Hãy luôn nhớ cảm ơn và vote 5*
nếu câu trả lời hữu ích nhé!
`c)` `int_(pi/6)^(pi/4) (tan^2x-cot^2x)dx`
`=int_(pi/6)^(pi/4) (1+tan^2x-1)dx-int_(pi/6)^(pi/4) (1+cot^2x-1)dx`
`=int_(pi/6)^(pi/4) (1+tan^2x)dx-int_(pi/6)^(pi/4) 1dx-[int_(pi/6)^(pi/4) (1+cot^2x)dx-int_(pi/6)^(pi/4)1dx]`
`=tanx|_(pi/6)^(pi/4)-pi/12+cotx|_(pi/6)^(pi/4)+pi/12`
`=tan\ pi/4-tan\ pi/6+cot\ pi/4-cot\ pi/6`
`=(6-4sqrt3)/3`
`=2-4/3*sqrt3`
`=>` `{(a=2),(b=4),(c=3):}`
`=>` `a+b+c=2+4+3=9`
`=>` Đúng
`d)` `int_0^(pi/3) |cos^2x-sinxcosx|dx`
Xét: `cos^2x-sinxcosx=0`
`<=>` `(1+cos 2x)/2=1/2sin 2x`
`<=>` `sin 2x-cos 2x=1`
`<=>` `sin(2x-pi/4)=sin\ pi/4`
`<=>` `[(2x-pi/4=pi/4+k2pi),(2x-pi/4=(3pi)/4+k2pi):}`
`<=>` `[(x=pi/4+kpi),(x=pi/2+kpi):} quad (k in ZZ)`
Vì `x in [0; pi/3]=>x=pi/4=>{( x in [0; pi/4]=>|cos^2x-sinxcosx|=cos^2x-sinxcosx),(x in [pi/4; pi/3]=>|cos^2x-sinxcosx=-(cos^2x-sinxcosx)):}`
Khi đó: `int_0^(pi/3) |cos^2x-sinxcosx|dx=int_0^(pi/4) |cos^2x-sinxcosx|dx+int_(pi/4)^(pi/3) |cos^2x-sinxcosx|dx`
`=int_0^(pi/4) (cos^2x-sinxcosx)dx-int_(pi/4)^(pi/3) (cos^2x-sinxcosx)dx`
`=pi/8+1/24(-9+3sqrt3+pi)`
`=3/8+pi/12-1/8*sqrt3`
`=>` `{(a=3/8),(b=1/12),(c=1/8):}`
`=>` `a-c+b=1/3`
`=>` Sai
Hãy giúp mọi người biết câu trả lời này thế nào?
Bảng tin