Question

求定积分的值。

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Answer 1

积分过程及结果如下

Answer 2

(1). ∫<-1,1>sinxtanxdx=∫<-1,1>(sin²x/cosx)dx=∫<-1,1>[(1-cos²x)/cosx]dx
=∫<-1,1>secxdx-∫<-1,1>cosxdx=[ln∣secx+tanx∣-sinx]∣<-1,1>
=[ln(sec1+tan1)-sin1]-[ln(sec1-tan1)+sin1]=ln[(sec1+tan1)/(sec1-tan1)]-2sin1;
(2). ∫<-1,1>x²√(1-x)dx【√(1-x)=u；1-x=u²；x=1-u²；dx=-2udu；x=-1，u=√2;
x=1，u=0；】=-2∫<√2，0>(1-u²)²•u²du=2∫<0,√2>(1-2u²+u^4)•u²du
=2∫<0,√2>(u²-2u^4+u^6)du=2[(1/3)u³-(2/5)u^5+(1/7)u^7]<0,√2>
=2[(2/3)√2-(8/5)√2+(8/7)√2]=(44/105)√2;