Question

Вычислить интеграл:

 

Answer 1

$\int limits_{frac{pi}{18}}^{frac{pi}{4}} {cos 4x}, dx=(*)\ t=4x\ dt=4, dx\ int limits_{frac{pi}{18}}^{frac{pi}{4}} frac{1}{4}{cos t}, dt=\ frac{1}{4}int limits_{frac{pi}{18}}^{frac{pi}{4}} {cos t}, dt=\ frac{1}{4}Big[sin tBig]_{frac{pi}{18}}^{frac{pi}{4}}=\ (*)=frac{1}{4}Big[sin 4xBig]_{frac{pi}{18}}^{frac{pi}{4}}=\ frac{1}{4}(sin (4cdotfrac{pi}{4})-sin (4cdotfrac{pi}{18}))=\frac{1}{4}(sin pi-sin frac{2pi}{9})=\frac{1}{4}(0-sin frac{2pi}{9})=\-frac{1}{4}sin frac{2pi}{9}approx-0,16$