फलन `int(dx)/(5+4 sin x)` का x के सापेक्ष समाकलन कीजिए ।

Explanation

माना `I=int(dx)/(5+4 sinx)`
`=int(dx)/(5(sin^(2)x//2+cos^(2)x//2)+8sin x//2 cos x//2)`
अंश हर को `cos^(2)x//2` से भाग देने पर ,
`I=int(sec^(2)x//2dx)/(5(1+tan^(2)x//2)+8tanx//2)`
`=(1)/(5)int(sec^(2)x//2dx)/(1+tan^(2)x//2+(8)/(5)tanx//2)`
माना `tan x//2 = t therefore sec^(2)x//2 dx=2dt`
`therefore" "I=(2)/(5)int(dt)/(1+t^(2)+(8)/(5)t)=(2)/(5)int(dt)/(1+(t^(2)+(8)/(5)t+(16)/(25)-(16)/(25)))`
`" "` (पूर्ण वर्ग बनाने पर )
`=(2)/(5)int(dt)/((1-(16)/(25))+(t+(4)/(5))^(2))=(2)/(5)int(dt)/(((3)/(5))^(2)+(t+(4)/(5))^(2))`
माना `t+(4)/(5)=u" "rArr" "dt=du" तथा " (3)/(5)=a`
`therefore" "I=(2)/(5)int(du)/(a^(2)+u^(2))=(2)/(5a)tan^(-1)((u)/(a))`
`=(2)/(5xx(3)/(5))tan^(-1)[(5t+4)/(3)]=(2)/(3)tan^(-1)[(5tan x//2 +4)/(3)]`
द्वितीय विधि -
`I=int(dx)/(5+4sinx)=int(dx)/(5+(4xx2tanx//2)/(1+tan^(2)x//2))`
`=int((1+tan^(2)x//2)dx)/(5(1+tan^(2)x//2)+8 tan x//2)`
`=(1)/(5)int(sec^(2)x//2dx)/(1+tan^(2)x//2+(8)/(5)tanx//2)`
`=(1)/(5)int(sec^(2)x//2dx)/(1+(tan^(2)x//2+(8)/(5)tanx//2+(16)/(25)-(16)/(25)))`
`" "` (पूर्ण वर्ग बनाने पर )
`=(1)/(5)int(sec^(2)x//2dx)/(((3)/(5))^(2)+(tanx//2+(4)/(5))^(2))`
माना `" "(3)/(5)=a` तथा `tan x//2+(4)/(5)=u" "therefore" "sec^(2).(x)/(2)dx=2du`
`therefore" "I=(2)/(5)int(du)/(a^(2)+u^(2))=(2)/(5a)tan^(-1)((u)/(a))`
`=(2)/(3)tan^(-1)[(5tanx//2+4)/(3)]`