Q.Integrate the following function: $\int x^2 \left(1 - \frac{1}{x^2}\right) dx$

Showing the 12 most recent of 19 on this concept.

1. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{x^4-16x^2+2x+8}{x^3-4x^2+2}\,dx=$ (A) $\dfrac{x^2+8x+c}{2}$ (B) $x^2+8x+c$ (C) $x^3-4x+c$ (D) $\dfrac{x^2-8x+c}{2}$

   ›Reveal solutionSolution

   The rational integrand is actually a polynomial in disguise — long division of the numerator by the denominator leaves zero remainder, collapsing the integral to a simple polynomial integral.

   Concept and Intuition

   When a numerator's degree exceeds the denominator's by 1, always try polynomial long division first; if the division is exact, the "hard" rational integral vanishes into an easy polynomial one.

   Step-by-Step Solution

   1. Divide $x^4+0x^3-16x^2+2x+8$ by $x^3-4x^2+0x+2$.
   2. First term of quotient: $x$. Multiply: $x^4-4x^3+0x^2+2x$. Subtract: $4x^3-16x^2+0x+8$.
   3. Next term of quotient: $4$. Multiply: $4x^3-16x^2+0x+8$. Subtract: remainder $=0$.
   4. So the integrand simplifies exactly to $x+4$. …

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2. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
   Q.$\displaystyle\int (1 - \cos x)\,\text{cosec}^2 x \; dx =$ (A) $\tan\left(\dfrac{x}{2}\right) + c$ (B) $-\tan\left(\dfrac{x}{2}\right) + c$ (C) $2\tan\left(\dfrac{x}{2}\right) + c$ (D) $-2\tan\left(\dfrac{x}{2}\right) + c$

   ›Reveal solutionSolution

   Splitting the integrand into two standard integrals gives $\csc x-\cot x$, which the half-angle identity simplifies exactly to $\tan(x/2)$.

   Concept and Intuition

   $(1-\cos x)\csc^2x = \csc^2x - \cos x\csc^2x = \csc^2x-\cot x\csc x$, a sum of two integrals with well-known antiderivatives ($-\cot x$ and $-\csc x$ respectively). The result $\csc x-\cot x$ then simplifies via the half-angle identity $1-\cos x=2\sin^2(x/2)$ and $\sin x = 2\sin(x/2)\cos(x/2)$.

   Step-by-Step Solution

   1. Expand: $(1-\cos x)\csc^2x=\csc^2x-\cos x\csc^2 x=\csc^2x-\cot x\csc x$.
   2. $\displaystyle\int\csc^2x\,dx=-\cot x+C_1$.
   3. $\displaystyle\int\cot x\csc x\,dx=-\csc x+C_2$.
   4. So $\displaystyle\int(1-\cos x)\csc^2x\,dx=-\cot x-(-\csc x)=\csc x-\cot x+C$.
   5. Write as a single fraction: $\csc x-\cot x=\dfrac{1-\cos x}{\sin x}$.
   6. Use half-angle identities: $1-\cos x=2\sin^2\left(\dfrac x2\right)$ and $\sin x=2\sin\left(\dfrac x2\right)\cos\left(\dfrac x2\right)$. …

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3. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
   Q.If $\displaystyle\int_0^1 f(x)\, dx = 1$, $\displaystyle\int_0^1 x f(x)\, dx = a$, $\displaystyle\int_0^1 x^2 f(x)\, dx = a^2$ then $\displaystyle\int_0^1 (x-a)^2 f(x)\, dx =$ ______ (A) $a^2$ (B) $a^2 + 1$ (C) $a^2 - 1$ (D) $0$

   ›Reveal solutionSolution

   Expand the square inside the integral and split it into the three given integrals — everything cancels to zero.

   Concept and Intuition

   This is really testing that $\int_0^1(x-a)^2f(x)\,dx$ is the "variance-like" expression built from the given moments, and with the specific values given ($\int x f=a$ and $\int x^2f=a^2$), the expansion collapses exactly to zero — a clean algebraic identity, not a coincidence of these particular numbers.

   Step-by-Step Solution

   1. Expand: $(x-a)^2 = x^2 - 2ax + a^2$.
   2. So $\displaystyle\int_0^1(x-a)^2f(x)\,dx = \int_0^1 x^2f(x)\,dx - 2a\int_0^1 xf(x)\,dx + a^2\int_0^1 f(x)\,dx$. …

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4. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
   Q.$\displaystyle\int (x+1)(x+2)^4(x+3)\, dx =$ (A) $\dfrac{(x+1)^2}{2} + \dfrac{(x+2)^2}{5} + \dfrac{(x+3)^2}{2} + c$ (B) $\dfrac{(x+2)^7}{7} - \dfrac{(x+2)^5}{5} + c$ (C) $\dfrac{(x+2)^7}{7} + \dfrac{(x+2)^5}{5} + c$ (D) $\dfrac{(x+3)^7}{7} - \dfrac{(x+3)^5}{5} + c$

   ›Reveal solutionSolution

   Shift the variable to center it on the repeated factor $(x+2)$, turning the product into a simple polynomial in $u$.

   Concept and Intuition

   When a product involves three linear factors symmetric about a middle value (here $x+1$ and $x+3$ are symmetric about $x+2$), substituting $u$ equal to the middle factor converts the outer two into a difference of squares, dramatically simplifying the integral.

   Step-by-Step Solution

   1. Let $u = x+2 \Rightarrow du = dx$. Then $x+1 = u-1$ and $x+3 = u+1$.
   2. $(x+1)(x+3) = (u-1)(u+1) = u^2-1$.
   3. So the integrand $(x+1)(x+2)^4(x+3) = (u^2-1)u^4 = u^6-u^4$.
   4. $\displaystyle\int (u^6-u^4)\,du = \dfrac{u^7}{7}-\dfrac{u^5}{5}+c$.
   5. Substitute back $u=x+2$: $\dfrac{(x+2)^7}{7}-\dfrac{(x+2)^5}{5}+c$. …

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5. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.$\displaystyle\sum_{n=1}^{\infty} \int_0^1 x^3(1 - x^2)^n\,dx =$ (A) $\dfrac{1}{4}$ (B) $\dfrac{1}{2}$ (C) $1$ (D) $2$

   ›Reveal solutionSolution

   Substituting $u=x^2$ turns the inner integral into a Beta-function value

   $\frac{1}{2(n+1)(n+2)}$; summing that telescoping series over $n\ge1$ gives $1/4$.

   Concept and Intuition

   $\int_0^1u^p(1-u)^q\,du=B(p+1,q+1)=\dfrac{p!\,q!}{(p+q+1)!}$ for non-negative integers

   $p,q$. Once the per-$n$ integral is reduced to this closed form, summing over $n$ often

   telescopes because $\frac{1}{(n+1)(n+2)}=\frac1{n+1}-\frac1{n+2}$.

   Step-by-Step Solution

   1. Substitute $u=x^2$, $du=2x\,dx$. Then $x^3\,dx=x^2\cdot x\,dx=u\cdot\dfrac{du}{2}$.
   2. $\displaystyle\int_0^1x^3(1-x^2)^n\,dx=\dfrac12\int_0^1u(1-u)^n\,du$.
   3. $\displaystyle\int_0^1u(1-u)^n\,du=B(2,n+1)=\dfrac{1!\,n!}{(n+2)!}=\dfrac{1}{(n+1)(n+2)}$.
   4. So the per-$n$ integral equals $\dfrac{1}{2(n+1)(n+2)}$. …

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6. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.$\displaystyle\int \sin^3 x \cos^2 x\, dx =$ (A) $\dfrac{\sin^4 x \cos x}{5} - \dfrac{\sin^2 x \cos x}{15} - \dfrac{2\cos x}{15} + c$ (B) $-\dfrac{\sin^4 x \cos x}{5} - \dfrac{\sin^2 x \cos x}{15} + \dfrac{2\cos x}{15} + c$ (C) $\dfrac{\sin^4 x \cos x}{5} - \dfrac{\sin^2 x \cos x}{15} + \dfrac{2x}{15} + c$ (D) $\dfrac{\sin^4 x \cos x}{5} + \dfrac{\sin^2 x \cos x}{3} - \dfrac{2x}{15} + c$

   ›Reveal solutionSolution

   Splitting off one $\sin x$ as $d(\cos x)$ and using $\sin^2x=1-\cos^2x$ reduces the integral to powers of $\cos x$, which re-expressed in $\sin x,\cos x$ mixed form matches option (A).

   Concept and Intuition

   For $\int\sin^{\text{odd}}x\cos^{\text{even}}x\,dx$, peel off one factor of $\sin x$ to pair with $dx$ (since $d(\cos x)=-\sin x\,dx$), turning the rest into a polynomial in $\cos x$ via $\sin^2x=1-\cos^2x$.

   Step-by-Step Solution

   1. $\sin^3x\cos^2x=\sin x(1-\cos^2x)\cos^2x=\sin x\cos^2x-\sin x\cos^4x$.
   2. $\displaystyle\int\sin x\cos^2x\,dx=-\dfrac{\cos^3x}3$, and $\displaystyle\int\sin x\cos^4x\,dx=-\dfrac{\cos^5x}5$.
   3. So the integral is $-\dfrac{\cos^3x}3-\left(-\dfrac{\cos^5x}5\right)=\dfrac{\cos^5x}5-\dfrac{\cos^3x}3+c$.
   4. Rewrite using $\cos^5x=(1-\sin^2x)^2\cos x$ and $\cos^3x=(1-\sin^2x)\cos x$: $\dfrac15(1-2\sin^2x+\sin^4x)\cos x-\dfrac13(1-\sin^2x)\cos x.$ …

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7. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.$\displaystyle\int_0^2 x^2(2-x)^5\,dx =$ (A) $\dfrac{128}{21}$ (B) $\dfrac{64}{7}$ (C) $\dfrac{32}{21}$ (D) $\dfrac{16}{7}$

   ›Reveal solutionSolution

   This tests the standard Beta-integral shortcut $\int_0^a x^m(a-x)^n\,dx=a^{m+n+1}\dfrac{m!\,n!}{(m+n+1)!}$, which avoids expanding $(2-x)^5$ by hand.

   Concept and Intuition

   Any integral of the form $\int_0^a x^m(a-x)^n\,dx$ is a scaled Beta function: substituting $x=at$ turns it into $a^{m+n+1}\int_0^1 t^m(1-t)^n\,dt=a^{m+n+1}B(m+1,n+1)$, and $B(p,q)=\dfrac{(p-1)!(q-1)!}{(p+q-1)!}$ for positive integers. This is far quicker than binomially expanding $(2-x)^5$ and integrating term by term.

   Step-by-Step Solution

   1. Identify $a=2,\ m=2,\ n=5$, so $m+n+1=8$.
   2. Apply the formula: $I=a^{8}\cdot\dfrac{m!\,n!}{8!}=2^8\cdot\dfrac{2!\,5!}{8!}$.
   3. Compute: $2^8=256$, $2!=2$, $5!=120$, $8!=40320$.
   4. $I=256\cdot\dfrac{2\cdot120}{40320}=256\cdot\dfrac{240}{40320}=\dfrac{61440}{40320}$. …

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8. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.If $\displaystyle\int \dfrac{x^4 + 1}{x^2 + 1}\,dx = Ax^3 + Bx^2 + Cx + D\tan^{-1}x + E$, then $A + B + C + D =$ (A) $\dfrac{3}{2}$ (B) $\dfrac{4}{3}$ (C) $\dfrac{1}{3}$ (D) $\dfrac{2}{3}$

   ›Reveal solutionSolution

   Polynomial division turns $\frac{x^4+1}{x^2+1}$ into $x^2-1+\frac{2}{x^2+1}$, whose integral matches the given form with $A+B+C+D=\frac43$.

   Concept and Intuition

   When the numerator's degree is higher than the denominator's, divide first (or find a clever algebraic split) to reduce the integral to standard forms — here a polynomial plus a $\tan^{-1}$ term.

   Step-by-Step Solution

   1. Write $x^4+1 = (x^2+1)(x^2-1)+2$ — check: $(x^2+1)(x^2-1)=x^4-1$, plus $2$ gives $x^4+1$. ✓
   2. So $\dfrac{x^4+1}{x^2+1} = x^2-1+\dfrac{2}{x^2+1}$.
   3. Integrate term by term: $\displaystyle\int(x^2-1)dx + \int\frac{2}{x^2+1}dx = \frac{x^3}{3}-x+2\tan^{-1}x+E$. …

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9. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{1-\cos x}{\cos x(1+\cos x)}\,dx=$ (A) $\log|\sec x+\tan x|-2(\csc x-\cot x)+C$ (B) $\log|\sec x+\tan x|-2(\csc x+\cot x)+C$ (C) $\log|\sec x+\tan x|+2(\csc x-\cot x)+C$ (D) $\log|\sec x+\tan x|+2(\csc x+\cot x)+C$

   ›Reveal solutionSolution

   Splitting the integrand by partial fractions in $\cos x$ reduces it to $\sec x-\sec^2(x/2)$, whose integral, rewritten via the half-angle identity $\tan(x/2)=\csc x-\cot x$, matches option (A).

   Concept and Intuition

   Treating $c=\cos x$ as the variable lets us do ordinary partial fractions; then the half-angle substitution $1+\cos x=2\cos^2(x/2)$ turns the resulting term into a standard $\sec^2$ integral.

   Step-by-Step Solution

   1. Let $c=\cos x$. Write $\dfrac{1-c}{c(1+c)}=\dfrac{A}{c}+\dfrac{B}{1+c}$: $1-c=A(1+c)+Bc$. At $c=0$: $A=1$. At $c=-1$: $2=-B\Rightarrow B=-2$.
   2. So the integrand $=\dfrac1{\cos x}-\dfrac{2}{1+\cos x}=\sec x-\dfrac{2}{2\cos^2(x/2)}=\sec x-\sec^2(x/2)$.
   3. $\int\sec x\,dx=\log|\sec x+\tan x|+C$.
   4. $\int\sec^2(x/2)\,dx=2\tan(x/2)$, so $-2\int\sec^2(x/2)\cdot\tfrac12\,dx$... directly: $\int\dfrac{2}{1+\cos x}dx=2\tan(x/2)$. …

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10. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
    Q.$\displaystyle\int \dfrac{x^8+4}{x^4-2x^2+2}\,dx = Ax^5 + Bx^3 + Cx + K$, then $5A + 3B + C =$ (A) $7$ (B) $5$ (C) $3$ (D) $1$

    ›Reveal solutionSolution

    The rational integrand simplifies via exact polynomial long division before integrating termwise. Answer: $5A+3B+C=5$.

    Concept and Intuition

    Before reaching for partial fractions or a substitution, it's always worth checking whether a "rational function" integrand is actually a disguised polynomial — i.e., whether the denominator divides the numerator exactly. Here, $x^4-2x^2+2$ turns out to divide $x^8+4$ with zero remainder, collapsing the whole problem to a simple polynomial integral.

    Step-by-Step Solution

    1. Divide $x^8+4$ by $x^4-2x^2+2$. First term of quotient: $x^4$ (since $x^8/x^4=x^4$). Multiply back: $x^4(x^4-2x^2+2)=x^8-2x^6+2x^4$. Subtract: $(x^8+4)-(x^8-2x^6+2x^4)=2x^6-2x^4+4$.
    2. Next term: $2x^2$ (since $2x^6/x^4=2x^2$). Multiply back: $2x^2(x^4-2x^2+2)=2x^6-4x^4+4x^2$. Subtract: $(2x^6-2x^4+4)-(2x^6-4x^4+4x^2)=2x^4-4x^2+4$.
    3. Next term: $2$ (since $2x^4/x^4=2$). Multiply back: $2(x^4-2x^2+2)=2x^4-4x^2+4$. Subtract: exactly $0$ remainder. …

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11. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
    Q.If $\displaystyle\int \dfrac{1+\sqrt{\tan x}}{\sin 2x}dx = A\log\tan x + B\tan x + C$ then $4A - 2B =$ (A) -1 (B) 2 (C) 1 (D) -2

    ›Reveal solutionSolution

    Substituting $t=\tan x$ turns the integral into a simple sum of $\frac{1}{2t}$ and $\frac12$, giving $A=B=\frac12$ and hence $4A-2B=1$.

    Concept and Intuition

    Whenever an integral involves $\sin 2x$ together with $\tan x$, writing everything in terms of $t=\tan x$ (using $\sin2x=\frac{2t}{1+t^2}$ and $dx=\frac{dt}{1+t^2}$) usually collapses the algebra dramatically, since the $(1+t^2)$ factors cancel.

    Step-by-Step Solution

    1. Let $t=\tan x$, so $dt = \sec^2x\,dx = (1+t^2)\,dx \Rightarrow dx = \dfrac{dt}{1+t^2}$.
    2. $\sin2x = 2\sin x\cos x = 2\tan x\cos^2x = \dfrac{2t}{1+t^2}$ (using $\cos^2x=\frac1{1+t^2}$).
    3. The integral becomes $\displaystyle\int \dfrac{1+t}{\dfrac{2t}{1+t^2}}\cdot\dfrac{dt}{1+t^2} = \int \dfrac{(1+t)(1+t^2)}{2t}\cdot\dfrac{dt}{1+t^2} = \int\dfrac{1+t}{2t}\,dt$.
    4. Split: $\displaystyle\int\left(\dfrac1{2t}+\dfrac12\right)dt = \dfrac12\log|t| + \dfrac t2 + C = \dfrac12\log(\tan x) + \dfrac12\tan x + C$. …

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12. AP EAPCET 2021Set eng-2021-08-20-AN1 markMCQ
    Q.$\displaystyle\int \dfrac{\sec x}{\sqrt{\sin(2x+\theta) + \sin\theta}}\, dx =$ ______ (A) $\sqrt{(\tan x + \tan\theta)\sec\theta} + c$ (B) $\sqrt{2(\tan x + \tan\theta)\sec\theta} + c$ (C) $\sqrt{2(\sin x + \tan\theta)\sec\theta} + c$ (D) $\sqrt{2(\cos x + \tan\theta)\sec\theta} + c$

    ›Reveal solutionSolution

    This tests the sum-to-product identity followed by a substitution to reduce the integral to a simple power form. The answer is $\sqrt{2(\tan x+\tan\theta)\sec\theta}+c$.

    Concept and Intuition

    The key move is recognizing $\sin(2x+\theta)+\sin\theta$ as a sum of two sines that factors via $\sin A+\sin B=2\sin\left(\tfrac{A+B}2\right)\cos\left(\tfrac{A-B}2\right)$, which produces a $\cos^2x$ factor that cancels nicely against $\sec x$ and the square root, leaving a simple substitution.

    Step-by-Step Solution

    1. $\sin(2x+\theta)+\sin\theta=2\sin(x+\theta)\cos x$ (sum-to-product with $A=2x+\theta, B=\theta$).
    2. Expand $\sin(x+\theta)=\sin x\cos\theta+\cos x\sin\theta$, so the expression $=2\cos x(\sin x\cos\theta+\cos x\sin\theta)=2\cos^2x(\tan x\cos\theta+\sin\theta)$.
    3. So $\sqrt{\sin(2x+\theta)+\sin\theta}=\cos x\sqrt{2(\tan x\cos\theta+\sin\theta)}$ (taking $\cos x>0$).
    4. The integrand $\dfrac{\sec x}{\sqrt{\ldots}}=\dfrac{1}{\cos^2x\sqrt{2(\tan x\cos\theta+\sin\theta)}}=\dfrac{\sec^2x}{\sqrt{2(\tan x\cos\theta+\sin\theta)}}$.
    5. Let $t=\tan x\cos\theta+\sin\theta$, so $dt=\sec^2x\cos\theta\,dx$, i.e. $\sec^2x\,dx=dt/\cos\theta$. …

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