Question

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

$01. \int 1*2*3* ....n^x dx$

$02. \int sec^2x.cosec^2x dx$

$04. \int \frac{1 + cos^2x}{1 + cos2x} dx$

$05. \int \frac{1+tan^2x}{1-tan^2x}dx$

Q6. $\int sinx . sin(\pi/3 - x) . sin(\pi/3 + x).dx$

$\int tan^2xdx$

Q8. $\int sin^2xdx$

Q9. $\int sin^3xdx$

$010. \int sin^4xdx$

Q11. $\int \frac{x^{2023} + x^{2025}}{(1 + x + x^2)(1 - x + x^2) - 1}dx$

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

Q13. $\int \frac{1 + x^{4048} + x^{8096}}{1 + x^{2024} + x^{4048}}dx$

Q14. $\int \frac{1 + 4^x + 16^x}{1 + 2^x + 4^x}dx$

Q16. $\int (\frac{sinx + sin2x + sin3x}{cosx + cos2x + cos3x})^2dx$

Q17. $\int \frac{x^4}{x^2+1}dx$

Q19. $\int \frac{cos8x - cos7x}{1 + 2cos5x}dx$

Q20. $\int \frac{dx}{x^4-1}$

chnique to get the answer.

$\int tan^4xdx$

Q3. $\int tan^5xdx$

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

Answer 1

Here are the solutions to the given indefinite integration problems using basic integration formulas and algebraic/trigonometric manipulations.

Q1. $\int 1 \cdot 2 \cdot 3 \cdot \dots \cdot n^x dx$ The product $1 \cdot 2 \cdot 3 \cdot \dots \cdot n$ is $n!$. So the integrand is $(n!)^x$. This is a standard integral of the form $\int a^x dx = \frac{a^x}{\ln a} + C$. Here, $a = n!$. $\int (n!)^x dx = \frac{(n!)^x}{\ln(n!)} + C$

Q2. $\int \sec^2x \cdot \csc^2x dx$ Rewrite $\sec^2x \cdot \csc^2x$ as $\frac{1}{\cos^2x \sin^2x}$. Use the identity $\sin^2x + \cos^2x = 1$ in the numerator: $\int \frac{\sin^2x + \cos^2x}{\cos^2x \sin^2x} dx = \int \left(\frac{\sin^2x}{\cos^2x \sin^2x} + \frac{\cos^2x}{\cos^2x \sin^2x}\right) dx$ $= \int \left(\frac{1}{\cos^2x} + \frac{1}{\sin^2x}\right) dx = \int (\sec^2x + \csc^2x) dx$ Integrate term by term: $= \tan x - \cot x + C$

Q4. $\int \frac{1 + \cos^2x}{1 + \cos2x} dx$ Use the trigonometric identity $1 + \cos2x = 2\cos^2x$ in the denominator: $\int \frac{1 + \cos^2x}{2\cos^2x} dx = \int \left(\frac{1}{2\cos^2x} + \frac{\cos^2x}{2\cos^2x}\right) dx$ $= \int \left(\frac{1}{2}\sec^2x + \frac{1}{2}\right) dx$ Integrate term by term: $= \frac{1}{2}\tan x + \frac{1}{2}x + C$

Q5. $\int \frac{1+\tan^2x}{1-\tan^2x}dx$ Use the identity $1+\tan^2x = \sec^2x$ in the numerator. The denominator $1-\tan^2x = \frac{\cos^2x - \sin^2x}{\cos^2x} = \frac{\cos2x}{\cos^2x}$. Substitute these into the integral: $\int \frac{\sec^2x}{\frac{\cos2x}{\cos^2x}} dx = \int \frac{1/\cos^2x}{\cos2x/\cos^2x} dx = \int \frac{1}{\cos2x} dx = \int \sec2x dx$ This is a standard integral: $= \frac{1}{2}\ln|\sec2x + \tan2x| + C$

Q6. $\int \sin x \cdot \sin(\pi/3 - x) \cdot \sin(\pi/3 + x) dx$ Use the trigonometric identity $\sin A \sin(60^\circ - A) \sin(60^\circ + A) = \frac{1}{4}\sin(3A)$. Here, $A=x$ and $60^\circ = \pi/3$. So, $\sin x \sin(\pi/3 - x) \sin(\pi/3 + x) = \frac{1}{4}\sin(3x)$. The integral becomes: $\int \frac{1}{4}\sin(3x) dx = \frac{1}{4} \left(-\frac{\cos(3x)}{3}\right) + C$ $= -\frac{1}{12}\cos(3x) + C$

Q7. $\int \tan^2x dx$ Use the trigonometric identity $\tan^2x = \sec^2x - 1$: $\int (\sec^2x - 1) dx$ Integrate term by term: $= \tan x - x + C$

Q8. $\int \sin^2x dx$ Use the trigonometric identity $\sin^2x = \frac{1-\cos2x}{2}$: $\int \frac{1-\cos2x}{2} dx = \frac{1}{2} \int (1-\cos2x) dx$ $= \frac{1}{2} \left(x - \frac{\sin2x}{2}\right) + C$ $= \frac{x}{2} - \frac{\sin2x}{4} + C$

Q9. $\int \sin^3x dx$ Rewrite $\sin^3x$ as $\sin^2x \cdot \sin x = (1-\cos^2x)\sin x$. Let $u = \cos x$, then $du = -\sin x dx$. So $\sin x dx = -du$. $\int (1-u^2)(-du) = \int (u^2-1) du$ $= \frac{u^3}{3} - u + C$ Substitute back $u = \cos x$: $= \frac{\cos^3x}{3} - \cos x + C$

Q10. $\int \sin^4x dx$ Rewrite $\sin^4x = (\sin^2x)^2$. Use $\sin^2x = \frac{1-\cos2x}{2}$: $\int \left(\frac{1-\cos2x}{2}\right)^2 dx = \int \frac{1}{4}(1 - 2\cos2x + \cos^22x) dx$ Use $\cos^2\theta = \frac{1+\cos2\theta}{2}$ for $\cos^22x$: $\cos^22x = \frac{1+\cos(2 \cdot 2x)}{2} = \frac{1+\cos4x}{2}$ Substitute this back: $\int \frac{1}{4}\left(1 - 2\cos2x + \frac{1+\cos4x}{2}\right) dx = \int \frac{1}{4}\left(1 - 2\cos2x + \frac{1}{2} + \frac{1}{2}\cos4x\right) dx$ $= \int \frac{1}{4}\left(\frac{3}{2} - 2\cos2x + \frac{1}{2}\cos4x\right) dx = \int \left(\frac{3}{8} - \frac{1}{2}\cos2x + \frac{1}{8}\cos4x\right) dx$ Integrate term by term: $= \frac{3}{8}x - \frac{1}{2}\frac{\sin2x}{2} + \frac{1}{8}\frac{\sin4x}{4} + C$ $= \frac{3x}{8} - \frac{\sin2x}{4} + \frac{\sin4x}{32} + C$

Q11. $\int \frac{x^{2023} + x^{2025}}{(1 + x + x^2)(1 - x + x^2) - 1}dx$ First, simplify the denominator: $(1 + x + x^2)(1 - x + x^2) = ((1+x^2)+x)((1+x^2)-x)$ This is of the form $(A+B)(A-B) = A^2 - B^2$, where $A = 1+x^2$ and $B = x$. So, $(1+x^2)^2 - x^2 = (1+2x^2+x^4) - x^2 = 1+x^2+x^4$. The denominator becomes $(1+x^2+x^4) - 1 = x^2+x^4 = x^2(1+x^2)$. Now simplify the numerator: $x^{2023} + x^{2025} = x^{2023}(1+x^2)$. Substitute these into the integral: $\int \frac{x^{2023}(1+x^2)}{x^2(1+x^2)} dx$ Cancel out the common term $(1+x^2)$ and $x^2$: $\int x^{2021} dx$ Using the power rule $\int x^n dx = \frac{x^{n+1}}{n+1} + C$: $= \frac{x^{2022}}{2022} + C$

Q12. $\int \frac{1+x^2+x^4}{1+x+x^2}dx$ Factor the numerator $1+x^2+x^4$. It can be written as $(x^2+1)^2 - x^2$, which is a difference of squares: $(x^2+1-x)(x^2+1+x)$. $\int \frac{(x^2-x+1)(x^2+x+1)}{x^2+x+1} dx$ Cancel out the common factor $(x^2+x+1)$: $\int (x^2-x+1) dx$ Integrate term by term: $= \frac{x^3}{3} - \frac{x^2}{2} + x + C$

Q13. $\int \frac{1 + x^{4048} + x^{8096}}{1 + x^{2024} + x^{4048}}dx$ Let $y = x^{2024}$. Then $x^{4048} = (x^{2024})^2 = y^2$ and $x^{8096} = (x^{2024})^4 = y^4$. The integral expression becomes $\frac{1+y^2+y^4}{1+y+y^2}$. As shown in Q12, this simplifies to $1-y+y^2$. Substitute back $y = x^{2024}$: $\int (1 - x^{2024} + (x^{2024})^2) dx = \int (1 - x^{2024} + x^{4048}) dx$ Integrate term by term: $= x - \frac{x^{2024+1}}{2024+1} + \frac{x^{4048+1}}{4048+1} + C$ $= x - \frac{x^{2025}}{2025} + \frac{x^{4049}}{4049} + C$

Q14. $\int \frac{1 + 4^x + 16^x}{1 + 2^x + 4^x}dx$ Let $y = 2^x$. Then $4^x = (2^x)^2 = y^2$ and $16^x = (2^x)^4 = y^4$. The integral expression becomes $\frac{1+y^2+y^4}{1+y+y^2}$. As shown in Q12, this simplifies to $1-y+y^2$. Substitute back $y = 2^x$: $\int (1 - 2^x + (2^x)^2) dx = \int (1 - 2^x + 4^x) dx$ Integrate term by term using $\int a^x dx = \frac{a^x}{\ln a} + C$: $= x - \frac{2^x}{\ln2} + \frac{4^x}{\ln4} + C$ Since $\ln4 = \ln(2^2) = 2\ln2$: $= x - \frac{2^x}{\ln2} + \frac{4^x}{2\ln2} + C$

Q16. $\int \left(\frac{\sin x + \sin2x + \sin3x}{\cos x + \cos2x + \cos3x}\right)^2dx$ Simplify the numerator using sum-to-product formulas: $\sin x + \sin3x + \sin2x = 2\sin\left(\frac{x+3x}{2}\right)\cos\left(\frac{x-3x}{2}\right) + \sin2x$ $= 2\sin2x\cos(-x) + \sin2x = 2\sin2x\cos x + \sin2x = \sin2x(2\cos x + 1)$. Simplify the denominator using sum-to-product formulas: $\cos x + \cos3x + \cos2x = 2\cos\left(\frac{x+3x}{2}\right)\cos\left(\frac{x-3x}{2}\right) + \cos2x$ $= 2\cos2x\cos(-x) + \cos2x = 2\cos2x\cos x + \cos2x = \cos2x(2\cos x + 1)$. The expression inside the parenthesis simplifies to: $\frac{\sin2x(2\cos x + 1)}{\cos2x(2\cos x + 1)} = \frac{\sin2x}{\cos2x} = \tan2x$ The integral becomes: $\int (\tan2x)^2 dx = \int \tan^2(2x) dx$ Use the identity $\tan^2\theta = \sec^2\theta - 1$: $\int (\sec^2(2x) - 1) dx$ Integrate term by term: $= \frac{\tan(2x)}{2} - x + C$

Q17. $\int \frac{x^4}{x^2+1}dx$ Use polynomial long division or algebraic manipulation: $\frac{x^4}{x^2+1} = \frac{x^4-1+1}{x^2+1} = \frac{(x^2-1)(x^2+1)+1}{x^2+1}$ $= \frac{(x^2-1)(x^2+1)}{x^2+1} + \frac{1}{x^2+1} = (x^2-1) + \frac{1}{x^2+1}$ The integral becomes: $\int \left(x^2-1 + \frac{1}{x^2+1}\right) dx$ Integrate term by term: $= \frac{x^3}{3} - x + \tan^{-1}x + C$

Q19. $\int \frac{\cos8x - \cos7x}{1 + 2\cos5x}dx$ Simplify the numerator using the sum-to-product identity $\cos A - \cos B = -2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$: $\cos8x - \cos7x = -2\sin\left(\frac{8x+7x}{2}\right)\sin\left(\frac{8x-7x}{2}\right) = -2\sin\left(\frac{15x}{2}\right)\sin\left(\frac{x}{2}\right)$. Simplify the denominator using the identity $1+2\cos(2\theta) = \frac{\sin(3\theta)}{\sin\theta}$. Let $2\theta = 5x$, so $\theta = 5x/2$: $1+2\cos5x = \frac{\sin(3 \cdot 5x/2)}{\sin(5x/2)} = \frac{\sin(15x/2)}{\sin(5x/2)}$. Substitute these into the integral: $\int \frac{-2\sin\left(\frac{15x}{2}\right)\sin\left(\frac{x}{2}\right)}{\frac{\sin\left(\frac{15x}{2}\right)}{\sin\left(\frac{5x}{2}\right)}} dx = \int -2\sin\left(\frac{x}{2}\right)\sin\left(\frac{5x}{2}\right) dx$ Use the product-to-sum identity $-2\sin A \sin B = \cos(A+B) - \cos(A-B)$: Here $A = x/2$ and $B = 5x/2$. So $A+B = 3x$ and $A-B = -2x$. $\int (\cos(3x) - \cos(-2x)) dx = \int (\cos(3x) - \cos(2x)) dx$ Integrate term by term: $= \frac{\sin(3x)}{3} - \frac{\sin(2x)}{2} + C$

Q20. $\int \frac{dx}{x^4-1}$ 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 (as shown in thought process), we get $A = 1/4$, $B = -1/4$, $C = 0$, $D = -1/2$. $\int \left(\frac{1}{4(x-1)} - \frac{1}{4(x+1)} - \frac{1}{2(x^2+1)}\right) dx$ Integrate term by term: $= \frac{1}{4}\ln|x-1| - \frac{1}{4}\ln|x+1| - \frac{1}{2}\tan^{-1}x + C$ $= \frac{1}{4}\ln\left|\frac{x-1}{x+1}\right| - \frac{1}{2}\tan^{-1}x + C$

Q21. $\int \tan^4x dx$ (Assuming this is the unnumbered integral after Q20) Rewrite $\tan^4x = \tan^2x \cdot \tan^2x$. Use $\tan^2x = \sec^2x - 1$: $\int \tan^2x (\sec^2x - 1) dx = \int (\tan^2x \sec^2x - \tan^2x) dx$ Substitute $\tan^2x = \sec^2x - 1$ for the second term: $= \int (\tan^2x \sec^2x - (\sec^2x - 1)) dx = \int (\tan^2x \sec^2x - \sec^2x + 1) dx$ Integrate term by term: For $\int \tan^2x \sec^2x dx$, let $u = \tan x$, then $du = \sec^2x dx$. So $\int u^2 du = \frac{u^3}{3} = \frac{\tan^3x}{3}$. For $\int \sec^2x dx = \tan x$. For $\int 1 dx = x$. $= \frac{\tan^3x}{3} - \tan x + x + C$

Q3. $\int \tan^5x dx$ (Assuming this is the integral labeled Q3 after the unnumbered ones) Rewrite $\tan^5x = \tan^3x \cdot \tan^2x$. Use $\tan^2x = \sec^2x - 1$: $\int \tan^3x (\sec^2x - 1) dx = \int (\tan^3x \sec^2x - \tan^3x) dx$ Rewrite $\tan^3x = \tan x \cdot \tan^2x = \tan x (\sec^2x - 1)$: $= \int (\tan^3x \sec^2x - \tan x (\sec^2x - 1)) dx = \int (\tan^3x \sec^2x - \tan x \sec^2x + \tan x) dx$ Integrate term by term: For $\int \tan^3x \sec^2x dx$, let $u = \tan x$, then $du = \sec^2x dx$. So $\int u^3 du = \frac{u^4}{4} = \frac{\tan^4x}{4}$. For $\int \tan x \sec^2x dx$, let $u = \tan x$, then $du = \sec^2x dx$. So $\int u du = \frac{u^2}{2} = \frac{\tan^2x}{2}$. For $\int \tan x dx = \ln|\sec x|$. $= \frac{\tan^4x}{4} - \frac{\tan^2x}{2} + \ln|\sec x| + C$

Q23. $\int \frac{\sin x}{x + \cos x}dx$ (Assuming this is the last unnumbered integral) This integral does not have an elementary antiderivative that can be found using the basic integration techniques and addition as specified. It is possible there is a typo in the problem statement.

Answer 2

Here are the solutions to the given indefinite integration problems using basic integration formulas and algebraic/trigonometric manipulations.

Q1: Recognize $1*2*...*n$ as $n!$. Apply the standard integral formula for $a^x$. Q2: Rewrite in terms of $\sin x$ and $\cos x$. Use $\sin^2x + \cos^2x = 1$ in the numerator and split the fraction. Integrate $\sec^2x$ and $\csc^2x$. Q4: Use the identity $1+\cos2x = 2\cos^2x$. Split the fraction and integrate. Q5: Use $1+\tan^2x = \sec^2x$ and $1-\tan^2x = \frac{\cos2x}{\cos^2x}$. Simplify to $\sec2x$ and integrate. Q6: Apply the identity $\sin A \sin(60^\circ - A) \sin(60^\circ + A) = \frac{1}{4}\sin(3A)$. Integrate the resulting term. Q7: Use the identity $\tan^2x = \sec^2x - 1$. Integrate. Q8: Use the identity $\sin^2x = \frac{1-\cos2x}{2}$. Integrate. Q9: Rewrite $\sin^3x = (1-\cos^2x)\sin x$. Use substitution $u = \cos x$. Q10: Rewrite $\sin^4x = (\sin^2x)^2$. Apply $\sin^2x = \frac{1-\cos2x}{2}$ twice. Simplify and integrate. Q11: Simplify the denominator using $(A+B)(A-B)=A^2-B^2$ and then factor. Factor the numerator. Cancel common terms. Integrate the power function. Q12: Factor the numerator $1+x^2+x^4 = (x^2-x+1)(x^2+x+1)$. Cancel common terms. Integrate the resulting polynomial. Q13: Substitute $y=x^{2024}$. The form becomes identical to Q12. Simplify and substitute back to integrate. Q14: Substitute $y=2^x$. The form becomes identical to Q12. Simplify and substitute back to integrate exponential terms. Q16: Use sum-to-product formulas for numerator and denominator. Simplify to $\tan2x$. Integrate $\tan^2(2x)$ using $\tan^2\theta = \sec^2\theta - 1$. Q17: Use polynomial long division or algebraic manipulation (add and subtract 1 in the numerator) to simplify the fraction. Integrate the resulting polynomial and standard inverse tangent term. Q19: Use sum-to-product for numerator $\cos A - \cos B$. Use a specific identity for denominator $1+2\cos(5x)$. Simplify the expression and then apply product-to-sum for the remaining sine terms. Integrate. Q20: Use partial fraction decomposition for $\frac{1}{x^4-1}$. Integrate the resulting logarithmic and inverse tangent terms. Q21 (unnumbered $\int \tan^4x dx$): Rewrite $\tan^4x = \tan^2x \cdot \tan^2x$. Use $\tan^2x = \sec^2x - 1$ repeatedly. Use substitution for $\int \tan^2x \sec^2x dx$. Q3 (labeled $\int \tan^5x dx$): Rewrite $\tan^5x = \tan^3x \cdot \tan^2x$. Use $\tan^2x = \sec^2x - 1$ repeatedly. Use substitution for terms like $\int \tan^3x \sec^2x dx$ and $\int \tan x \sec^2x dx$. Q23 (unnumbered $\int \frac{\sin x}{x + \cos x}dx$): This integral does not have an elementary antiderivative.