你在运算不定积分时，对待积分项的分子和分母转化成都含有cost-1的因式，然后将相同因式约掉，是完全可以的，因为待积分函数具有乘法性质的交换律和分配律的。
=cos(x-1)*x^2/2+∫sin(x-1)*x^2/2)+c =cos(x-1)*x^2/2+sin(x-1)*x^2/2-∫xcos(x-1)dx+c 到此方程两边都有∫xcos(x-1)dx，移项得：2∫xcos(x-1)dx=cos(x-1)*x^2/2+sin(x-1)*x^2/2+c 化简得∫xcos(x-1)dx=√2/4[sin(x-1+π/4)]+c ...
∫cos^(-1)x dx =∫secxdx =ln|secx+tanx|+c ∫cos√x dx 令√x=t,x=t²,dx=2tdt 原式=2∫tcostdt =2∫tdsint =2tsint-2∫sintdt =2tsint+2cost+c =2√xsin√x+2cos√x+c
= (1/2)∫[1/(cosx-1) - 1/(cosx+1)]dcosx = (1/2){∫[1/(cosx-1)]d(cosx-1) - ∫[1/(cosx+1)]d(cosx+1)} = (1/2)[ln|cosx-1| - ln|cosx+1|] + C = (1/2)ln|(cosx-1)/(cosx+1)| + C = (1/2)ln|(cosx-1)^2/[(cosx)^2-1)]| + C = (1...
1/cosx dx = cosx / (1-sin^2(x)) dx = 1/2(1-sinx) + 1/2(1+sinx) dsinx = ln( (1+sinx)/(1-sinx) ) / 2
∫xcos(x-1)dx=∫cos(x-1)d(x^2/2)=cos(x-1)*x^2/2-∫xd(cos(x-1)+c=cos(x-1)*x^2/2+∫sin(x-1)d(x^2/2)+c=cos(x-1)*x^2/2+∫sin(x-1)*x^2/2)+c=cos(x-1)*x^2/2+sin(x-1)*x^2/2-∫xcos(x-1)dx+c到此方程两边都有∫xc... 解析看不懂?免费查看同类...
求一道积分 ∫(cos(x))^-1 dx 答案 用万能代换 cosx=[1-(tanx/2)^2]/[1+(tanx/2)^2] 令t=tanx/2 则dx=2/(1+t^2)dt∫(cos(x))^-1 dx =∫[(1+t^2)/(1-t^2)]*[2/(1+t^2)]dt =∫2/(1+t)(1-t)dt =∫[1/(t+1)]-[1/(t-1)]dt =ln(t+1)-ln(t-1)+C =...
【答案】：∫cos(1-x)dx=-∫cos(1-x)d(1-x)=-sin(1-x)+c
∫cos^(-1)x dx =∫secxdx =ln|secx+tanx|+c ∫cos√x dx 令√x=t,x=t²,dx=2tdt 原式=2∫tcostdt =2∫tdsint =2tsint-2∫sintdt =2tsint+2cost+c =2√xsin√x+2cos√x+c 分析总结。 xdx∫cos√xdx扫码下载作业帮拍照答疑一拍即得答案解析查看更多优质解析举报∫cos结果...
用万能代换 cosx=[1-(tanx/2)^2]/[1+(tanx/2)^2] 令t=tanx/2 则dx=2/(1+t^2)dt∫(cos(x))^-1 dx =∫[(1+t^2)/(1-t^2)]*[2/(1+t^2)]dt =∫2/(1+t)(1-t)dt =∫[1/(t+1)]-[1/(t-1)]dt =ln(t+1)-ln(t-1)+C =ln[(t+1)/(t-... 解析看不懂?免费查...