sin-2x-pi-4-dx-

Question Number 146285 by mathdanisur last updated on 12/Jul/21

\int s i n (2 x - \frac{π}{4}) d x = ?

Commented by gsk2684 last updated on 12/Jul/21

w h a t a r e y o u s t u d y i n g n o w ?

Commented by mathdanisur last updated on 12/Jul/21

t h e s e

Answered by gsk2684 last updated on 12/Jul/21

u s e t h e f o r m u l a

\int \sin (a x + b) d x = - \frac{\cos (a x + b)}{a}

Commented by Ar Brandon last updated on 12/Jul/21

I = \int \sin (ax + b) dx

Let u = ax + b \Rightarrow du = adx

I = \int \sin (ax + b) dx = \frac{1}{a} \int \sin (u) du

= - \frac{\cos (u)}{a} + C = - \frac{\cos (ax + b)}{a} + C

Commented by mathdanisur last updated on 13/Jul/21

t h a n k s S e r

Answered by puissant last updated on 12/Jul/21

= \int \sin (2 x) \cos \frac{π}{4} - \cos (2 x) \sin \frac{π}{4} dx

= \frac{\sqrt{2}}{2} \int (\sin (2 x) - \cos (2 x)) dx

= \frac{\sqrt{2}}{2} [- \frac{1}{2} \cos (2 x) - \frac{1}{2} \sin (2 x)] + c

I = - \frac{\sqrt{2}}{4} (\cos (2 x) + \sin (2 x)) + c . .

Commented by mathdanisur last updated on 12/Jul/21

T h a n k y o u S e r

Answered by Ar Brandon last updated on 12/Jul/21

I = \int \sin (2 x + \frac{π}{4}) dx

= - \frac{1}{2} \cos (2 x + \frac{π}{4}) + C

Commented by mathdanisur last updated on 13/Jul/21

t h a n k s S e r