by equating the coefficients of sin x and cos x , or otherwise, find constants A and B satisfying the identity.
A(2sinx + cosx) + B(2cosx - sinx) = sinx + 8cosx
I got A = 2, B = 3, which the answers said were correct.
However, another part of the question asked
hence integrate
sinx + 8cosx
-------------dx
2sinx + cosx
and the answers were 2x + 3Ln (2sinx + cosx) + C
Could you explain how to get this?
Thanks in advance !!! :)
use what the first part showed. Using c for cos and s for sin, you showed that
s+8c = 2(2s+c) + 3(2c-s)
so,
(s+8c)/(2s+c)
= [2(2s+c) + 3(2c-s)]/(2s+c)
= 2 + 3(2c-s)/(2s+c)
now, you have ∫2 dx + ∫3du/u
integral is
2x + 3lnu
= 2x + 3ln(2s+c) + C
To find constants A and B in the given expression, A(2sinx + cosx) + B(2cosx - sinx) = sinx + 8cosx, you need to equate the coefficients of sinx and cosx on both sides of the equation.
On the left side, you have 2A for the coefficient of sinx and A - B for the coefficient of cosx. On the right side, you have 1 for the coefficient of sinx and 8 for the coefficient of cosx.
Equating the coefficients, you get two equations:
2A = 1 ---(1)
A - B = 8 ---(2)
From equation (1), you can solve for A by dividing both sides by 2:
A = 1/2
Substituting this value of A in equation (2), you can solve for B:
(1/2) - B = 8
- B = 8 - 1/2
- B = 15/2
B = -15/2
Therefore, A = 1/2 and B = -15/2.
Now, let's move on to integrating the expression sinx + 8cosx divided by 2sinx + cosx.
Integrating the numerator, sinx + 8cosx, requires simple integration rules:
∫ (sinx + 8cosx) dx = -cosx + 8sinx + C1 ----(3)
where C1 is the constant of integration.
Next, let's integrate the denominator, 2sinx + cosx, by using a substitution. Let u = 2sinx + cosx:
Differentiating both sides with respect to x:
du/dx = 2cosx - sinx
Rearranging the equation, we get:
dx = du / (2cosx - sinx)
Substituting this back into equation (3):
∫ (sinx + 8cosx) / (2sinx + cosx) dx = ∫ (-cosx + 8sinx) / (2cosx - sinx) du / (2cosx - sinx)
Simplifying the expression:
∫ (-cosx + 8sinx) du / (2cosx - sinx)
Now, we can see that -cosx + 8sinx matches the numerator of the expression we need to integrate. Therefore, we can rewrite the integral as:
∫ du
which is simply u + C2, where C2 is another constant of integration.
Substituting back the value of u = 2sinx + cosx:
u + C2 = 2sinx + cosx + C2
Finally, the integral of sinx + 8cosx / 2sinx + cosx is given by:
2x + 3Ln (2sinx + cosx) + C
where C is the constant of integration.