shengq_1021
幼苗
共回答了25个问题采纳率:92% 举报
(5+4cosx)/[ (2+cosx)^2 *sinx]
=-3/[(2+cosx)^2 *sinx] + 4/[ (2+cosx) *sinx]
∫dx/[ (2+cosx) *sinx]
=∫sinxdx/[ (2+cosx) *sinx^2]
=-∫dcosx/[ (2+cosx) *(1-cosx^2)]
cosx=t
=-∫dt/[(2+t)*(1-t^2)]
=-1/3 *∫[(2+t)+(1-t)]dt/[(2+t)*(1-t)*(1+t)]
=-1/3 *{∫dt/[(1-t)(1+t) + ∫dt/[(2+t)*(1+t)]
∫dt/[(1-t)(1+t)]
=1/2* {∫dt/(1+t) + ∫dt/(1-t)
=1/2* ln(1+t)-ln(1-t)
∫dt/[(2+t)*(1+t)]
=∫dt/(1+t) - ∫dt/(2+t)
=ln(1+t)-ln(2+t)
1/[(2+cosx)^2 *sinx]
=sinx/[(2+cosx)^2 *sinx^2]
∫sinxdx/[(2+cosx)^2*(1-cosx^2)
=-∫dcosx/[(2+cosx)^2*(1-cosx^2)]
cosx=t
-∫dt/[(2+t)^2 * (1-t)(1+t)]
=-∫[(2+t)-(1+t)]dt/[(2+t)^2 * (1-t)(1+t)]
=-∫dt/[(2+t)*(1-t^2) + ∫dt/[(2+t)^2 *(1-t)]
∫dt/[(2+t)^2 *(1-t)]
=1/3 * ∫[(2+t)+(1-t)]dt/[(2+t)^2 *(1-t)]
=1/3* ∫dt/[(2+t) *(1-t)] + 1/3* ∫dt/[(2+t)^2 ]
∫dt/[(2+t) *(1-t)]
=1/3 * ∫dt/(1-t) + 1/3* ∫dt/(2+t)
=1/3*[ln(2+t)-ln(1-t) ]
其他的请自己完成
1年前
2