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求x∧2Cos3√x的不定积分

令x=2sint ∫x^3√(4-x^2)dx =∫(2sint)^3×2cost×2costdt =32∫sin^3tcos^2tdt =-32∫sin^2tcos^2tdcost =-32∫(1-cos^2t)cos^2tdcost =-32∫(cos^2t-cos^4t)dcost =-32/3cos^3t+32/5cos^5t+C =-32/3[√(4-x^2)]^3+32/5[√(4-x^2)]^5+C =-32/3(4-x^2)√(4-...

∫x^2[cos(x/2)]^2dx =(1/2)∫x^2(1+cosx)dx =(1/2)∫x^2dx+(1/2)∫x^2cosxdx =(1/6)x^3+(1/2)∫x^2d(sinx) =(1/6)x^3+(1/2)x^2sinx-(1/2)∫sinxd(x^2) =(1/6)x^3+(1/2)x^2sinx-∫xsinxdx =(1/6)x^3+(1...

(x+sin(x))/(4cos(x)-4) 首先改写降次: xcos^4(x/2)/sin^3(x) = 1/8 xcot(x/2)csc^2(x/2) ... 注:csc(x)=1/sin(x) 换元积分:令u=x/2: dx=2du, ∫ 1/8 xcot(u)csc^2(u) 2du =1/8 * 2 * 2 ∫ ucot(u)csc^2(u) du 然后分部积分, =1/2 u(-1/2cot^2...

答: 设x=sint,-π/2

第一种思路比较好算 ∫ x • cos³x dx = ∫ x • (1 - sin²x) dsinx = ∫ x dsinx - ∫ x • sin²x dsinx = xsinx - ∫ sinx - (1/3)∫ x dsin³x = xsinx + cosx - (1/3)xsin³x + (1/3)∫ sin³x dx = xsi...

∫xcos(x/3)dx =3∫xdsin(x/3) =3xsin(x/3)-3∫sin(x/3)dx =3xsin(x/3)+9cos(x/3)+c

用万能代换,令t=tan(x/2),则cosx=(1-t²)/(1+t²),dx=2dt/(1+t²) ∫dx/(3+cosx) =2∫dt/(2t²+4) =∫dt/(t²+2) =√2/2arctan(t/√2)+C =√2/2arctan(tan(x/2)/√2)+C

可以用凑微分法如图计算。经济数学团队帮你解答,请及时采纳。谢谢!

(cos(x))^3*dx=(cos(x))^2*cosxdx=[1-(sin(x))^2]d(sinx(x))==> inf[(cos(x))^3,x]=sin(x)-(sin(x))^3/3+C

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