Giải bài 25 trang 15

Đề bài

Tìm:

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

b) $\int {\sin }^{2}2xdx+\int {\cos }^{2}2xdx$;

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) $\int {\sin }^{2}2xdx+\int {\cos }^{2}2xdx$

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

$=\int \left(\frac{1-\cos 4x}{2}+\frac{1+\cos 4x}{2}\right)dx$

$=\int 1dx=x+C$.

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

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

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

$=\int \left(1+\sin x\right)dx=x-\cos x+C$.

e) $\int {\cos }^4\frac{x}{2}dx-\int {\sin }^4\frac{x}{2}dx=\int \left({\cos }^4\frac{x}{2}-{\sin }^4\frac{x}{2}\right)dx$

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

$=\int \cos x.1dx=\int \cos xdx=\sin x+C$.

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

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