Integration using exponent

What could be the techniques we need to use to solve this integration

$\displaystyle \int\tan^2\theta\frac{\sin^2(\sec\theta\tan\theta)}{\sec^2\theta}d\theta \tag1$?

How do I convert this in to a solvable standard form ?

Solutions Collecting From Web of "Integration using exponent"

$\int\tan^2\theta\dfrac{\sin^2(\sec\theta\tan\theta)}{\sec^2\theta}d\theta$

$=\int\sin^2\theta~\sin^2(\sec\theta\tan\theta)~d\theta$

$=\int\dfrac{(1-\cos^2\theta)(1-\cos(2\sec\theta\tan\theta))}{2}d\theta$

$=\int\dfrac{(\cos^2\theta-1)}{2}\sum\limits_{n=1}^\infty\dfrac{(-1)^n4^n\sec^{2n}\theta\tan^{2n}\theta}{(2n)!}d\theta$

$=\int\sum\limits_{n=1}^\infty\dfrac{(-1)^n2^{2n-1}\sec^{2n-2}\theta\tan^{2n}\theta}{(2n)!}d\theta-\int\sum\limits_{n=1}^\infty\dfrac{(-1)^n2^{2n-1}\sec^{2n}\theta\tan^{2n}\theta}{(2n)!}d\theta$

$=-\int\tan^2\theta~d\theta+\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n2^{2n+3}\sec^{2n+2}\theta\tan^{2n+4}\theta}{(2n+4)!}d\theta+\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n2^{2n+1}\sec^{2n+2}\theta\tan^{2n+2}\theta}{(2n+2)!}d\theta$

$=\int(1-\sec^2\theta)~d\theta+\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n2^{2n+3}\sec^{2n}\theta\tan^{2n+4}\theta}{(2n+4)!}d(\tan\theta)+\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n2^{2n+1}\sec^{2n}\theta\tan^{2n+2}\theta}{(2n+2)!}d(\tan\theta)$

$=\int(1-\sec^2\theta)~d\theta+\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n2^{2n+3}(1+\tan^2\theta)^n\tan^{2n+4}\theta}{(2n+4)!}d(\tan\theta)+\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n2^{2n+1}(1+\tan^2\theta)^n\tan^{2n+2}\theta}{(2n+2)!}d(\tan\theta)$

$=\int(1-\sec^2\theta)~d\theta+\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^n2^{2n+3}C_k^n\tan^{2k}\theta\tan^{2n+4}\theta}{(2n+4)!}d(\tan\theta)+\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^n2^{2n+1}C_k^n\tan^{2k}\theta\tan^{2n+2}\theta}{(2n+2)!}d(\tan\theta)$

$=\int(1-\sec^2\theta)~d\theta+\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^n2^{2n+3}n!\tan^{2n+2k+4}\theta}{(2n+4)!k!(n-k)!}d(\tan\theta)+\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^n2^{2n+1}n!\tan^{2n+2k+2}\theta}{(2n+2)!k!(n-k)!}d(\tan\theta)$

$=\theta-\tan\theta+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^n2^{2n+3}n!\tan^{2n+2k+5}\theta}{(2n+4)!k!(n-k)!(2n+2k+5)}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^n2^{2n+1}n!\tan^{2n+2k+3}\theta}{(2n+2)!k!(n-k)!(2n+2k+3)}+C$