Question

Integrate $\int -{\cos }^{2}t$ dt

Answer 1

$\int -{\cos }^{2}tdt$

$-\int {\cos }^{2}tdt$

[${\cos }^2\theta =\frac{1+\cos 2\theta }{2}$]

$-\int \frac{1+\cos 2t}{2}dt$

$-\frac{1}{2}\int dt-\frac{1}{2}\int \cos 2tdt$

$2t=u$

$2dt=du$

$\cos 2t=2\sin 2t$

So, using the above value for $\cos 2t$

$-\frac{t}{2}-\frac{1}{4}\sin 2t+C$(Ans)