JAMB - Mathematics (2011 - No. 36)
Find \(\int\) cos4 x dx
\(\frac{3}{4}\) sin 4x + k
-\(\frac{1}{4}\) sin 4x + k
-\(\frac{3}{4}\) sin 4x + k
\(\frac{1}{4}\) sin 4x + k
Giải thích
\(\int\) cos4 x dx
let u = 4x
\(\frac{dy}{dx}\) = 4
dx = \(\frac{dy}{4}\)
\(\int\)cos u. \(\frac{dy}{4}\) = \(\frac{1}{4}\)\(\int\)cos u du
= \(\frac{1}{4}\) sin u + k
= \(\frac{1}{4}\) sin4x + k
let u = 4x
\(\frac{dy}{dx}\) = 4
dx = \(\frac{dy}{4}\)
\(\int\)cos u. \(\frac{dy}{4}\) = \(\frac{1}{4}\)\(\int\)cos u du
= \(\frac{1}{4}\) sin u + k
= \(\frac{1}{4}\) sin4x + k