (8/3)[sint-cost]<0,π> = 16/3.
I2 = ∫<0,π/2>dt∫<0,2cost>r(cost+sint)rdr
= (8/3)∫<0,π/2>(cost+sint)(cost)^3dt
= (8/3)[∫<0,π/2>(cost)^4dt+∫<0,π/2>sint(cost)^3dt]
其中
I3 = (8/3)∫<0,π/2>(cost)^4dt = (2/3)∫<0,π/2>(1+cos2t)^2dt
= (1/3)∫<0,π/2>[2+4cos2t+2(cos2t)^2]dt
= (1/3)∫<0,π/2>[3+4cos2t+cos4t]dt
= (1/3)[3t+2sin2t+(1/4)sin4t]<0,π/2> = π/2;
I4 = (8/3)∫<0,π/2>sint(cost)^3dt]
= (-8/3)∫<0,π/2>(cost)^3d(cost)
= (-2/3)[(cost)^4]<0,π/2> = 2/3.
I2 = I3+I4 = π/2+2/3.
I = I1-I2 = 16/3-π/2-2/3 = 14/3-π/2.
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