∫dt16+6sint
=12∫dt8+3sint
Let sint=1−tan2t21+tan2t2
=12∫1+tan2t28+8tan2t2+3−3tan2t2dt
=12∫sec2t28+8tan2t2+3−3tan2t2dt
Let u=tant2⇒du=12sec2t2dt
⇒2du=sec2t2dt
=12∫2du11+5u2
=12∫2du15((√55)2+u2)
=5√55tan−1(u√55)+c where c is the constant of integration.
=5√55tan−1⎛⎜
⎜⎝tant2√55⎞⎟
⎟⎠+c where u=tant2