1. $intsin x dx$ 2. $intcos5x dx$ 3. $intcos xsin^{4}x dx$ 4. $intcsc^{2}x dx$ 5. $int\tan x dx$ 6. $intcot…

1. For (\int\sin xdx):

Explanation:

Step1: Recall integral formula

The integral of (\sin x) with respect to (x) is (-\cos x + C) according to the basic - integral rules. (\int\sin xdx=-\cos x + C)

Answer:

(-\cos x + C)

2. For (\int\cos(5x)dx):

Explanation:

Step1: Use substitution

Let (u = 5x), then (du=5dx), and (dx=\frac{1}{5}du). (\int\cos(5x)dx=\frac{1}{5}\int\cos(u)du)

Step2: Integrate (\cos(u))

We know that (\int\cos(u)du=\sin(u)+C). (\frac{1}{5}\int\cos(u)du=\frac{1}{5}\sin(u)+C)

Step3: Substitute back (u = 5x)

(\frac{1}{5}\sin(5x)+C)

Answer:

(\frac{1}{5}\sin(5x)+C)

3. For (\int\cos x\sin^{4}x dx):

Explanation:

Step1: Use substitution

Let (u=\sin x), then (du = \cos xdx). (\int\cos x\sin^{4}x dx=\int u^{4}du)

Step2: Integrate (u^{4})

Using the power - rule (\int u^{n}du=\frac{u^{n + 1}}{n+1}+C) ((n\neq - 1)), for (n = 4), we have (\int u^{4}du=\frac{u^{5}}{5}+C).

Step3: Substitute back (u=\sin x)

(\frac{\sin^{5}x}{5}+C)

Answer:

(\frac{\sin^{5}x}{5}+C)

4. For (\int\csc^{2}x dx):

Explanation:

Step1: Recall integral formula

The integral of (\csc^{2}x) with respect to (x) is (-\cot x + C) according to the basic - integral rules. (\int\csc^{2}x dx=-\cot x + C)

Answer:

(-\cot x + C)

5. For (\int\tan xdx):

Explanation:

Step1: Rewrite (\tan x)

(\tan x=\frac{\sin x}{\cos x}), let (u = \cos x), then (du=-\sin xdx). (\int\tan xdx=-\int\frac{du}{u})

Step2: Integrate (\frac{1}{u})

We know that (\int\frac{1}{u}du=\ln|u|+C). (-\int\frac{du}{u}=-\ln|\cos x|+C=\ln|\sec x|+C)

Answer:

(\ln|\sec x|+C)

6. For (\int\cot xdx):

Explanation:

Step1: Rewrite (\cot x)

(\cot x=\frac{\cos x}{\sin x}), let (u=\sin x), then (du = \cos xdx). (\int\cot xdx=\int\frac{du}{u})

Step2: Integrate (\frac{1}{u})

We know that (\int\frac{1}{u}du=\ln|u|+C). (\int\frac{du}{u}=\ln|\sin x|+C)

Answer:

(\ln|\sin x|+C)

7. For (\int\sec xdx):

Explanation:

Step1: Multiply by (\frac{\sec x+\tan x}{\sec x+\tan x})

(\int\sec xdx=\int\frac{\sec x(\sec x + \tan x)}{\sec x+\tan x}dx=\int\frac{\sec^{2}x+\sec x\tan x}{\sec x+\tan x}dx)

Step2: Use substitution

Let (u=\sec x+\tan x), then (du = (\sec x\tan x+\sec^{2}x)dx). (\int\frac{\sec^{2}x+\sec x\tan x}{\sec x+\tan x}dx=\int\frac{du}{u}=\ln|\sec x+\tan x|+C)

Answer:

(\ln|\sec x+\tan x|+C)

8. For (\int\csc xdx):

Explanation:

Step1: Multiply by (\frac{\csc x-\cot x}{\csc x-\cot x})

(\int\csc xdx=\int\frac{\csc x(\csc x - \cot x)}{\csc x-\cot x}dx=\int\frac{\csc^{2}x-\csc x\cot x}{\csc x-\cot x}dx)

Step2: Use substitution

Let (u=\csc x-\cot x), then (du = (-\csc x\cot x+\csc^{2}x)dx). (\int\frac{\csc^{2}x-\csc x\cot x}{\csc x-\cot x}dx=\int\frac{du}{u}=\ln|\csc x-\cot x|+C)

Answer:

(\ln|\csc x - \cot x|+C)

9. For (\int\sin^{2}x dx):

Explanation:

Step1: Use the double - angle formula (\sin^{2}x=\frac{1 - \cos(2x)}{2})

(\int\sin^{2}x dx=\int\frac{1-\cos(2x)}{2}dx=\frac{1}{2}\int(1 - \cos(2x))dx)

Step2: Integrate term - by - term

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

Answer:

(\frac{x}{2}-\frac{1}{4}\sin(2x)+C)

10. For (\int\sin^{3}x dx):

Explanation:

Step1: Rewrite (\sin^{3}x) as (\sin x(1 - \cos^{2}x))

(\int\sin^{3}x dx=\int\sin x(1 - \cos^{2}x)dx)

Step2: Use substitution

Let (u = \cos x), then (du=-\sin xdx). (\int\sin x(1 - \cos^{2}x)dx=-\int(1 - u^{2})du)

Step3: Integrate term - by - term

(-\int(1 - u^{2})du=-\left(u-\frac{u^{3}}{3}\right)+C=-\cos x+\frac{\cos^{3}x}{3}+C)

Answer:

(-\cos x+\frac{\cos^{3}x}{3}+C)

11. For (\int\sin^{4}x dx):

Explanation:

Step1: Use (\sin^{2}x=\frac{1 - \cos(2x)}{2})

(\sin^{4}x=\left(\sin^{2}x\right)^{2}=\left(\frac{1 - \cos(2x)}{2}\right)^{2}=\frac{1 - 2\cos(2x)+\cos^{2}(2x)}{4})

Step2: Use (\cos^{2}(2x)=\frac{1+\cos(4x)}{2})

(\sin^{4}x=\frac{1 - 2\cos(2x)+\frac{1 + \cos(4x)}{2}}{4}=\frac{2-4\cos(2x)+1+\cos(4x)}{8}=\frac{3 - 4\cos(2x)+\cos(4x)}{8})

Step3: Integrate term - by - term

(\int\sin^{4}x dx=\frac{1}{8}\int(3 - 4\cos(2x)+\cos(4x))dx=\frac{1}{8}\left(3x-2\sin(2x)+\frac{1}{4}\sin(4x)\right)+C)

Answer:

(\frac{3x}{8}-\frac{\sin(2x)}{4}+\frac{\sin(4x)}{32}+C)

12. For (\int\sin^{5}x dx):

Explanation:

Step1: Rewrite (\sin^{5}x=\sin x(1 - \cos^{2}x)^{2})

Let (u = \cos x), then (du=-\sin xdx). (\int\sin^{5}x dx=-\int(1 - u^{2})^{2}du)

Step2: Expand ((1 - u^{2})^{2})

((1 - u^{2})^{2}=1 - 2u^{2}+u^{4})

Step3: Integrate term - by - term

(-\int(1 - 2u^{2}+u^{4})du=-\left(u-\frac{2u^{3}}{3}+\frac{u^{5}}{5}\right)+C=-\cos x+\frac{2\cos^{3}x}{3}-\frac{\cos^{5}x}{5}+C)

Answer:

(-\cos x+\frac{2\cos^{3}x}{3}-\frac{\cos^{5}x}{5}+C)

13. For (\int\frac{1}{a^{2}+b^{2}x^{2}}dx):

Explanation:

Step1: Rewrite the integrand

(\int\frac{1}{a^{2}+b^{2}x^{2}}dx=\frac{1}{a^{2}}\int\frac{1}{1+\left(\frac{bx}{a}\right)^{2}}dx)

Step2: Use substitution

Let (u=\frac{bx}{a}), then (du=\frac{b}{a}dx) and (dx=\frac{a}{b}du). (\frac{1}{a^{2}}\int\frac{1}{1 + u^{2}}\cdot\frac{a}{b}du=\frac{1}{ab}\arctan(u)+C=\frac{1}{ab}\arctan\left(\frac{bx}{a}\right)+C)

Answer:

(\frac{1}{ab}\arctan\left(\frac{bx}{a}\right)+C)

14. For (\int\frac{1}{\sqrt{a^{2}-x^{2}}}dx):

Explanation:

Step1: Use substitution

Let (x = a\sin\theta), then (dx=a\cos\theta d\theta) and (\sqrt{a^{2}-x^{2}}=a\cos\theta). (\int\frac{1}{\sqrt{a^{2}-x^{2}}}dx=\int\frac{a\cos\theta}{a\cos\theta}d\theta=\int d\theta=\theta + C) Since (x = a\sin\theta), (\theta=\arcsin\left(\frac{x}{a}\right))

Answer:

(\arcsin\left(\frac{x}{a}\right)+C)

15. For (\int\frac{\sec(\sqrt{x})}{\sqrt{x}}dx):

Explanation:

Step1: Use substitution

Let (u = \sqrt{x}), then (x = u^{2}) and (dx = 2udu). (\int\frac{\sec(\sqrt{x})}{\sqrt{x}}dx=2\int\sec(u)du)

Step2: Integrate (\sec(u))

We know that (\int\sec(u)du=\ln|\sec(u)+\tan(u)|+C) (2\int\sec(u)du = 2\ln|\sec(\sqrt{x})+\tan(\sqrt{x})|+C)

Answer:

(2\ln|\sec(\sqrt{x})+\tan(\sqrt{x})|+C)

16. For (\int_{0}^{\frac{\pi}{3}}\sqrt{1 + \cos(2x)}dx):

Explanation:

Step1: Use the double - angle formula (\cos(2x)=2\cos^{2}x - 1), then (1+\cos(2x)=2\cos^{2}x)

(\sqrt{1 + \cos(2x)}=\sqrt{2}|\cos x|). Since (x\in[0,\frac{\pi}{3}]), (\cos x\geq0), so (\sqrt{1 + \cos(2x)}=\sqrt{2}\cos x) (\int_{0}^{\frac{\pi}{3}}\sqrt{1 + \cos(2x)}dx=\sqrt{2}\int_{0}^{\frac{\pi}{3}}\cos xdx)

Step2: Integrate (\cos x)

(\sqrt{2}\int_{0}^{\frac{\pi}{3}}\cos xdx=\sqrt{2}[\sin x]_{0}^{\frac{\pi}{3}}=\sqrt{2}\left(\sin\frac{\pi}{3}-\sin0\right)=\sqrt{2}\times\frac{\sqrt{3}}{2}=\frac{\sqrt{6}}{2})

Answer:

(\frac{\sqrt{6}}{2})

17. For (\int_{0}^{\frac{\pi}{2}}\sqrt{1-\cos x}dx):

Explanation:

Step1: Use the half - angle formula (1-\cos x = 2\sin^{2}\frac{x}{2})

(\sqrt{1-\cos x}=\sqrt{2}\left|\sin\frac{x}{2}\right|). Since (x\in[0,\frac{\pi}{2}]), (\sin\frac{x}{2}\geq0), so (\sqrt{1 - \cos x}=\sqrt{2}\sin\frac{x}{2}) (\int_{0}^{\frac{\pi}{2}}\sqrt{1-\cos x}dx=\sqrt{2}\int_{0}^{\frac{\pi}{2}}\sin\frac{x}{2}dx)

Step2: Use substitution

Let (u=\frac{x}{2}), then (du=\frac{1}{2}dx) and (dx = 2du) (\sqrt{2}\int_{0}^{\frac{\pi}{2}}\sin\frac{x}{2}dx = 2\sqrt{2}[-\cos\frac{x}{2}]_{0}^{\frac{\pi}{2}}=2\sqrt{2}\left(1-\frac{\sqrt{2}}{2}\right)=2\sqrt{2}-2)

Answer:

(2\sqrt{2}-2)

18. For (\int\tan^{3}x dx):

Explanation:

Step1: Rewrite (\tan^{3}x=\tan x\tan^{2}x=\tan x(\sec^{2}x - 1))

(\int\tan^{3}x dx=\int\tan x(\sec^{2}x - 1)dx=\int\tan x\sec^{2}x dx-\int\tan xdx)

Step2: Integrate (\tan x\sec^{2}x) and (\tan x)

For (\int\tan x\sec^{2}x dx), let (u = \tan x), then (du=\sec^{2}x dx), (\int\tan x\sec^{2}x dx=\frac{\tan^{2}x}{2}) We know that (\int\tan xdx=\ln|\sec x|+C) (\int\tan^{3}x dx=\frac{\tan^{2}x}{2}-\ln|\sec x|+C)

Answer:

(\frac{\tan^{2}x}{2}-\ln|\sec x|+C)

19. For (\int\sin(7x)\cos(3x)dx):

Explanation:

Step1: Use the product - to - sum formula (\sin A\cos B=\frac{1}{2}[\sin(A + B)+\sin(A - B)])

(\sin(7x)\cos(3x)=\frac{1}{2}[\sin(10x)+\sin(4x)]) (\int\sin(7x)\cos(3x)dx=\frac{1}{2}\int[\sin(10x)+\sin(4x)]dx)

Step2: Integrate term - by - term

(\frac{1}{2}\int[\sin(10x)+\sin(4x)]dx=\frac{1}{2}\left(-\frac{1}{10}\cos(10x)-\frac{1}{4}\cos