a) \(\int {\left( {2\cos x - \frac{3}{{{{\sin }^2}x}}} \right)} dx\)
\(= 2\int {\cos x\, dx} - 3\int {\frac{1}{{{{\sin }^2}x}}\, dx}\)
\(= 2\sin x + 3\cot x + C\)
b) Từ công thức nhân đôi \(\cos 2x = 1 - 2\sin^2 x\), thay \(x\) bằng \(\frac{x}{2}\):
\(\cos x = 1 - 2\sin^2\frac{x}{2}\)
\(\Rightarrow 4\sin^2\frac{x}{2} = 2(1 - \cos x)\)
Do đó:
\(\int {4{{\sin }^2}\frac{x}{2}\, dx} = \int {2(1 - \cos x)\, dx}\)
\(= 2\int dx - 2\int {\cos x\, dx}\)
\(= 2x - 2\sin x + C\)
c) Khai triển bình phương:
\(\left(\sin\frac{x}{2} - \cos\frac{x}{2}\right)^2 = \sin^2\frac{x}{2} + \cos^2\frac{x}{2} - 2\sin\frac{x}{2}\cos\frac{x}{2}\)
\(= 1 - \sin x\)
Do đó:
\(\int {{{\left( {\sin \frac{x}{2} - \cos \frac{x}{2}} \right)}^2}} dx = \int {(1 - \sin x)\, dx}\)
\(= \int dx - \int {\sin x\, dx}\)
\(= x + \cos x + C\)
d) Viết lại \(\tan^2 x = \frac{1}{\cos^2 x} - 1\), suy ra:
\(\int {\left( {x + {{\tan }^2}x} \right)} dx = \int {x\, dx} + \int {\left( {\frac{1}{{{{\cos }^2}x}} - 1} \right)dx}\)
\(= \frac{x^2}{2} + \tan x - x + C\)
