Question

Integral of 2cos^2 (2x) sin 4x dx

Answer 1

To Find

$∫(2 {cos }^{2}(2x).sin(4x)) \: dx$

.

Solution

$= 2∫ {cos}^{2} (2x).sin(4x)dx$

$= 2∫ {cos}^{2} (2x).2sin(2x)cos(2x)dx$

$= 4∫ {cos}^{3} (2x).sin(2x)dx$

NOW Let; cos(2x) = t

(-2)sin(2x)dx = dt

and; sin(2x)dx = dt ÷ (-2)

$= - \frac{4}{2} ∫t ^{3} dt$

$= - 2 \frac{ {t}^{4} }{4}$

$= - \frac{ {t}^{3} }{2}$

$= - \frac{cos {}^{4} (2x)}{2} + C ; C ∈ ℝ$