Question

Интеграл cosx/2*cosx/3dx

Answer 1

Решение на фотографии:

Answer 2

To find the integral of cos(x/2) * cos(x/3) with respect to x, we can use the trigonometric identity:

cos(A) * cos(B) = (1/2) * [cos(A + B) + cos(A - B)]

In our case, A = x/2 and B = x/3, so we can rewrite the integral as follows:

∫(cos(x/2) * cos(x/3)) dx = ∫((1/2) * [cos(x/2 + x/3) + cos(x/2 - x/3])) dx

Now, let's integrate each term separately:

∫cos(x/2 + x/3) dx:

To integrate cos(x/2 + x/3), we can use the substitution method. Let u = x/2 + x/3, then du = (1/2 + 1/3) dx = (5/6) dx.

∫cos(x/2 + x/3) dx = ∫cos(u) * (6/5) du = (6/5) * ∫cos(u) du

The integral of cos(u) with respect to u is sin(u). So, the above becomes:

(6/5) * sin(u) = (6/5) * sin(x/2 + x/3)

∫cos(x/2 - x/3) dx:

To integrate cos(x/2 - x/3), we can use the substitution method again. Let v = x/2 - x/3, then dv = (1/2 - 1/3) dx = (1/6) dx.

∫cos(x/2 - x/3) dx = ∫cos(v) * (6/1) dv = 6 * ∫cos(v) dv

The integral of cos(v) with respect to v is sin(v). So, the above becomes:

6 * sin(v) = 6 * sin(x/2 - x/3)

Now, putting it all together:

∫(cos(x/2) * cos(x/3)) dx = (1/2) * [cos(x/2 + x/3) + cos(x/2 - x/3)] dx = (1/2) * [(6/5) * sin(x/2 + x/3) + 6 * sin(x/2 - x/3)]

So, the integral of cos(x/2) * cos(x/3) with respect to x is:

(1/2) * [(6/5) * sin(x/2 + x/3) + 6 * sin(x/2 - x/3)] + C

where C is the constant of integration.