基础题，求不定积分 微积分 不定积分的一些基础题，求大神给个过程，非常感谢！
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以上，请采纳。
∵cos(3x+2x)=cos3xcos2x-sin3xsin2x
cos(3x-2x)=cos3xcos2x+sin3xsin2x
∴cos(3x+2x)+cos(3x-2x)=2cos3xcos2x
即cos5x+cosx=2cos3xcos2x
∴cos3xcos2x=1/2*(cosx+cos5x)
∴∫cos3xcos2xdx
=1/2*∫(cosx+cos5x)dx
=1/2*sinx+1/10*sin5x+C
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绛旓細锛1锛変护x=tant锛屽垯dx=sec^2tdt 鍘熷紡=鈭玪n(tant+sect)/sect*sec^2tdt =鈭玪n(tant+sect)*sectdt =鈭玪n(tant+sect)d[ln(tant+sect)]=(1/2)*[ln(tant+sect)]^2+C =(1/2)*[ln(x+鈭(1+x^2)]^2+C锛屽叾涓瑿鏄换鎰忓父鏁 锛2锛夊師寮=鈭玡^(sin2x-2x)*sin^2xdx =(1/2)*鈭玡...
绛旓細1.鈭玸ec³xdx锛濃埆secxd(tanx)锛漷anxsecx锛嶁埆tan²xsecxdx 锛漷anxsecx锛嶁埆(sec²x-1)secxdx 锛漷anxsecx锛嶁埆sec³xdx锛嬧埆secxdx 鏁寸悊寰 鈭玸ec³xdx锛漑tanxsecx锛嬧埆secxdx]/2锛婥 鍏朵腑鏄撴眰鈭玸ecxdx.(浠庣暐)2.杩炵画浣跨敤鍒嗛儴绉垎娉(鎶奵os2x鍑戝井鍒,浣垮緱x²鍦ㄦ眰瀵间笅闄...
绛旓細瑙ｏ細璁綟(x)=cosx/x鏄痜(x)鐨勪竴涓鍘熷嚱鏁,鍒檉(x)=(-x*sinx-cosx)/x^2,f(x/2)=[(-x/2)*sin(x/2)-cos(x/2)]/(x^2/4)鈭玿f鈥(x/2)dx=2鈭玿df(x/2)=2xf(x/2)-2鈭玣(x/2)dx =2xf(x/2)-4鈭玣(x/2)d(x/2)=2xf(x/2)-4F(x/2)+C =-8[(x/2)*sin(x/...
绛旓細鈭1/(1+e^x)路dx =鈭(1+e^x-e^x)/(1+e^x)路dx =鈭1路dx-鈭玡^x/(1+e^x)路dx =x-鈭1/(1+e^x)路d(1+e^x)=x-ln(1+e^x)+C
绛旓細瑙 鈭玿cos(x²+1)dx =1/2鈭玞os(x²+1)d(x²+1)=1/2鈭玞osudu =1/2sinu+C =1/2sin(x²+1)+C
绛旓細鈭玸inxlntanxdx 璁綰=lntanx dv锛漵inxdx 鍒檇u=1/tanx*sec^2x,v=-cosx.鏍规嵁鍒嗛儴绉垎鍏紡:鍘熷紡=-cosx*lntanx+鈭玞osx*1/tanx*sec^2xdx =-cosx*lntanx+鈭玞scxdx =-cosx*lntanx-ln(cscx+cotx)+C 娉:ln(cscx+cotx)涓殑鎷彿瀹為檯涓婃槸缁濆鍊肩鍙,鏃犲洜娉曟墦鍑烘潵,鏁呯敤鎷彿琛ㄧず....
绛旓細瑙:鈭玿lnxdx=(1/2)*鈭玪nxdx^2 (姝ら鑰冭檻鍒嗛儴绉垎锛鍏堢Н骞傚嚱鏁)=1/2*[(x^2)*(lnx)-鈭玿^2*1/xdx]=1/2*[x^2*lnx-鈭玿dx]=1/2*x^2*lnx- 1/4*x^2+C锛孋涓轰换鎰忓父鏁般傗埆e^xcosxdx=鈭玞osxde^x (姝ら鑰冭檻鍒嗛儴绉垎锛屽厛绉寚鏁板嚱鏁)=cosx*e^x+鈭玡^x*sinxdx=cosx*e^x+鈭...
