∫(cos√x)^2dx

∫(cos√x)^2dx

Let t = root x, x = T ^ 2, DX = 2tdt
Original formula = ∫ (cost) ^ 2 * 2tdt = ∫ (cos2t + 1) / 2 * 2tdt
=∫tcos2tdt+∫tdt
=1/2t*sin2t-1/2∫sin2tdt+t^2
=t/2*sin2t+1/4cos2t+t^2+C
Finally, replace T with root X
I can't remember the formula. I don't know if it's right. Please correct it

Calculus to solve ∫ e ^ x cos (e ^ x) DX=

∫e^x cos(e^x)dx
=∫cos(e^x)d(e^x)
=sin(e^x)+C

Find ∫ arctan (e ^ x) / (e ^ x) DX?

a=e^x
x=lna
dx=da/a
So the original formula = ∫ arctana * DA / A ²
=-∫arctanad(1/a)
=-arctana/a+∫1/a*darctana
=-arctana/a+∫1/a*da/(1+a ²)
∫1/a*da/(1+a ²)
=∫(1+a ²- a ²)/ a(a ²+ 1)da
=∫[1/a-a/(a ²+ 1)]da
=∫1/ada-∫a/(a ²+ 1)da
=lna-1/2∫d(a ²+ 1)/(a ²+ 1)
=lna-1/2*ln(a ²+ 1)+C
So the original formula = - arctana / A + LNA-1 / 2 * ln (a ²+ 1)+C
=-arctan(e^x)/e^x+x-1/2*ln(e^2x+1)+C

On calculus problems ∫ DX / (e ^ x + e ^ - x)

dx／(e^x+e^-x)
Multiply both up and down by e ^ X
e^xdx/[(e^x)^2+1]
And e ^ xdx = D (e ^ x)
So it becomes
d(e^x)/[(e^x)^2+1]
Let t = e ^ x
dt/(t^2+1)
Original integral = arctan (T) + C = arctan (e ^ x) + C

∫arctan(1+√x)dx

∫ arctan (1 + √ x) DX conversion t = arctan (1 + √ x), (tant - 1) ^ 2 = x = ∫ t d (tant-1) ^ 2 = t (tant-1) ^ 2 - ∫ (tant-1) ^ 2 DT = t (tant-1) ^ 2 - ∫ (sint cost) ^ 2 / cos ^ 2T DT = t (tant-1) ^ 2 - ∫ (1-2sintcost) / cos ^ 2T DT = t (tant-1) ^ 2 - ∫ 1

Find the integral of X * sin (a * x) DX and X * cos (a * x) DX respectively Although I already know the answer, I want to know the process of seeking Again, can this integral be found in MATLAB? I can't seem to work it out with mathmatica

int('x*sin(a*x)','x')
ans =
1/a^2*(sin(a*x)-a*x*cos(a*x))
int('x*cos(a*x) ','x')
ans =
1/a^2*(cos(a*x)+a*x*sin(a*x))