`intsqrt(tanx)dx` का मान ज्ञात कीजिए ।

Explanation

माना `tanx=t^(2) rArr sec^(2)xdx=2tdt`
`therefore" "dx=(2t)/(sec^(2)x)dt=(2t)/(1+tan^(2)x)dt=(2t)/(1+t^(4))dt`
`therefore" "intsqrt(tanx )dx=intt.(2t)/(1+t^(4))dt=int(2t^(2))/(1+t^(4))dt`
`=int((t^(2)+1)+(t^(2)-1))/(1+t^(4))dt`
`=int(t^(2)+1)/(1+t^(4))dt+int(t^(2)-1)/(1+t^(4))dt`
`=int(1+(1)/(t^(2)))/(t^(2)+(1_/(t^(2)))dt+int(1-(1)/(t^(2)))/(t^(2)+(1)/(t^(2)))dt`
`=int((1+(1)/(t^(2))))/((t-(1)/(t))^(2)+2)dt+nt(1-(1)/(t^(2)))/((t+(1)/(t))^(2)-2)dt`
प्रथम समाकलन में `t-(1)/(t)=u` अर्थात `(1+(1)/(t^(2)))dt=du` तथा
द्वितीय समाकलन में `t+(1)/(t)=v` अर्थात `(1-(1)/(t^(2)))dt=dv` रखने
पर
`intsqrt(tanx)dx=int(du)/(u^(2)+(sqrt2)^(2))+int(dv)/(v^(2)-(sqrt2)^(2))`
`=(1)/(sqrt2)tan^(-1)((u)/(sqrt2))+(1)/(2sqrt2)log((v-sqrt2)/(v+sqrt2))+c`
`=(1)/(sqrt2)tan^(-1)((t-1//t)/(sqrt2))+(1)/(2sqrt2)log((t+(1)/(t)-sqrt2)/(t+(1)/(t)+sqrt2))+c`
`=(1)/(sqrt2)tan^(-1)((t^(2)-1)/(sqrtt))+(1)/(2sqrt2)log((t^(2)-sqrt2t+1)/(t^(2)+sqrt2t+1))+c`
`=(1)/(sqrt2)tan^(-1)((tanx-1)/(sqrt(2 tanx)))+(1)/(2sqrt2)log((tanx-sqrt(2 tanx)+1)/(tanx+sqrt(2 tanx)+1))+c`