`int((tan theta +tan^(3) theta))/((1+ tan^(3) theta))d theta ` का मान ज्ञात कीजिए ।

Explanation

`int((tan theta +tan^(3) theta))/((1+ tan^(3) theta))d theta `
अब `(tan theta +tan^(3) theta)/(1+tan^(3) theta)=(tan theta(1+ tan^(2) theta))/((1+tan^(3) theta))=(tan theta sec^(2) theta)/((1+ tan^(3) theta))`
तब `int((tan theta+ tan^(3) theta))/((1+tan^(3) theta))d theta= int(tan theta sec^(2) theta)/( (1+tan^(3) theta))d theta" ...(1)"`
यदि `tan theta = t rArr sec^(2) theta d theta = dt`
अब समीकरण (1 ) से
`=int(t)/((1+t^(3)))dt=int(t)/((1+t)(1-t+t^(2)))dt`
आंशिक भिन्नो में व्यक्त करने पर
`(t)/((1+t)(1-t+t^(2)))=(A)/((1+t))+((Bt+C))/((1-t+t^(2)))`
`t=A(1-t+t^(2))+(Bt+c)(1+t)" ...(2)"`
यदि `" "t=-1" तब "A=-(1)/(3)`
व अब समीकरण (2 ) के दोनों पक्षों में t की घातों की तुलना करने पर
`A+B=0 rArr B=(1)/(3)`
व `A+C=0 rArr C=(1)/(3)`
अब `(t)/((1+t)(1-t+t^(2)))=(-1)/(3(1+t))+(((1)/(3)t+(1)/(3)))/((1-t+t^(2)))`
`int(t)/((1+t)(1-t+t^(2)))dt`
`=-(1)/(3)int(dt)/((1+t))+(1)/(6)int(2t)/((t^(2)-t+1))dt+(1)/(3)int(dt)/((t^(2)-t+1))`
`=-(1)/(3)int(dt)/((1+t))+(1)/(6)int((2t-1)+1)/((t^(2)-t+1))dt+(1)/(3)int(dt)/((t^(2)-t+1))`
`=-(1)/(3)int(dt)/((1+t))+(1)/(6)int((2t-1))/((t^(2)-t+1))dt+(1)/(2)int(dt)/((t^(2)-t+1))`
`=-(1)/(3)log(1+t)+(1)/(6)log(t^(2)-t+1)+(1)/(2)int(dt)/((t^(2)-t+(1)/(4))+(3)/(4))`
`=-(1)/(3)log(1+t)+(1)/(6)log(t^(2)-t+1)+(1)/(2)int(dt)/((t-(1)/(2))^(2)+((sqrt3)/(2))^(2))`
`=-(1)/(3)log(1+t)+(1)/(6)log(t^(2)-t+1)+(1)/(2).(2)/(sqrt3)tan^(-1).((t-(1)/(2)))/(((sqrt3)/(2)))+c`
`=-(1)/(3)log(1+t)+(1)/(6)log(t^(2)-t+1)+(1)/(sqrt3)tan^(-1)((2t-1)/(sqrt3))+c`
`=-(1)/(3)log(1+tan theta)+(1)/(6)log (tan^(2)theta - tan theta+1)+(1)/(sqrt3)tan^(-1)((2 tan theta-1)/(sqrt3))+c`