MHB Can Trig Substitution be used to Solve Trigonometric Integrals?

[karush]

{8.7.4 whit} nmh{962}
$$\displaystyle
I=\int \sin\left({t}\right) \cos\left({2t}\right) \ dt $$
substitution

$u=\cos\left({t}\right)
\ \ \ du=-\sin\left({t}\right) \ dt
\ \ \ \cos\left({2t}\right) =2\cos^2 \left({t}\right)-1 $
$\displaystyle
I=\int \left(1-2u^2 \right) du
\implies u-\frac{2u^3}{3}+C$
Back substittute u

$\displaystyle \cos\left({t}\right)-\frac{2\cos^3(t) }{3}+C $
TI gave a different answer but might be alternative form

 

[MarkFL]

Differentiating your anti-derivative w.r.t $t$ gets you back to the original integrand, so it is correct. (Yes)

 

[karush]

Guess I'm slowly getting out of square one with
Integral substitutions