# 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 | (Level 1) Valedictorian

Posted on

`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`

``

pramodpandey | College Teacher | (Level 3) Valedictorian

Posted on

Given

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

`Let`

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

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

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

`=-intsqrt(9-t^2)dt+int9/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=-(9/2)(sin(x)cos(x)+x)+9x+c`

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

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

``

``