Giải giúp mk vs ạ

1) \(\int \frac{\cos^2 x}{1 - \sin x} \, dx\) Ta có: \[ \frac{\cos^2 x}{1 - \sin x} = \frac{1 - \sin^2 x}{1 - \sin x} = 1 + \sin x \] Do đó: \[ \int \frac{\cos^2 x}{1 - \sin x} \, dx = \int (1 + \sin x) \, dx = x - \cos x + C \] 2) \(\int (1 + 3 \sin^2 \frac{x}{2}) \, dx\) Ta có: \[ \sin^2 \frac{x}{2} = \frac{1 - \cos x}{2} \] Do đó: \[ 1 + 3 \sin^2 \frac{x}{2} = 1 + 3 \cdot \frac{1 - \cos x}{2} = \frac{5}{2} - \frac{3}{2} \cos x \] Vậy: \[ \int (1 + 3 \sin^2 \frac{x}{2}) \, dx = \int \left( \frac{5}{2} - \frac{3}{2} \cos x \right) \, dx = \frac{5}{2} x - \frac{3}{2} \sin x + C \] 3) \(\int \frac{2 \cos^3 x + 3}{\cos^2 x} \, dx\) Ta có: \[ \frac{2 \cos^3 x + 3}{\cos^2 x} = 2 \cos x + \frac{3}{\cos^2 x} = 2 \cos x + 3 \sec^2 x \] Do đó: \[ \int \frac{2 \cos^3 x + 3}{\cos^2 x} \, dx = \int (2 \cos x + 3 \sec^2 x) \, dx = 2 \sin x + 3 \tan x + C \] 4) \(\int (5 \sin x - 6 \cos x) \, dx\) Ta có: \[ \int (5 \sin x - 6 \cos x) \, dx = 5 \int \sin x \, dx - 6 \int \cos x \, dx = -5 \cos x - 6 \sin x + C \] 5) \(\int \sin^2 2x \, dx + \int \cos^2 2x \, dx\) Ta có: \[ \sin^2 2x + \cos^2 2x = 1 \] Do đó: \[ \int \sin^2 2x \, dx + \int \cos^2 2x \, dx = \int 1 \, dx = x + C \] 6) \(\int \sin^2 \frac{x}{2} \, dx\) Ta có: \[ \sin^2 \frac{x}{2} = \frac{1 - \cos x}{2} \] Do đó: \[ \int \sin^2 \frac{x}{2} \, dx = \int \frac{1 - \cos x}{2} \, dx = \frac{1}{2} \int (1 - \cos x) \, dx = \frac{1}{2} \left( x - \sin x \right) + C = \frac{x}{2} - \frac{\sin x}{2} + C \] 7) \(\int (\sin \frac{x}{2} + \cos \frac{x}{2})^2 \, dx\) Ta có: \[ (\sin \frac{x}{2} + \cos \frac{x}{2})^2 = \sin^2 \frac{x}{2} + 2 \sin \frac{x}{2} \cos \frac{x}{2} + \cos^2 \frac{x}{2} = 1 + \sin x \] Do đó: \[ \int (\sin \frac{x}{2} + \cos \frac{x}{2})^2 \, dx = \int (1 + \sin x) \, dx = x - \cos x + C \] 8) \(\int \cos^4 \frac{x}{2} \, dx - \int \sin^4 \frac{x}{2} \, dx\) Ta có: \[ \cos^4 \frac{x}{2} - \sin^4 \frac{x}{2} = (\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2})(\cos^2 \frac{x}{2} + \sin^2 \frac{x}{2}) = \cos x \] Do đó: \[ \int \cos^4 \frac{x}{2} \, dx - \int \sin^4 \frac{x}{2} \, dx = \int \cos x \, dx = \sin x + C \] 9) \(\int \tan^2 x \, dx\) Ta có: \[ \tan^2 x = \sec^2 x - 1 \] Do đó: \[ \int \tan^2 x \, dx = \int (\sec^2 x - 1) \, dx = \tan x - x + C \] 10) \(\int (-\cos x) \, dx\) Ta có: \[ \int (-\cos x) \, dx = - \int \cos x \, dx = - \sin x + C \] 11) \(\int \frac{1 - \cos^2 x}{\cos^2 x} \, dx\) Ta có: \[ \frac{1 - \cos^2 x}{\cos^2 x} = \frac{\sin^2 x}{\cos^2 x} = \tan^2 x \] Do đó: \[ \int \frac{1 - \cos^2 x}{\cos^2 x} \, dx = \int \tan^2 x \, dx = \tan x - x + C \] 12) \(\int (3 \sin x - 4 \cos x) \, dx\) Ta có: \[ \int (3 \sin x - 4 \cos x) \, dx = 3 \int \sin x \, dx - 4 \int \cos x \, dx = -3 \cos x - 4 \sin x + C \] 13) \(\int (7 + 5 \cot^2 x) \, dx\) Ta có: \[ 5 \cot^2 x = 5 (\csc^2 x - 1) \] Do đó: \[ 7 + 5 \cot^2 x = 7 + 5 \csc^2 x - 5 = 2 + 5 \csc^2 x \] Vậy: \[ \int (7 + 5 \cot^2 x) \, dx = \int (2 + 5 \csc^2 x) \, dx = 2x - 5 \cot x + C \] 14) \(\int 2 \sin x \, dx\) Ta có: \[ \int 2 \sin x \, dx = 2 \int \sin x \, dx = -2 \cos x + C \] 15) \(\int (\cos x + x^3) \, dx\) Ta có: \[ \int (\cos x + x^3) \, dx = \int \cos x \, dx + \int x^3 \, dx = \sin x + \frac{x^4}{4} + C \] 16) \(\int \left( \frac{-x^4}{2} - 3 \cos x \right) \, dx\) Ta có: \[ \int \left( \frac{-x^4}{2} - 3 \cos x \right) \, dx = \int \frac{-x^4}{2} \, dx - 3 \int \cos x \, dx = -\frac{x^5}{10} - 3 \sin x + C \] 17) \(\int \left( 2 \cos x + \frac{3}{\sqrt{x}} \right) \, dx\) Ta có: \[ \int \left( 2 \cos x + \frac{3}{\sqrt{x}} \right) \, dx = 2 \int \cos x \, dx + 3 \int x^{-\frac{1}{2}} \, dx = 2 \sin x + 6 \sqrt{x} + C \] 18) \(\int (3 \sqrt{x} - 4 \sin x) \, dx\) Ta có: \[ \int (3 \sqrt{x} - 4 \sin x) \, dx = 3 \int x^{\frac{1}{2}} \, dx - 4 \int \sin x \, dx = 2 x^{\frac{3}{2}} + 4 \cos x + C \] 19) \(\int \left( x + \sin^2 \frac{x}{2} \right) \, dx\) Ta có: \[ \sin^2 \frac{x}{2} = \frac{1 - \cos x}{2} \] Do đó: \[ x + \sin^2 \frac{x}{2} = x + \frac{1 - \cos x}{2} = x + \frac{1}{2} - \frac{\cos x}{2} \] Vậy: \[ \int \left( x + \sin^2 \frac{x}{2} \right) \, dx = \int \left( x + \frac{1}{2} - \frac{\cos x}{2} \right) \, dx = \frac{x^2}{2} + \frac{x}{2} - \frac{\sin x}{2} + C \] 20) \(\int (2 \tan x + \cot x)^2 \, dx\) Ta có: \[ (2 \tan x + \cot x)^2 = 4 \tan^2 x + 4 \tan x \cot x + \cot^2 x = 4 \tan^2 x + 4 + \cot^2 x \] Do đó: \[ \int (2 \tan x + \cot x)^2 \, dx = \int (4 \tan^2 x + 4 + \cot^2 x) \, dx = 4 \int \tan^2 x \, dx + 4 \int 1 \, dx + \int \cot^2 x \, dx \] \[ = 4 (\tan x - x) + 4x + (-\cot x - x) + C = 4 \tan x - 4x + 4x - \cot x - x + C = 4 \tan x - \cot x - x + C \] 21) \(\int 3 \cos x \, dx\) Ta có: \[ \int 3 \cos x \, dx = 3 \int \cos x \, dx = 3 \sin x + C \] 22) \(\int (\sin x + \cos x) \, dx\) Ta có: \[ \int (\sin x + \cos x) \, dx = \int \sin x \, dx + \int \cos x \, dx = -\cos x + \sin x + C \] 23) \(\int \left( \frac{1}{\sin^2 x} - \frac{1}{\cos^2 x} \right) \, dx\) Ta có: \[ \frac{1}{\sin^2 x} - \frac{1}{\cos^2 x} = \csc^2 x - \sec^2 x \] Do đó: \[ \int \left( \frac{1}{\sin^2 x} - \frac{1}{\cos^2 x} \right) \, dx = \int \csc^2 x \, dx - \int \sec^2 x \, dx = -\cot x - \tan x + C \] 24) \(\int (1 + \tan^2 x) \, dx\) Ta có: \[ 1 + \tan^2 x = \sec^2 x \] Do đó: \[ \int (1 + \tan^2 x) \, dx = \int \sec^2 x \, dx = \tan x + C \] 25) \(\int 2 \sin \frac{x}{2} \cos \frac{x}{2} \, dx\) Ta có: \[ 2 \sin \frac{x}{2} \cos \frac{x}{2} = \sin x \] Do đó: \[ \int 2 \sin \frac{x}{2} \cos \frac{x}{2} \, dx = \int \sin x \, dx = -\cos x + C \]