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

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#

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#.

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