hello, I need a step by step help in this question integral sec^2 theta tan^2 theta / sqrt( 9-tan^2 theta) dtheta thanks in advance

oldnick | Student

`int (sec^2 theta tan^2 theta)/(sqrt(9-tan^2 theta)) d theta`  `=int (tan^2theta)/(sqrt(9-tan^2theta)) d tan theta`


developing by parts:


`= int tan theta xx (tan theta)/(sqrt(9-tan^2 theta)) d tan theta=`

`=-tan theta sqrt(9- tan^2theta)+int (sqrt(9-tan^2theta)) d tan theta =`

`-tan theta sqrt(9-tan^2 theta)+9/2arcsin((tan theta)/3)+1/2 tan theta sqrt(9-tan^2 theta)+c=`

`= 1/2[9arcsin((tan theta)/3)-tan theta sqrt(9-tan^2 theta)] +c`


thak you so much
so much easier to understand!
thanks a millions!

pramodpandey | Student


`I=int(sec^2(theta)tan^2(theta))/sqrt(9-tan^2(theta))d theta`


`tan(theta)=t, sec^2(theta)d theta=dt`

`I=int t^2/sqrt(9-t^2)dt`



`If `

we substitute  t=3sinx ,dt =3cosx , so first integrand become

9cos^2(x)=9{cos(2x)+1}/2  and  second integral 9 so  we can intgrate


`I=-(1/2)(t sqrt(9-t^2)+(9/2)sin^(-1)(t/3)+c`