求不定积分∫x.sinx^2.cosx^2dx 求不定积分∫x^5sinx^2dx
\u222bsinx^2cosx^2dx \u600e\u4e48\u89e3\u7b54\u222bsinx^2cosx^2dx
=(1/4) * \u222b4sinx^2cosx^2dx
=(1/4) * \u222b(2sinxcosx)^2dx (\u6839\u636e\u6b63\u5f26\u500d\u89d2\u516c\u5f0f)
=(1/4) * \u222b(sin2x)^2dx \uff08\u6839\u636e\u4f59\u5f26\u500d\u89d2\u516c\u5f0f\uff09
=(1/8) *\u222b(1-cos4x)dx
=(1/8) *x - (1/8) * \u222bcos4xdx + C (C\u662f\u4e0d\u5b9a\u79ef\u5206\u4efb\u610f\u5e38\u6570)
=(1/8) *x - (1/32) * \u222bcos4xd4x + C (C\u662f\u4e0d\u5b9a\u79ef\u5206\u4efb\u610f\u5e38\u6570)
=(1/8) *x - (1/32)sin4x + C (C\u662f\u4e0d\u5b9a\u79ef\u5206\u4efb\u610f\u5e38\u6570)
=x/8 - sin4x/32 + C (C\u662f\u4e0d\u5b9a\u79ef\u5206\u4efb\u610f\u5e38\u6570)
x^5sinx^2dx =x^2*sinx^2-1/2*x^4*cosx^2+cosx^2
\u5982\u679c\u4f60\u8ba4\u53ef\u6211\u7684\u56de\u7b54\uff0c\u656c\u8bf7\u53ca\u65f6\u91c7\u7eb3\uff0c
~\u5982\u679c\u4f60\u8ba4\u53ef\u6211\u7684\u56de\u7b54\uff0c\u8bf7\u53ca\u65f6\u70b9\u51fb\u3010\u91c7\u7eb3\u4e3a\u6ee1\u610f\u56de\u7b54\u3011\u6309\u94ae
~~\u624b\u673a\u63d0\u95ee\u7684\u670b\u53cb\u5728\u5ba2\u6237\u7aef\u53f3\u4e0a\u89d2\u8bc4\u4ef7\u70b9\u3010\u6ee1\u610f\u3011\u5373\u53ef\u3002
~\u4f60\u7684\u91c7\u7eb3\u662f\u6211\u524d\u8fdb\u7684\u52a8\u529b
~~O(\u2229_\u2229)O\uff0c\u8bb0\u5f97\u597d\u8bc4\u548c\u91c7\u7eb3\uff0c\u4e92\u76f8\u5e2e\u52a9
∫x(sinx)^2 (cosx)^2dx
=1/4 ∫x(sin2x)^2dx
=1/8 ∫x(1-cos4x)dx
=1/8 (∫xdx-∫xcos4xdx)
=1/16 x^2 - 1/64 xsin4x + 1/64 ∫sin4xdx
= 1/16 x^2 - 1/64 xsin4x - 1/256 cos4x + C
∫x.sinx^2.cosx^2dx
=∫x.sinx^2.cosx^2dx
=∫x/4*(sin2x)^2dx
=1/8∫x(1-(cos2x)^2)dx
=1/8[∫xdx-∫x(cos2x)^2dx]
=1/8[∫xdx-∫x(1/2+1/2cos4x)dx]
=1/8∫xdx-1/16∫xdx-1/16∫xco4xdx
=1/16∫xdx-1/64∫xd(sin4x)
=1/16∫xdx-1/64(xsin4x-∫sin4xdx)
=1/16∫xdx-1/64xsin4x-1/256∫d(cos4x)
=1/32x^2-1/64xsin4x-1/256cos4x+c
先把x移到微元中
变成dx^2
结果-1/8cos2x^2+c
∫x.sinx^2.cosx^2dx
=(1/2)∫xsin2x^2dx
令u=2x^2
du=4x
原式=(1/8)∫sinudu
=-(1/8)cosu+C
=-(1/8)cos2x^2+C
绛旓細鈭玿.sinx^2.cosx^2dx =(1/2)鈭玿sin2x^2dx 浠=2x^2 du=4x 鍘熷紡=(1/8)鈭玸inudu =-(1/8)cosu+C =-(1/8)cos2x^2+C
绛旓細鍏堝埄鐢ㄥ嶈鍏紡锛岀劧鍚庡埄鐢ㄥ垎閮绉垎娉曞強绗竴鎹㈠厓绉垎娉曪細鈭玿(sinx)^2 (cosx)^2dx =1/4 鈭玿(sin2x)^2dx =1/8 鈭玿(1-cos4x)dx =1/8 (鈭玿dx-鈭玿cos4xdx)=1/16 x^2 - 1/64 xsin4x + 1/64 鈭玸in4xdx = 1/16 x^2 - 1/64 xsin4x - 1/256 cos4x + C ...
绛旓細鎵浠(x^2/2)/dx=x 鎵浠(x^2/2)=dx 鎵浠モ埆xsinx^2dx=鈭玸in(x^2)d(x^2/2)=鈭玸in(x^2)d(x^2)/2=-cos(x^2)/2+C 1/2鏄庝箞涓嶆槸鍦╠鍚庨潰鎬庝箞鎻愬嚭鏉ョ殑?甯告暟鐨勪綅缃彲浠ユ嬁鍒板墠闈㈠晩 绉垎鐨勬ц川 鈭玨f(x)dx=k鈭玣(x)dx ...
绛旓細x*(-1/2*cos(x)*sin(x)+1/2*x)-1/4*cos(x)^2-1/4*x^2+C
绛旓細绛旓細鈭玿sinx^2dx =鈭玿(1-cos2x)/2dx =1/2鈭玿dx-1/2鈭玿cos2xdx =1/2鈭玿dx-1/4鈭玿 d(sin2x)=1.4x^2-1/4xsin2x-1/8cos2x+c
绛旓細涓嶅畾绉垎鍘熷紡=1/2鈭玸inx²dx²=-1/2cosx²+c涓昏杩愮敤鍑戝井鍒嗘硶
绛旓細x(sinx)^2 =x/2(1-cos2x)鐒跺悗鐢ㄥ垎姝绉垎娉曞氨鍙互浜
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绛旓細鈭玿sin(x^2)dx =(1/2)鈭玸in(x^2) dx^2 =-(1/2)cos(x^2) + C 涓涓嚱鏁帮紝鍙互瀛樺湪涓嶅畾绉垎锛岃屼笉瀛樺湪瀹氱Н鍒嗭紝涔熷彲浠ュ瓨鍦ㄥ畾绉垎锛岃屾病鏈変笉瀹氱Н鍒嗐傝繛缁嚱鏁帮紝涓瀹氬瓨鍦ㄥ畾绉垎鍜屼笉瀹氱Н鍒嗐
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