Giải bài tập 25 trang 15 sách bài tập toán 12 - Cánh diều — Không quảng cáo

Đề bài

Tìm:

a) $\int {\left( {5\sin x - 6\cos x} \right)dx}$;

b) $\int {{{\sin }^2}2{\rm{x}}dx}  + \int {{{\cos }^2}2{\rm{x}}dx}$;

c) $\int {{{\sin }^2}\frac{x}{2}dx}$;

d) $\int {{{\left( {\sin \frac{x}{2} + \cos \frac{x}{2}} \right)}^2}dx}$;

e) $\int {{{\cos }^4}\frac{x}{2}dx}  - \int {{{\sin }^4}\frac{x}{2}dx}$;

g) $\int {{{\tan }^2}xdx}$.

Lời giải chi tiết

a) $\int {\left( {5\sin x - 6\cos x} \right)dx}  = 5\left( { - \cos x} \right) - 6\sin x + C =  - 5\cos x - 6\sin x + C$.

b) $\int {{{\sin }^2}2{\rm{x}}dx}  + \int {{{\cos }^2}2{\rm{x}}dx}  = \int {\left( {{{\sin }^2}2{\rm{x}} + {{\cos }^2}2{\rm{x}}} \right)dx}  = \int {\left( {\frac{{1 - \cos 4{\rm{x}}}}{2} + \frac{{1 + \cos 4{\rm{x}}}}{2}} \right)dx}  = \int {1dx}  = x + C$.

c) $\int {{{\sin }^2}\frac{x}{2}dx}  = \int {\frac{{1 - \cos x}}{2}dx}  = \int {\left( {\frac{1}{2} - \frac{1}{2}\cos x} \right)dx}  = \frac{1}{2}x - \frac{1}{2}\sin x + C = \frac{{x - \sin x}}{2} + C$.

d)

$\begin{array}{l}\int {{{\left( {\sin \frac{x}{2} + \cos \frac{x}{2}} \right)}^2}dx}  = \int {\left( {{{\sin }^2}\frac{x}{2} + 2.\sin \frac{x}{2}.\cos \frac{x}{2} + {{\cos }^2}\frac{x}{2}} \right)dx} \\ = \int {\left( {\frac{{1 - \cos x}}{2} + \sin x + \frac{{1 + \cos x}}{2}} \right)dx}  = \int {\left( {1 + \sin x} \right)dx}  = x - \cos x + C\end{array}$

e)

$\begin{array}{l}\int {{{\cos }^4}\frac{x}{2}dx}  - \int {{{\sin }^4}\frac{x}{2}dx}  = \int {\left( {{{\cos }^4}\frac{x}{2} - {{\sin }^4}\frac{x}{2}} \right)dx}  = \int {\left( {{{\cos }^2}\frac{x}{2} - {{\sin }^2}\frac{x}{2}} \right)\left( {{{\cos }^2}\frac{x}{2} + {{\sin }^2}\frac{x}{2}} \right)dx} \\ = \int {\cos x.1dx}  = \int {\cos xdx}  = \sin x + C\end{array}$

g) $\int {{{\tan }^2}xdx}  = \int {\left( {{{\tan }^2}x + 1 - 1} \right)dx}  = \int {\left( {\frac{1}{{{{\cos }^2}x}} - 1} \right)dx}  = \tan x - x + C$.