Question

Solve using trig sub: `\int(\sec^2\theta\tan^2\theta)/\sqrt(9-\tan^2\theta)d\theta`?

Symbolab has it with u-sub, so...

(Unless you have to do u-sub to get the trig-sub?)

Answer 1

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

Explanation:

Substitute `tan theta = 3 sin phi`, to get `sec^2 theta\ d theta = 3 cos phi\ d phi` so that

`\int(\sec^2\theta\tan^2\theta)/\sqrt(9-\tan^2\theta)d\theta = int {(3 sin phi)^2 3 cos phi\ d phi}/{3 cos phi}`
`qquad = 9 int\ sin^2 phi\ d phi = 9/2 int \ (1-cos(2 phi))\ d phi`
`qquad = 9/2 phi-9/4 sin(2 phi)+C`
`qquad = 9/2 phi -9/2 sin phi cos phi=C`
`qquad = 9/2 sin^-1(1/3 tan theta ) - 9/2 1/3 tan theta sqrt(1-(1/3 tan theta)^2)+C`
`qquad = 9/2 sin^-1(1/3 tan theta ) - 1/2 tan theta sqrt(9-tan^2 theta)+C`

Answer 2

`-1/2tanthetasqrt(9-tan^2theta)+9/2arc sin(1/3*tantheta)+C`.

Explanation:

As Second Method, let us solve this Problem without using

any Trigonometric Substitution.

If we take `tantheta=t, then, sec^2thetad theta=dt`.

`:. I=int(tan^2thetasec^2theta)/sqrt(9-tan^2theta)d theta`,

`=intt^2/sqrt(9-t^2)dt`,

`=-int{(9-t^2)-9}/sqrt(9-t^2)dt`,

`=-int{(9-t^2)/sqrt(9-t^2)-9/sqrt(9-t^2)}dt`,

`=-intsqrt(3^2-t^2)dt+9int1/sqrt(9-t^2)dt`,

`=-{t/2sqrt(9-t^2)+9/2arc sin(t/3)}+9arcsin(t/3)`,

`=-t/2sqrt(9-t^2)+9/2arc sin(t/3)`.

`because t=tantheta,`

`I=-1/2tanthetasqrt(9-tan^2theta)+9/2arc sin(1/3*tantheta)+C`.

Feel the Joy of Maths.!