ให้ $$u=\sin\ 2x$$  ต้

ต้องใช้ความสัมพันธ์ $$\sin^2x-\cos^2x=1$$  

ได้คำตอบ คือ $$-\frac{1}{2}\left[\cos\ 2x-\frac{\cos^22x}{2}\right]+C$$  

ต้องใช้ความรู้เรื่อง $$\int\ \frac{1}{u}du=\ln\left|u\right|+C$$  

ต้องจัดรูปโดยให้ $$\sin^mx\cos^nx=\sin^mx\cos^{n-1}x\cos x$$

$$\int\cos3x\sin2xdx=\frac{1}{2}\int\left[\sin5x-\sin x\right]dx$$  

$$\int\sin2x\sin5xdx=\frac{1}{2}\int\left[\cos7x-\cos3x\right]dx$$  

$$\int\sin2x\cos7xdx=\frac{1}{2}\int\left[\sin5x+\sin5x\right]dx$$  

$$\int\cos3x\cos5xdx=\frac{1}{2}\int\left[-\cos2x+\cos8x\right]dx$$  

$$\int\sin3x\cos7xdx=\frac{1}{2}\int\left[\sin4x+\sin9x\right]dx$$  

ให้ $$u=5-x^4$$  

$$a=5$$  

ต้องใช้สูตร $$\int\ \frac{1}{\sqrt{u}}du=\int\ u^{-\frac{1}{2}}du=2u^{\frac{1}{2}}+C$$  

ได้คำตอบ คือ $$\sin^{-1}\ \left(\frac{x^2}{5}\right)+C$$  

$$du=2xdx$$  

$$\frac{x^2}{3}e^{3x}-\frac{2}{9}xe^{3x}+\frac{2}{27}e^{3x}+C$$  

$$x^2e^{3x}-2xe^{3x}+2e^{3x}+C$$  

$$\frac{x^2}{3}e^{3x}-\frac{x}{9}e^{3x}+\frac{1}{27}e^{3x}+C$$  

$$3x^2e^{3x}-18xe^{3x}+54e^{3x}+C$$  

$$3x^2e^{3x}-9xe^{3x}+27e^{3x}+C$$  

$$-\frac{5}{x^2\left(x^2+1\right)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^2+1}$$  

$$-\frac{3}{x^2\left(x^2+2\right)}=\frac{Ax+B}{x^2}+\frac{Cx+D}{x^2+2}$$  

$$-\frac{2}{x^2\left(x^2+3\right)}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+3}$$  

$$-\frac{1}{x^2\left(x^2+4\right)}=\frac{A}{x}+\frac{Bx+C}{x^2}+\frac{Dx+E}{x^2+4}$$  

$$-\frac{3}{x^2\left(x^2+5\right)}=\frac{Ax}{x^2}+\frac{Bx+C}{x^2+5}$$  

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6

-6

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ให้ $$u=x^2+4$$  

$$dx=\sec^2\theta\ d\theta$$  

$$\sqrt{x^2+4}=2\sec\ \theta$$  

มีคำตอบ คือ $$\ln\ \left|\frac{\sqrt{x^{2\ }+4}}{2}+\frac{x}{2}\right|+C$$  

$$\int\sqrt{x^2+4}dx=\int\sqrt{\tan^{2\ }\theta+4}\cdot\sec^2\theta d\theta$$  

$$\int\sin x\ \cos xdx=\frac{\sin^2x}{2}+C$$  

$$\int\sin x\ \cos xdx=-\frac{\cos^2x}{2}+C$$  

$$\int\sec^2x\ \tan xdx=\frac{\sec^2x}{2}+C$$  

$$\int\csc^2x\ \cot xdx=\frac{\csc^2x}{2}+C$$  

$$\int\tan x\sec^2xdx=\frac{\tan^2x}{2}+C$$  

$$\int\ \frac{1}{4+x^2}dx=\frac{1}{2}\tan^{-1}\ \frac{x}{2}+C$$  

$$\int\ \frac{1}{x\sqrt{x^4-3}}dx=\frac{1}{2\sqrt{3}}\sec^{-1}\ \frac{x^2}{\sqrt{3}}+C$$  

$$\int\ \frac{x}{\sqrt{4-x^4}}dx=\frac{1}{2}\sin^{-1}\ \frac{x^2}{2}+C$$  

$$\int\ \frac{x}{x^4+4}dx=\frac{1}{2}\tan^{-1}\ \frac{x^2}{2}+C$$  

$$\int\ \frac{1}{x\sqrt{x^2-3}}dx=\frac{1}{\sqrt{3}}\sec^{-1}\ \frac{x}{\sqrt{3}}+C$$  

$$\int\ \sin\ 2x\ dx=\frac{1}{2}\cos2x+C$$  

$$\int\ \cos\ 2x\ dx=\frac{1}{2}\sin2x+C$$  

$$\int\ \tan\ 2x\ dx=\frac{1}{2}\ln\left|\sec2x\right|+C$$  

$$\int\ \csc\ 2x\ dx=\frac{1}{2}\ln\left|\csc2x-\cot2x\right|+C$$  

$$\int\ \sec\ 2x\ dx=\frac{1}{2}\ln\left|\sec2x+\tan2x\right|+C$$