`int sin(t)sec^2(cos(t)) dt` Evaluate the indefinite integral.

You need to use the following substitution  `cos t=u,` such that:

`cos t=u=>-sin t dt= du => sin t dt = -du`

`int sin t*sec^2(cos t)dt = - int sec^2 u du`

`- int sec^2 u du = -tan u + c`

Replacing back  cos t for u yields:

`int sin t*sec^2(cos t)dt = -tan (cos t) + c`

Hence, evaluating the indefinite integral, yields `int sin t*sec^2(cos t)dt = -tan (cos t) + c.`