Question

Find following integrations. (Find them By 12 basic integrations and addition technique)

$\int 1*2*3*....n^xdx$

$\int \frac{4^x+8^x}{3^x+6^x}dx$

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

$\int tan^2xdx$

$\int \frac{1+x^2+x^4}{1+x+x^2}dx$

$\int sec^2x.cosec^2xdx$

$\int \frac{1+cos^2x}{1+cos2x}dx$

$\int sinx.sin(\pi/3-x).sin(\pi/3+x).dx$

$\int sin^2xdx$

$\int sin^3xdx$

$\int \frac{x^{2023}+x^{2025}}{(1+x+x^2)(1-x+x^2)-1}dx$

$\int \frac{1+x^{4048}+x^{8096}}{1+x^{2024}+x^{4048}}dx$

$\int (\frac{sinx+sin2x+sin3x}{cosx+cos2x+cos3x})^2dx$

$\int \frac{1+4^x+16^x}{1+2^x+4^x}dx$

$\int \frac{x^4}{x^2+1}dx$

$\int \frac{cos8x-cos7x}{1+2cos5x}dx$

$\int \frac{dx}{x^4-1}$

$\int tan^4xdx$

$\int tan^5xdx$

$\int \frac{sinx}{x+cosx}dx$

Answer 1

1. $\frac{(n!)^x}{\ln(n!)} + C$
2. $\frac{(4/3)^x}{\ln(4/3)} + C$
3. $\frac{1}{2}\ln|\sec2x + \tan2x| + C$
4. $\tan x - x + C$
5. $x - \frac{x^2}{2} + \frac{x^3}{3} + C$
6. $\tan x - \cot x + C$
7. $\frac{1}{2}\tan x + \frac{1}{2}x + C$
8. $-\frac{1}{12}\cos(3x) + C$
9. $\frac{x}{2} - \frac{\sin2x}{4} + C$
10. $\frac{\cos^3x}{3} - \cos x + C$
11. $\frac{x^{2022}}{2022} + C$
12. $x - \frac{x^{2025}}{2025} + \frac{x^{4049}}{4049} + C$
13. $\frac{\tan(2x)}{2} - x + C$
14. $x - \frac{2^x}{\ln 2} + \frac{4^x}{\ln 4} + C$
15. $\frac{x^3}{3} - x + \tan^{-1}x + C$
16. $\frac{\sin 3x}{3} - \frac{\sin 2x}{2} + C$
17. $\frac{1}{4}\ln\left|\frac{x-1}{x+1}\right| - \frac{1}{2}\tan^{-1}x + C$
18. $\frac{\tan^3x}{3} - \tan x + x + C$
19. $\frac{\tan^4x}{4} - \frac{\tan^2x}{2} + \ln|\sec x| + C$
20. This integral is not solvable by elementary methods.

Answer 2

Here are the solutions to the given indefinite integrals:

1. $\int 1*2*3*....n^xdx$

The product $1 \cdot 2 \cdot 3 \cdot \dots \cdot n$ is $n!$. So the integral is $\int (n!)^x dx$. This is of the form $\int a^x dx = \frac{a^x}{\ln a} + C$. Solution: $\frac{(n!)^x}{\ln(n!)} + C$

2. $\int \frac{4^x+8^x}{3^x+6^x}dx$

Factor the numerator and denominator: Numerator: $4^x+8^x = 4^x(1+2^x)$ Denominator: $3^x+6^x = 3^x(1+2^x)$ The expression simplifies to $\frac{4^x(1+2^x)}{3^x(1+2^x)} = \frac{4^x}{3^x} = \left(\frac{4}{3}\right)^x$. Solution: $\frac{(4/3)^x}{\ln(4/3)} + C$

3. $\int \frac{1+tan^2x}{1-tan^2x}dx$

Use the identity $1+\tan^2x = \sec^2x$. Rewrite the denominator: $1-\tan^2x = 1 - \frac{\sin^2x}{\cos^2x} = \frac{\cos^2x-\sin^2x}{\cos^2x} = \frac{\cos2x}{\cos^2x}$. The integrand becomes $\frac{\sec^2x}{\frac{\cos2x}{\cos^2x}} = \frac{1/\cos^2x}{\cos2x/\cos^2x} = \frac{1}{\cos2x} = \sec2x$. Solution: $\frac{1}{2}\ln|\sec2x + \tan2x| + C$

4. $\int tan^2xdx$

Use the identity $\tan^2x = \sec^2x - 1$. Solution: $\tan x - x + C$

5. $\int \frac{1+x^2+x^4}{1+x+x^2}dx$

Factor the numerator using the identity $a^4+a^2b^2+b^4 = (a^2-ab+b^2)(a^2+ab+b^2)$. Here $a=1, b=x$. So, $1+x^2+x^4 = (1-x+x^2)(1+x+x^2)$. The integrand simplifies to $\frac{(1-x+x^2)(1+x+x^2)}{1+x+x^2} = 1-x+x^2$. Solution: $x - \frac{x^2}{2} + \frac{x^3}{3} + C$

6. $\int sec^2x.cosec^2xdx$

Rewrite in terms of $\sin x$ and $\cos x$: $\frac{1}{\cos^2x \sin^2x}$. Use the identity $1 = \sin^2x + \cos^2x$ in the numerator: $\frac{\sin^2x + \cos^2x}{\cos^2x \sin^2x}$. Split the fraction: $\frac{\sin^2x}{\cos^2x \sin^2x} + \frac{\cos^2x}{\cos^2x \sin^2x} = \frac{1}{\cos^2x} + \frac{1}{\sin^2x} = \sec^2x + \csc^2x$. Solution: $\tan x - \cot x + C$

7. $\int \frac{1+cos^2x}{1+cos2x}dx$

Use the identity $1+\cos2x = 2\cos^2x$. The integrand becomes $\frac{1+\cos^2x}{2\cos^2x} = \frac{1}{2\cos^2x} + \frac{\cos^2x}{2\cos^2x} = \frac{1}{2}\sec^2x + \frac{1}{2}$. Solution: $\frac{1}{2}\tan x + \frac{1}{2}x + C$

8. $\int sinx.sin(\pi/3-x).sin(\pi/3+x).dx$

Use the trigonometric identity $\sin\theta \sin(60^\circ-\theta) \sin(60^\circ+\theta) = \frac{1}{4}\sin(3\theta)$. Here $\theta=x$. So, $\sin x \sin(\pi/3-x) \sin(\pi/3+x) = \frac{1}{4}\sin(3x)$. Solution: $\int \frac{1}{4}\sin(3x) dx = -\frac{1}{12}\cos(3x) + C$

9. $\int sin^2xdx$

Use the identity $\sin^2x = \frac{1-\cos2x}{2}$. Solution: $\int \frac{1-\cos2x}{2} dx = \frac{x}{2} - \frac{\sin2x}{4} + C$

10. $\int sin^3xdx$

Rewrite $\sin^3x = \sin x \cdot \sin^2x = \sin x (1-\cos^2x)$. Let $u = \cos x$, then $du = -\sin x dx$. The integral becomes $\int (1-u^2)(-du) = \int (u^2-1)du = \frac{u^3}{3} - u + C$. Solution: $\frac{\cos^3x}{3} - \cos x + C$

11. $\int \frac{x^{2023}+x^{2025}}{(1+x+x^2)(1-x+x^2)-1}dx$

Simplify the denominator: $(1+x+x^2)(1-x+x^2) = (1+x^2)^2 - x^2 = 1+2x^2+x^4-x^2 = 1+x^2+x^4$. So, the denominator is $(1+x^2+x^4)-1 = x^2+x^4 = x^2(1+x^2)$. Factor the numerator: $x^{2023}+x^{2025} = x^{2023}(1+x^2)$. The integrand becomes $\frac{x^{2023}(1+x^2)}{x^2(1+x^2)} = x^{2021}$. Solution: $\frac{x^{2022}}{2022} + C$

12. $\int \frac{1+x^{4048}+x^{8096}}{1+x^{2024}+x^{4048}}dx$

Let $y = x^{2024}$. The integrand becomes $\frac{1+y^2+y^4}{1+y+y^2}$. As in integral 5, this simplifies to $1-y+y^2$. Substitute back $y=x^{2024}$: $1-x^{2024}+(x^{2024})^2 = 1-x^{2024}+x^{4048}$. Solution: $x - \frac{x^{2025}}{2025} + \frac{x^{4049}}{4049} + C$

13. $\int (\frac{sinx+sin2x+sin3x}{cosx+cos2x+cos3x})^2dx$

Simplify the terms inside the parenthesis using sum-to-product identities: Numerator: $(\sin x + \sin 3x) + \sin 2x = 2\sin(2x)\cos(-x) + \sin 2x = 2\sin 2x \cos x + \sin 2x = \sin 2x (2\cos x + 1)$. Denominator: $(\cos x + \cos 3x) + \cos 2x = 2\cos(2x)\cos(-x) + \cos 2x = 2\cos 2x \cos x + \cos 2x = \cos 2x (2\cos x + 1)$. The expression inside the square becomes $\frac{\sin 2x (2\cos x + 1)}{\cos 2x (2\cos x + 1)} = \tan 2x$. The integral is $\int (\tan 2x)^2 dx = \int \tan^2(2x) dx$. Use the identity $\tan^2\theta = \sec^2\theta - 1$. Solution: $\int (\sec^2(2x) - 1) dx = \frac{\tan(2x)}{2} - x + C$

14. $\int \frac{1+4^x+16^x}{1+2^x+4^x}dx$

Let $y = 2^x$. Then $4^x = y^2$ and $16^x = y^4$. The integrand becomes $\frac{1+y^2+y^4}{1+y+y^2}$. As in integral 5 and 12, this simplifies to $1-y+y^2$. Substitute back $y=2^x$: $1-2^x+(2^x)^2 = 1-2^x+4^x$. Solution: $\int (1-2^x+4^x) dx = x - \frac{2^x}{\ln 2} + \frac{4^x}{\ln 4} + C$

15. $\int \frac{x^4}{x^2+1}dx$

Use polynomial long division or algebraic manipulation: $\frac{x^4}{x^2+1} = \frac{x^4-1+1}{x^2+1} = \frac{(x^2-1)(x^2+1)+1}{x^2+1} = x^2-1 + \frac{1}{x^2+1}$. Solution: $\frac{x^3}{3} - x + \tan^{-1}x + C$

16. $\int \frac{cos8x-cos7x}{1+2cos5x}dx$

Numerator: Use $\cos A - \cos B = -2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$. $\cos 8x - \cos 7x = -2\sin\left(\frac{15x}{2}\right)\sin\left(\frac{x}{2}\right)$. Denominator: Use the identity $1+2\cos(2\theta) = \frac{\sin(3\theta)}{\sin(\theta)}$ or derive it as follows: Multiply $1+2\cos5x$ by $2\sin(5x/2)$: $2\sin(5x/2)(1+2\cos5x) = 2\sin(5x/2) + 4\sin(5x/2)\cos(5x)$. Use $2\sin A \cos B = \sin(A+B) + \sin(A-B)$: $4\sin(5x/2)\cos(5x) = 2[\sin(5x/2+5x) + \sin(5x/2-5x)] = 2[\sin(15x/2) + \sin(-5x/2)] = 2\sin(15x/2) - 2\sin(5x/2)$. So, $2\sin(5x/2)(1+2\cos5x) = 2\sin(5x/2) + 2\sin(15x/2) - 2\sin(5x/2) = 2\sin(15x/2)$. Thus, $1+2\cos5x = \frac{\sin(15x/2)}{\sin(5x/2)}$. The integrand becomes $\frac{-2\sin(\frac{15x}{2})\sin(\frac{x}{2})}{\frac{\sin(15x/2)}{\sin(5x/2)}} = -2\sin\left(\frac{x}{2}\right)\sin\left(\frac{5x}{2}\right)$. Use product-to-sum identity: $-2\sin A \sin B = \cos(A+B) - \cos(A-B)$. $-2\sin(x/2)\sin(5x/2) = \cos(x/2+5x/2) - \cos(x/2-5x/2) = \cos(3x) - \cos(-2x) = \cos 3x - \cos 2x$. Solution: $\int (\cos 3x - \cos 2x) dx = \frac{\sin 3x}{3} - \frac{\sin 2x}{2} + C$

17. $\int \frac{dx}{x^4-1}$

Factor the denominator: $x^4-1 = (x^2-1)(x^2+1) = (x-1)(x+1)(x^2+1)$. Use partial fraction decomposition: $\frac{1}{x^4-1} = \frac{A}{x-1} + \frac{B}{x+1} + \frac{Cx+D}{x^2+1}$. Solving for A, B, C, D yields $A=1/4, B=-1/4, C=0, D=-1/2$. So, $\frac{1}{x^4-1} = \frac{1}{4(x-1)} - \frac{1}{4(x+1)} - \frac{1}{2(x^2+1)}$. Solution: $\frac{1}{4}\ln|x-1| - \frac{1}{4}\ln|x+1| - \frac{1}{2}\tan^{-1}x + C = \frac{1}{4}\ln\left|\frac{x-1}{x+1}\right| - \frac{1}{2}\tan^{-1}x + C$

18. $\int tan^4xdx$

Rewrite $\tan^4x = \tan^2x \cdot \tan^2x = \tan^2x (\sec^2x - 1) = \tan^2x \sec^2x - \tan^2x$. Further substitute $\tan^2x = \sec^2x - 1$: $\tan^2x \sec^2x - (\sec^2x - 1) = \tan^2x \sec^2x - \sec^2x + 1$. For $\int \tan^2x \sec^2x dx$, let $u=\tan x$, $du=\sec^2x dx$. Then $\int u^2 du = \frac{u^3}{3} = \frac{\tan^3x}{3}$. Solution: $\frac{\tan^3x}{3} - \tan x + x + C$

19. $\int tan^5xdx$

Rewrite $\tan^5x = \tan^3x \cdot \tan^2x = \tan^3x (\sec^2x - 1) = \tan^3x \sec^2x - \tan^3x$. For $\int \tan^3x \sec^2x dx$, let $u=\tan x$, $du=\sec^2x dx$. Then $\int u^3 du = \frac{u^4}{4} = \frac{\tan^4x}{4}$. For $\int \tan^3x dx$: $\int \tan^3x dx = \int \tan x (\sec^2x - 1) dx = \int (\tan x \sec^2x - \tan x) dx$. For $\int \tan x \sec^2x dx$, let $v=\tan x$, $dv=\sec^2x dx$. Then $\int v dv = \frac{v^2}{2} = \frac{\tan^2x}{2}$. For $\int \tan x dx = \ln|\sec x|$. So, $\int \tan^3x dx = \frac{\tan^2x}{2} - \ln|\sec x|$. Combining the results: Solution: $\frac{\tan^4x}{4} - \left(\frac{\tan^2x}{2} - \ln|\sec x|\right) + C = \frac{\tan^4x}{4} - \frac{\tan^2x}{2} + \ln|\sec x| + C$

20. $\int \frac{sinx}{x+cosx}dx$

This integral is not solvable using elementary functions (i.e., it cannot be expressed in terms of polynomials, rational functions, exponentials, logarithms, trigonometric functions, and their inverses). It does not fit the "12 basic integrations and addition technique" methods.

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Explanation of the solution:

Each integral was solved by applying basic integration formulas, algebraic manipulation, and trigonometric identities.

1. $\int (n!)^x dx$: Standard exponential integral $\int a^x dx = a^x/\ln a$.
2. $\int \frac{4^x+8^x}{3^x+6^x}dx$: Factored numerator and denominator to cancel common terms, resulting in $\int (4/3)^x dx$.
3. $\int \frac{1+tan^2x}{1-tan^2x}dx$: Used $1+\tan^2x=\sec^2x$ and $1-\tan^2x = \cos2x/\cos^2x$ to simplify to $\int \sec2x dx$.
4. $\int tan^2xdx$: Used $\tan^2x = \sec^2x-1$.
5. $\int \frac{1+x^2+x^4}{1+x+x^2}dx$: Factored the numerator $1+x^2+x^4 = (1-x+x^2)(1+x+x^2)$ and cancelled the common term.
6. $\int sec^2x.cosec^2xdx$: Rewrote in terms of $\sin x, \cos x$, used $\sin^2x+\cos^2x=1$ in numerator, and split into $\int (\sec^2x+\csc^2x)dx$.
7. $\int \frac{1+cos^2x}{1+cos2x}dx$: Used $1+\cos2x=2\cos^2x$ and split the fraction.
8. $\int sinx.sin(\pi/3-x).sin(\pi/3+x).dx$: Applied the identity $\sin\theta \sin(60^\circ-\theta) \sin(60^\circ+\theta) = \frac{1}{4}\sin(3\theta)$.
9. $\int sin^2xdx$: Used $\sin^2x = (1-\cos2x)/2$.
10. $\int sin^3xdx$: Rewrote as $\sin x(1-\cos^2x)$ and used substitution $u=\cos x$.
11. $\int \frac{x^{2023}+x^{2025}}{(1+x+x^2)(1-x+x^2)-1}dx$: Simplified the denominator to $x^2(1+x^2)$ and factored the numerator to $x^{2023}(1+x^2)$, leading to $\int x^{2021}dx$.
12. $\int \frac{1+x^{4048}+x^{8096}}{1+x^{2024}+x^{4048}}dx$: Used substitution $y=x^{2024}$ to reduce it to the form of integral 5.
13. $\int (\frac{sinx+sin2x+sin3x}{cosx+cos2x+cos3x})^2dx$: Applied sum-to-product formulas to simplify the fraction to $\tan2x$, then used $\tan^2\theta=\sec^2\theta-1$.
14. $\int \frac{1+4^x+16^x}{1+2^x+4^x}dx$: Used substitution $y=2^x$ to reduce it to the form of integral 5.
15. $\int \frac{x^4}{x^2+1}dx$: Used algebraic division $x^4/(x^2+1) = x^2-1 + 1/(x^2+1)$.
16. $\int \frac{cos8x-cos7x}{1+2cos5x}dx$: Used sum-to-product identities for numerator and a specific identity for the denominator, leading to $\int (\cos3x-\cos2x)dx$.
17. $\int \frac{dx}{x^4-1}$: Used partial fraction decomposition after factoring the denominator.
18. $\int tan^4xdx$: Rewrote as $\tan^2x(\sec^2x-1)$ and used substitution $u=\tan x$.
19. $\int tan^5xdx$: Rewrote as $\tan^3x(\sec^2x-1)$ and used substitution $u=\tan x$ for one part and further reduction for $\int \tan^3x dx$.
20. $\int \frac{sinx}{x+cosx}dx$: This integral is not expressible in terms of elementary functions.