Tính các nguyên hàm sau:

a) $\int \sin 2x\sin 5xdx$ b) $\int \cos x\sin 3xdx$ c) $\int ({\tan }^{2}x+{\tan }^{4}x)dx$

d) $\int \frac{\cos 2x}{1+\sin x\cos x}dx$ e) $\int \frac{dx}{2\sin x+1}$ f) $\int \frac{dx}{\cos x}$

g) $\int \frac{1-\sin x}{\cos x}dx$ h) $\int \frac{{\sin }^{3}x}{\cos x}dx$ i) $\int \frac{dx}{\cos x\cos \left(x+\frac{\pi }{4}\right)}$

k) $\int \cos x\cos 2x\cos 3xdx$ l) $\int {\cos }^{3}xdx$ m) $\int {\sin }^{4}xdx$

(AI giải đề cho tôi)
Bài 2 (Nguyên hàm), tính từng tích phân:

a) Dùng tích thành tổng: $\sin 2x\sin 5x=\frac{1}{2}(\cos 3x-\cos 7x)$. Vậy $\int \sin 2x\sin 5xdx=\frac{1}{2}\left(\frac{\sin 3x}{3}-\frac{\sin 7x}{7}\right)+C$.

b) $\cos x\sin 3x=\frac{1}{2}(\sin 4x+\sin 2x)$, nên nguyên hàm $=-\frac{\cos 4x}{8}-\frac{\cos 2x}{4}+C$.

c) ${\tan }^{2}x+{\tan }^{4}x={\tan }^{2}x(1+{\tan }^{2}x)=\frac{{\tan }^{2}x}{{\cos }^{2}x}$, đặt $t=\tan x$: nguyên hàm $=\frac{{\tan }^{3}x}{3}+C$.

d) Đặt $t=\sin x\cos x\Rightarrow dt=\cos 2xdx$. Khi đó $\int \frac{\cos 2x}{1+\sin x\cos x}dx=\int \frac{dt}{1+t}=\ln (1+\sin x\cos x)+C$ (vì $1+\sin x\cos x\ge \frac{1}{2}>0$).

e) Nhân với liên hợp: $\frac{1}{2\sin x+1}=\frac{2\sin x-1}{4{\sin }^{2}x-1}=\frac{2\sin x-1}{3-4{\cos }^{2}x}$.
- $\int \frac{2\sin xdx}{3-4{\cos }^{2}x}$ với $t=\cos x$, $dt=-\sin xdx$: $=-2\int \frac{dt}{3-4t^2}=-\frac{1}{2\sqrt{3}}\ln \left|\frac{\sqrt{3}+2\cos x}{\sqrt{3}-2\cos x}\right|$ (vì $\int \frac{dt}{3-4t^2}=\frac{1}{4\sqrt{3}}\ln \left|\frac{\sqrt{3}+2t}{\sqrt{3}-2t}\right|$).
- $\int \frac{dx}{3-4{\cos }^{2}x}=\int \frac{d(\tan x)}{3{\tan }^{2}x-1}=\frac{1}{2\sqrt{3}}\ln \left|\frac{\sqrt{3}\tan x-1}{\sqrt{3}\tan x+1}\right|$.
Vậy kết quả: $-\frac{1}{2\sqrt{3}}\ln \left|\frac{\sqrt{3}+2\cos x}{\sqrt{3}-2\cos x}\right|+\frac{1}{2\sqrt{3}}\ln \left|\frac{\sqrt{3}\tan x-1}{\sqrt{3}\tan x+1}\right|+C$.

f) $\int \frac{dx}{\cos x}=\ln |\sec x+\tan x|+C=\ln \left|\tan \left(\frac{x}{2}+\frac{\pi }{4}\right)\right|+C$.

g) $\frac{1-\sin x}{\cos x}=\sec x-\tan x$, nên nguyên hàm $=\ln |\sec x+\tan x|+\ln |\cos x|+C=\ln |1+\sin x|+C$.

h) Đặt $t=\cos x$, $dt=-\sin xdx$: $\int \frac{{\sin }^{3}x}{\cos x}dx=\int \frac{-(1-t^2)}{t}dt=\int \left(t-\frac{1}{t}\right)dt=\frac{{\cos }^{2}x}{2}-\ln |\cos x|+C$.

i) Vì $\sin \left((x+\frac{\pi }{4})-x\right)=\frac{\sqrt{2}}{2}$, ta có $\tan (x+\frac{\pi }{4})-\tan x=\frac{\frac{\sqrt{2}}{2}}{\cos x\cos (x+\frac{\pi }{4})}$, suy ra $\frac{1}{\cos x\cos (x+\frac{\pi }{4})}=\sqrt{2}\left[\tan (x+\frac{\pi }{4})-\tan x\right]$. Do đó nguyên hàm $=\sqrt{2}\ln \left|\frac{\cos x}{\cos (x+\frac{\pi }{4})}\right|+C$.

k) $\cos x\cos 3x=\frac{1}{2}(\cos 4x+\cos 2x)$, nhân $\cos 2x$: $\frac{1}{4}(\cos 6x+\cos 2x)+\frac{1}{2}{\cos }^{2}2x=\frac{1}{4}\cos 6x+\frac{1}{4}\cos 2x+\frac{1}{4}+\frac{1}{4}\cos 4x$. Nguyên hàm: $\frac{x}{4}+\frac{\sin 2x}{8}+\frac{\sin 4x}{16}+\frac{\sin 6x}{24}+C$.

l) ${\cos }^{3}x=\cos x(1-{\sin }^{2}x)$, đặt $t=\sin x$: nguyên hàm $=\sin x-\frac{{\sin }^{3}x}{3}+C$.

m) ${\sin }^{4}x={\left(\frac{1-\cos 2x}{2}\right)}^2=\frac{3}{8}-\frac{\cos 2x}{2}+\frac{\cos 4x}{8}$. Nguyên hàm: $\frac{3x}{8}-\frac{\sin 2x}{4}+\frac{\sin 4x}{32}+C$.

Kiểm tra: đạo hàm mỗi kết quả đều trả về hàm dưới dấu tích phân (ví dụ i): $\sqrt{2}\left(-\tan (x+\frac{\pi }{4})+\tan x\right)=\sqrt{2}\cdot \frac{\frac{\sqrt{2}}{2}}{\cos x\cos (x+\frac{\pi }{4})}=\frac{1}{\cos x\cos (x+\frac{\pi }{4})}$ ✓; ví dụ m): $\frac{3}{8}-\frac{\cos 2x}{2}+\frac{\cos 4x}{8}={\sin }^{4}x$ ✓).