Question

integrate(sin2x+cos2x) w.r.t.x

Answer 1

your ans is -1/2cos2x+1/2sin2x
..... tq.....
now step wise solution
=integration of sin2x dx+integration of cos2x
=-1/2sin2x+1/2cos2x+c


we know that
sin ax=-1/a cosax
incase of derivative
sin ax= a cos ax
and integration of cos ax= 1/a sin ax
incase of derivative
cos ax=-a sin ax
........
you want to expert in integration so you first practise all the formula and their tacknic
....... thank u

Answer 2

Hey There!!!

We can conclude some things as follows:   

$\frac{d}{dx} \sin ax = a\,\cos ax \,\,\implies \int \cos ax \, dx = \frac{\sin ax}{a} + c \\ \\ \\ \frac{d}{dx} \cos ax = -a\,\sin ax \,\,\implies \int \sin ax \, dx = -\frac{\cos ax}{a} + c$ 


Now, coming to your question: 

$\int \sin 2x + \cos 2x \,dx \\ \\ = \boxed{-\frac{\cos 2x}{2} + \frac{\sin 2x}{2} + c}$

Hope this helps
Purva
Brainly Community