Đáp án: I=∫limits (sin^4xcos^2x) dx - Lời giải Ta cósin^4xcos^2x=sin^2x(sinxcosx)^2=(1-cos2x/2)((1/2)sin2x)^2=(1/8)sin^22x-(1/8)sin^22xcos2x=(1/8)(1-cos4x/2)-(1/8)sin^22xcos2x=>I=(1/16)∫ dx-(1/16) ∫ (cos4x) dx-(1/8) ∫ (sin^2xcos2x) dx=(1/16)x-(1/64)sin4x-(1/48)sin^32x+C

Lời giải Ta cósin^4xcos^2x=sin^2x(sinxcosx)^2=(1-cos2x/2)((1/2)sin2x)^2=(1/8)sin^22x-(1/8)sin^22xcos2x=(1/8)(1-cos4x/2)-(1/8)sin^22xcos2x=>I=(1/16)∫ dx-(1/16) ∫ (cos4x) dx-(1/8) ∫ (sin^2xcos2x) dx=(1/16)x-(1/64)sin4x-(1/48)sin^32x+C

Giải thích các bước giải Ta cósin^4xcos^2x=sin^2x(sinxcosx)^2=(1-cos2x/2)((1/2)sin2x)^2=(1/8)sin^22x-(1/8)sin^22xcos2x=(1/8)(1-cos4x/2)-(1/8)sin^22xcos2x=>I=(1/16)∫ dx-(1/16) ∫ (cos4x) dx-(1/8) ∫ (sin^2xcos2x) dx=(1/16)x-(1/64)sin4x-(1/48)sin^32x+C

KimAnh859

2 năm trước

$I=$$\int\limits {sin^4x.cos^2x} \, dx$

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Phuc Nguyen

2 năm trước

Lời giải:

Ta có:
$sin^4x.cos^2x=sin^2x.(sinx.cosx)^2=\frac{1-cos2x}{2}.(\frac{1}{2}sin2x)^2$
$=\frac{1}{8}sin^22x-\frac{1}{8}sin^22x.cos2x$
$=\frac{1}{8}.\frac{1-cos4x}{2}-\frac{1}{8}.sin^22x.cos2x$
=>$I=\frac{1}{16}$$\int\limits {} \, dx-\frac{1}{16}$ $\int\limits {cos4x} \, dx-\frac{1}{8}$ $\int\limits {sin^2x.cos2x} \, dx$
$=\frac{1}{16}x-\frac{1}{64}sin4x-\frac{1}{48}sin^32x+C$