Question

$\int \frac{cos^2x + sin2x}{(2cosx - sinx)^2}dx$

Answer 1

$-\frac{2}{5}\ln|\tan x - 2| - \frac{1}{\tan x - 2} + \frac{2}{5}\ln|\sec x| - \frac{1}{5}x + C$

Answer 2

The given integral is $I = \int \frac{\cos^2x + \sin2x}{(2\cos x - \sin x)^2}dx$.

Step 1: Simplify the numerator using $\sin2x = 2\sin x \cos x$. Numerator $= \cos^2x + 2\sin x \cos x$.

Step 2: Divide both the numerator and the denominator by $\cos^2x$. $I = \int \frac{\frac{\cos^2x + 2\sin x \cos x}{\cos^2x}}{\frac{(2\cos x - \sin x)^2}{\cos^2x}}dx$ $I = \int \frac{1 + 2\tan x}{\left(\frac{2\cos x - \sin x}{\cos x}\right)^2}dx$ $I = \int \frac{1 + 2\tan x}{(2 - \tan x)^2}dx$.

Step 3: Substitute $t = \tan x$. Then $dt = \sec^2x \, dx = (1 + \tan^2x) \, dx = (1 + t^2) \, dx$. So, $dx = \frac{dt}{1 + t^2}$. Substituting these into the integral: $I = \int \frac{1 + 2t}{(2 - t)^2} \cdot \frac{dt}{1 + t^2} = \int \frac{2t + 1}{(t - 2)^2 (t^2 + 1)}dt$.

Step 4: Perform partial fraction decomposition for the integrand $\frac{2t + 1}{(t - 2)^2 (t^2 + 1)}$. Let $\frac{2t + 1}{(t - 2)^2 (t^2 + 1)} = \frac{A}{t - 2} + \frac{B}{(t - 2)^2} + \frac{Ct + D}{t^2 + 1}$. Multiplying by $(t - 2)^2 (t^2 + 1)$: $2t + 1 = A(t - 2)(t^2 + 1) + B(t^2 + 1) + (Ct + D)(t - 2)^2$.

To find coefficients:

1. Set $t = 2$: $2(2) + 1 = A(0) + B(2^2 + 1) + (Ct + D)(0)$ $5 = 5B \implies B = 1$.

2. Compare coefficients of $t^3$: LHS has $0t^3$. RHS has $A(t^3) + C(t)(t^2) = (A + C)t^3$. So, $A + C = 0 \implies C = -A$.

3. Compare coefficients of $t^2$: LHS has $0t^2$. RHS has $A(-2t^2) + B(t^2) + C(-4t^2) + D(t^2) = (-2A + B - 4C + D)t^2$. So, $-2A + B - 4C + D = 0$. Substitute $B = 1$ and $C = -A$: $-2A + 1 - 4(-A) + D = 0$ $-2A + 1 + 4A + D = 0$ $2A + D + 1 = 0$.

4. Compare constant terms (set $t = 0$): LHS has $1$. RHS has $A(-2)(1) + B(1) + D(-2)^2 = -2A + B + 4D$. So, $1 = -2A + B + 4D$. Substitute $B = 1$: $1 = -2A + 1 + 4D \implies 0 = -2A + 4D \implies A = 2D$.

Now we have a system of equations for A and D: (1) $2A + D + 1 = 0$ (2) $A = 2D$ Substitute (2) into (1): $2(2D) + D + 1 = 0$ $4D + D + 1 = 0 \implies 5D = -1 \implies D = -\frac{1}{5}$. From $A = 2D$, $A = 2(-\frac{1}{5}) = -\frac{2}{5}$. From $C = -A$, $C = -(-\frac{2}{5}) = \frac{2}{5}$.

So, the partial fraction decomposition is: $\frac{2t + 1}{(t - 2)^2 (t^2 + 1)} = \frac{-2/5}{t - 2} + \frac{1}{(t - 2)^2} + \frac{(2/5)t - 1/5}{t^2 + 1}$ $= -\frac{2}{5(t - 2)} + \frac{1}{(t - 2)^2} + \frac{1}{5} \frac{2t - 1}{t^2 + 1}$.

Step 5: Integrate each term. $I = \int \left( -\frac{2}{5(t - 2)} + \frac{1}{(t - 2)^2} + \frac{1}{5} \frac{2t - 1}{t^2 + 1} \right) dt$ $I = -\frac{2}{5} \int \frac{1}{t - 2} dt + \int (t - 2)^{-2} dt + \frac{1}{5} \int \frac{2t}{t^2 + 1} dt - \frac{1}{5} \int \frac{1}{t^2 + 1} dt$ $I = -\frac{2}{5} \ln|t - 2| - \frac{1}{t - 2} + \frac{1}{5} \ln(t^2 + 1) - \frac{1}{5} \arctan t + C$.

Step 6: Substitute back $t = \tan x$. $I = -\frac{2}{5} \ln|\tan x - 2| - \frac{1}{\tan x - 2} + \frac{1}{5} \ln(\tan^2 x + 1) - \frac{1}{5} \arctan(\tan x) + C$. Using $\tan^2 x + 1 = \sec^2 x$ and $\arctan(\tan x) = x$: $I = -\frac{2}{5} \ln|\tan x - 2| - \frac{1}{\tan x - 2} + \frac{1}{5} \ln(\sec^2 x) - \frac{1}{5} x + C$ $I = -\frac{2}{5} \ln|\tan x - 2| - \frac{1}{\tan x - 2} + \frac{2}{5} \ln|\sec x| - \frac{1}{5}x + C$.