Q.Integrate the function $\frac{x^3\sin(\tan^{-1}x^4)}{1+x^8}$

Showing the 12 most recent of 51 on this concept.

1. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{x^3 \tan^{-1} x^4}{1 + x^8}\,dx =$ (A) $\dfrac{(\tan^{-1}(x^4))^2}{8} + c$ (B) $\dfrac{(\tan^{-1}(x^4))^3}{3} + c$ (C) $\dfrac{(\tan^{-1}(x^4))^2}{4} + c$ (D) $\dfrac{(\tan^{-1}(x^4))^2}{2} + c$

   ›Reveal solutionSolution

   A double substitution (first $u=x^4$, then $v=\tan^{-1}u$) turns this into a trivial $\int v\,dv$. Answer: $\dfrac{(\tan^{-1}x^4)^2}{8}+c$.

   Concept and Intuition

   The presence of $x^3\,dx$ alongside $x^4$ inside the arctan and $x^8=(x^4)^2$ in the denominator is a strong signal to substitute $u=x^4$ first. After that, the structure $\tan^{-1}u \cdot \dfrac{du}{1+u^2}$ is exactly of the form "function times its own derivative," solved by a second substitution.

   Step-by-Step Solution

   1. Let $u=x^4$, so $du=4x^3\,dx \Rightarrow x^3\,dx = \dfrac{du}{4}$.
   2. The integral becomes $\displaystyle\int \frac{\tan^{-1}u}{1+u^2}\cdot\frac{du}{4} = \frac14\int \frac{\tan^{-1}u}{1+u^2}\,du$. …

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2. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{\cos^3 x}{(1 + \sin x)^4}\, dx =$ (A) $-\dfrac{\cos^4 x}{5(1 + \sin x)^5} + c$ (B) $\dfrac{\cos^4 x}{5(1 + \sin x)^5} + c$ (C) $\dfrac{\cos^4 x}{4(1 + \sin x)^4} + c$ (D) $-\dfrac{\cos^4 x}{4(1 + \sin x)^4} + c$

   ›Reveal solutionSolution

   Factor $\cos^3x$ using $\cos^2x=(1-\sin x)(1+\sin x)$ and substitute $t=\sin x$; the resulting antiderivative can equivalently be written in the $\cos^4x/(1+\sin x)^4$ form given in the options (they differ only by an added constant). Answer: $-\dfrac{\cos^4x}{4(1+\sin x)^4}+c$.

   Concept and Intuition

   Integrals of $\cos^{\text{odd}}x$ over powers of $(1+\sin x)$ are handled by peeling off one factor of $\cos x$ to pair with $dx$ (making $d(\sin x)$) and expressing the remaining even power of $\cos x$ in terms of $\sin x$. Since the MCQ options are phrased in terms of $\cos^4x$ rather than $\sin x$ directly, it is often faster (and safer against sign traps) to guess-and-check an antiderivative of that shape by differentiating a general form $A\cos^4x(1+\sin x)^{-n}$ and matching powers/coefficients — this is exactly how the printed option is confirmed.

   Step-by-Step Solution

   1. $\cos^3x=\cos x\cdot\cos^2x=\cos x(1-\sin^2x)=\cos x(1-\sin x)(1+\sin x)$.
   2. Integrand $=\dfrac{\cos x(1-\sin x)(1+\sin x)}{(1+\sin x)^4}=\dfrac{\cos x(1-\sin x)}{(1+\sin x)^3}$.
   3. Let $t=\sin x$, $dt=\cos x\,dx$: $I=\displaystyle\int\frac{1-t}{(1+t)^3}dt$. Writing $1-t=2-(1+t)$: $I=\displaystyle\int\left(\frac{2}{(1+t)^3}-\frac{1}{(1+t)^2}\right)dt=-\frac{1}{(1+t)^2}+\frac{1}{1+t}+c=\frac{t}{(1+t)^2}+c$.
   4. So $I=\dfrac{\sin x}{(1+\sin x)^2}+c$ is one valid closed form. …

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3. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{x^2(x^4+1)^{3/4}}=$ (A) $\left(1+\dfrac{1}{x^4}\right)^{3/4}+c$ (B) $\left(1+\dfrac{1}{x^6}\right)^{1/2}+c$ (C) $-\left(1+\dfrac{1}{x^4}\right)^{-1/4}+c$ (D) $-\left(1+\dfrac{1}{x^4}\right)^{1/4}+c$

   ›Reveal solutionSolution

   Pulling $x^4$ out from under the radical and substituting $t=1+x^{-4}$ reduces the integral to a simple power rule, giving $-(1+1/x^4)^{1/4}+c$.

   Concept and Intuition

   When the integrand has $(x^n+1)^p$, it often helps to factor out the highest power of $x$ from inside the bracket so that a substitution like $t = 1+x^{-n}$ produces a clean differential matching the rest of the integrand.

   Step-by-Step Solution

   1. Write $(x^4+1)^{3/4} = \left(x^4\left(1+\dfrac{1}{x^4}\right)\right)^{3/4} = x^3\left(1+\dfrac{1}{x^4}\right)^{3/4}$.
   2. So $\dfrac{1}{x^2(x^4+1)^{3/4}} = \dfrac{1}{x^2\cdot x^3\left(1+\frac{1}{x^4}\right)^{3/4}} = \dfrac{1}{x^5\left(1+\frac{1}{x^4}\right)^{3/4}} = x^{-5}\left(1+x^{-4}\right)^{-3/4}$.
   3. Let $t = 1+x^{-4}$. Then $dt = -4x^{-5}dx$, so $x^{-5}dx = -\dfrac{dt}{4}$. …

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4. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{\cos^4 x}{\left(\sin^2 x + \sin^{-3} x \cos^5 x\right)^{3}}\;dx =$ (A) $\dfrac{1}{5}\left(1 + \cot^5 x\right)^{-2} + C$ (B) $\dfrac{1}{10}\left(1 + \cot^2 x\right)^{-5} + C$ (C) $\dfrac{1}{10}\left(1 + \cot^5 x\right)^{-2} + C$ (D) $\dfrac{1}{5}\left(1 + \cot^5 x\right)^{-5} + C$

   ›Reveal solutionSolution

   Factoring $\sin^2x$ out of the denominator turns it into $\sin^2x(1+\cot^5x)$, and the substitution $t=1+\cot^5x$ makes the whole integral a simple power-rule integration, giving $\dfrac{1}{10}(1+\cot^5x)^{-2}+C$.

   Concept and Intuition

   Integrals with mixed powers of $\sin x$ and $\cos x$ in odd/negative combinations often simplify beautifully once you factor out a common power to expose a $(1+\cot^n x)$ or $(1+\tan^n x)$ structure — this is exactly the kind of expression whose derivative (via chain rule) reproduces $\cot^{n-1}x\csc^2x$, matching what's left over in the integrand.

   Step-by-Step Solution

   1. Denominator: $\sin^2x+\sin^{-3}x\cos^5x$. Factor out $\sin^2x$: $=\sin^2x\left[1+\dfrac{\cos^5x}{\sin^5x}\right]=\sin^2x\,(1+\cot^5x)$.
   2. So the full denominator cubed: $\left[\sin^2x(1+\cot^5x)\right]^3=\sin^6x\,(1+\cot^5x)^3$.
   3. Integrand: $\dfrac{\cos^4x}{\sin^6x(1+\cot^5x)^3} = \dfrac{\cos^4x}{\sin^4x}\cdot\dfrac{1}{\sin^2x}\cdot(1+\cot^5x)^{-3}=\cot^4x\,\csc^2x\,(1+\cot^5x)^{-3}$.
   4. Substitute $t=1+\cot^5x$. Then $\dfrac{dt}{dx}=5\cot^4x\cdot(-\csc^2x)=-5\cot^4x\csc^2x$, so $\cot^4x\csc^2x\,dx=-\dfrac{dt}{5}$. …

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5. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{\sin^2 x \cos^2 x}{\cos^6 x + \sin^6 x}\,dx =$ (A) $\dfrac{1}{2}\mathrm{Tan}^{-1}(\tan^2 x) + c$ (B) $\dfrac{1}{3}\mathrm{Tan}^{-1}(\tan^2 x) + c$ (C) $\dfrac{1}{3}\mathrm{Tan}^{-1}(\tan^3 x) + c$ (D) $\mathrm{Tan}^{-1}(\tan^3 x) + c$

   ›Reveal solutionSolution

   Dividing through by $\cos^6x$ to introduce $\tan x$, then a double substitution ($t=\tan x$, then $u=t^3$), reduces the integral to a standard arctangent form, giving $\dfrac13\mathrm{Tan}^{-1}(\tan^3x)+c$.

   Concept and Intuition

   When an integrand has $\sin$ and $\cos$ appearing only in even powers that can be grouped, dividing everything by the highest power of $\cos x$ converts the whole expression into a rational function purely of $\tan x$ — a very common trick that opens the door to the substitution $t=\tan x$. Here, since the denominator naturally forms $1+\tan^6x$, a further substitution $u=t^3$ turns it into the standard $\int\frac{du}{1+u^2}$ arctangent integral.

   Step-by-Step Solution

   1. Divide numerator and denominator of $\dfrac{\sin^2x\cos^2x}{\cos^6x+\sin^6x}$ by $\cos^6x$: numerator becomes $\dfrac{\sin^2x\cos^2x}{\cos^6x} = \dfrac{\sin^2x}{\cos^4x} = \tan^2x\sec^2x$; denominator becomes $1 + \tan^6x$.
   2. The integral is now $\displaystyle\int \dfrac{\tan^2x\sec^2x}{1+\tan^6x}\,dx$.
   3. Substitute $t=\tan x$, so $dt = \sec^2x\,dx$. The integral becomes $\displaystyle\int\dfrac{t^2}{1+t^6}\,dt$.
   4. Substitute $u = t^3$, so $du = 3t^2\,dt$, i.e. $t^2\,dt = \dfrac{du}{3}$. The integral becomes $\dfrac13\displaystyle\int\dfrac{du}{1+u^2}$. …

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6. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.$\displaystyle\int(\tan^7x+\tan x)\,dx=$ (A) $\dfrac{\tan^2 x}{12}(2\tan^4 x - 3\tan^2 x+6)+c$ (B) $\dfrac{\tan^2 x}{6}-\dfrac{\tan^5 x}{4}+\dfrac{\tan^4 x}{2}+c$ (C) $\dfrac{\tan^2 x}{6}(\tan^4 x+3\tan^2 x+4)+c$ (D) $\dfrac{\tan x}{12}(\tan^4 x-3\tan^2 x+6)+c$

   ›Reveal solutionSolution

   Factoring $\tan^7x+\tan x$ using the sum-of-like-terms identity $t^7+t=t(t^6+1)=t(t^2+1)(t^4-t^2+1)$ exposes the $\sec^2x$ needed for a clean $t=\tan x$ substitution. Answer: $\dfrac{\tan^2x}{12}(2\tan^4x-3\tan^2x+6)+c$.

   Concept and Intuition

   Whenever an integrand is built purely from powers of $\tan x$ together with an explicit or hidden $\sec^2x$, substituting $t=\tan x$ turns it into a polynomial integral — the key is recognizing that $t^6+1=(t^2+1)(t^4-t^2+1)$ supplies exactly the $\sec^2x=1+\tan^2x$ factor needed.

   Step-by-Step Solution

   1. $\tan^7x+\tan x=\tan x(\tan^6x+1)=\tan x(\tan^2x+1)(\tan^4x-\tan^2x+1)=\tan x\sec^2x(\tan^4x-\tan^2x+1)$.
   2. Let $t=\tan x$, $dt=\sec^2x\,dx$. Integral becomes $\displaystyle\int t(t^4-t^2+1)dt=\int(t^5-t^3+t)dt$.
   3. $=\dfrac{t^6}{6}-\dfrac{t^4}{4}+\dfrac{t^2}{2}+c=\dfrac{\tan^6x}{6}-\dfrac{\tan^4x}{4}+\dfrac{\tan^2x}{2}+c$. …

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7. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.If $\displaystyle\int \dfrac{dx}{2\cos x + 3\sin x + 4} = \dfrac{2}{\sqrt{3}}f(x) + c$, then $f\left(\dfrac{2\pi}{3}\right) =$ (A) $\dfrac{\pi}{12}$ (B) $\dfrac{\pi}{8}$ (C) $\dfrac{5\pi}{12}$ (D) $\dfrac{5\pi}{8}$

   ›Reveal solutionSolution

   This is a Weierstrass ($t=\tan(x/2)$) substitution problem for a linear combination of sine and cosine plus a constant in the denominator. Evaluating $f$ at the given point gives $\dfrac{5\pi}{12}$, option (C).

   Concept and Intuition

   Whenever the denominator mixes $\sin x$, $\cos x$ and a constant, the universal substitution $t=\tan(x/2)$ (with $\cos x=\frac{1-t^2}{1+t^2}$, $\sin x=\frac{2t}{1+t^2}$, $dx=\frac{2\,dt}{1+t^2}$) converts the trigonometric denominator into a plain quadratic in $t$, reducing the whole problem to a standard $\int\frac{dt}{\text{quadratic}}$ that integrates to an arctangent.

   Step-by-Step Solution

   1. Substitute: denominator becomes

   $2\cdot\frac{1-t^2}{1+t^2}+3\cdot\frac{2t}{1+t^2}+4=\frac{2-2t^2+6t+4+4t^2}{1+t^2}=\frac{2t^2+6t+6}{1+t^2}.$

   2. The integral becomes $\displaystyle\int\frac{2\,dt/(1+t^2)}{(2t^2+6t+6)/(1+t^2)}=\int\frac{2\,dt}{2t^2+6t+6}=\int\frac{dt}{t^2+3t+3}$.
   3. Complete the square: $t^2+3t+3=\left(t+\tfrac32\right)^2+\tfrac34$, so $\int\frac{dt}{\left(t+\frac32\right)^2+\frac34}=\frac{1}{\sqrt{3}/2}\arctan\!\left(\frac{t+3/2}{\sqrt3/2}\right)+c=\frac{2}{\sqrt3}\arctan\!\left(\frac{2t+3}{\sqrt3}\right)+c.$ …

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8. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{\sin^6 x}{\cos^8 x}\,dx =$ (A) $\tan 7x + c$ (B) $\dfrac{\tan^7 x}{7} + c$ (C) $\dfrac{\tan 7x}{7} + c$ (D) $\sec^7 x$

   ›Reveal solutionSolution

   $\int \tan^6x\sec^2x\,dx = \dfrac{\tan^7x}{7}+c$ via $u=\tan x$.

   Concept and Intuition

   $\sin^6x/\cos^8x = \tan^6x\cdot\sec^2x$, and $\sec^2x\,dx$ is exactly $d(\tan x)$ — a direct power-rule substitution.

   Step-by-Step Solution

   1. $\dfrac{\sin^6x}{\cos^8x} = \tan^6x\cdot\dfrac{1}{\cos^2x} = \tan^6x\sec^2x$.
   2. Let $u=\tan x \Rightarrow du=\sec^2x\,dx$.
   3. $\int u^6\,du = \dfrac{u^7}{7}+c = \dfrac{\tan^7x}{7}+c$.

   Common Mistakes …

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9. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.If $\displaystyle\int \dfrac{\cos^3 x}{\sin^2 x + \sin^4 x}dx = c - \mathrm{cosec}\,x - f(x)$, then $f\left(\dfrac{\pi}{2}\right)=$ (A) 1 (B) 0 (C) $\dfrac{\pi}{2}$ (D) $\pi$

   ›Reveal solutionSolution

   Substituting $s=\sin x$ and partial-fractioning gives $-\mathrm{cosec}\,x-2\mathrm{Tan}^{-1}(\sin x)$, so $f(x)=2\mathrm{Tan}^{-1}(\sin x)$ and $f(\pi/2)=\pi/2$.

   Concept and Intuition

   Writing $\cos^3x\,dx = \cos^2x\cdot\cos x\,dx=(1-\sin^2x)\,d(\sin x)$ converts a trig integral into an algebraic one in $s=\sin x$, which is then handled by ordinary partial fractions.

   Step-by-Step Solution

   1. Let $s=\sin x$, so $ds=\cos x\,dx$, and $\cos^3x\,dx=(1-s^2)\,ds$.
   2. The integral becomes $\displaystyle\int\frac{1-s^2}{s^2+s^4}\,ds=\int\frac{1-s^2}{s^2(1+s^2)}\,ds$.
   3. Partial fractions (in $u=s^2$): $\dfrac{1-u}{u(1+u)}=\dfrac1u-\dfrac2{1+u}$, so the integrand is $\dfrac1{s^2}-\dfrac2{1+s^2}$.
   4. Integrating: $\displaystyle\int\left(\frac1{s^2}-\frac2{1+s^2}\right)ds=-\frac1s-2\,\mathrm{Tan}^{-1}s+C=-\mathrm{cosec}\,x-2\,\mathrm{Tan}^{-1}(\sin x)+C$. …

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10. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
    Q.Find the value of $k$ if $\displaystyle\int \cos^k(x)\sin(x)\,dx = \dfrac{-1}{4}\cos^4(x) + c$ (A) $4$ (B) $3$ (C) $2$ (D) $1$

    ›Reveal solutionSolution

    Differentiate the given antiderivative and match powers of $\cos x$ to find $k=3$.

    Concept and Intuition

    When an integral is given in the form $\int \cos^k(x)\sin(x)\,dx = F(x)+c$, the fastest way to find $k$ is not to integrate but to differentiate the claimed answer $F(x)$ — the derivative must reproduce the original integrand exactly.

    Step-by-Step Solution

    1. Differentiate $F(x) = -\dfrac14\cos^4x$:

    $F'(x) = -\dfrac14\cdot 4\cos^3x\cdot(-\sin x) = \cos^3x\sin x.$

    2. This must equal the integrand $\cos^k x\sin x$.
    3. Comparing powers of $\cos x$: $k=3$. …

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11. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
    Q.$\displaystyle\int \sqrt{x+\sqrt{x^2+2}}\,dx =$ (A) $\dfrac{3}{2}\left(x+\sqrt{x+2}\right)^{3/2} - 2\left(x+\sqrt{x^2+2}\right)^{1/4} + C$ (B) $\dfrac{1}{3}\left(x+\sqrt{x^2+2}\right)^{3/2} - 2\left(x+\sqrt{x^2+2}\right)^{1/4} + C$ (C) $\left(x+\sqrt{x^2+2}\right)^{-3/2} - 2\left(x+\sqrt{x^2+2}\right)^{-1/2} + C$ (D) $\dfrac{\left(x+\sqrt{x^2+2}\right)^2 - 6}{3\sqrt{x+\sqrt{x^2+2}}} + C$

    ›Reveal solutionSolution

    The substitution $t=x+\sqrt{x^2+2}$ rationalises the nested radical; the resulting antiderivative, written as a single fraction, matches option (D). Answer: option (D).

    Concept and Intuition

    Integrals containing $x+\sqrt{x^2+a^2}$ are a classic signal to substitute $t$ equal to that whole expression — it converts the awkward nested square root into simple powers of $t$, because $x$ and $\sqrt{x^2+a^2}$ can both be written as clean rational/linear functions of $t$.

    Step-by-Step Solution

    1. Let $t=x+\sqrt{x^2+2}$. Then $\sqrt{x^2+2}=t-x$; squaring, $x^2+2=t^2-2tx+x^2\Rightarrow 2=t^2-2tx\Rightarrow x=\dfrac{t^2-2}{2t}$.
    2. Then $\sqrt{x^2+2}=t-x=t-\dfrac{t^2-2}{2t}=\dfrac{2t^2-t^2+2}{2t}=\dfrac{t^2+2}{2t}$.
    3. Differentiate $t$ w.r.t. $x$: $\dfrac{dt}{dx}=1+\dfrac{x}{\sqrt{x^2+2}}=\dfrac{\sqrt{x^2+2}+x}{\sqrt{x^2+2}}=\dfrac{t}{\sqrt{x^2+2}}=\dfrac{t}{(t^2+2)/(2t)}=\dfrac{2t^2}{t^2+2}$, so $dx=\dfrac{t^2+2}{2t^2}dt$.
    4. Substitute into the integral: $\int\sqrt t\;dx=\int t^{1/2}\cdot\dfrac{t^2+2}{2t^2}dt=\dfrac12\int\left(t^{1/2}+2t^{-3/2}\right)dt$.
    5. Integrate: $\dfrac12\left[\dfrac23t^{3/2}-4t^{-1/2}\right]+C=\dfrac13t^{3/2}-2t^{-1/2}+C$. …

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12. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
    Q.If $\displaystyle\int \dfrac{\sqrt{1 - x^4}}{x^7}\, dx = f(x)\left\{\sqrt{1 - x^4}\right\}^n + c$, then $(f(x))^n =$ ______ (A) $\dfrac{-1}{6x^6}$ (B) $\dfrac{-1}{216x^{18}}$ (C) $\dfrac{1}{36x^{12}}$ (D) $\dfrac{1}{216x^{18}}$

    ›Reveal solutionSolution

    A reduction-formula-style substitution $t=x^{-2}$ turns the integral into a simple power form; matching to the given answer form identifies $f(x)$ and $n$, then $(f(x))^n$ follows directly.

    Concept and Intuition

    Integrals of the form $\int x^m(a+bx^n)^p\,dx$ can often be simplified by substituting $t=x^{-n}$ when $(m+1)/n$ isn't an integer but $(m+1)/n+p$ is — exactly the situation here with $m=-7$, $n=4$, $p=1/2$.

    Step-by-Step Solution

    1. Write the integral as $\displaystyle\int x^{-7}(1-x^4)^{1/2}\,dx$.
    2. Substitute $t = x^{-2}$, so $dt = -2x^{-3}dx \Rightarrow dx = -\dfrac{x^3}{2}dt$, and $x^{-4}=t^2$.
    3. $x^{-7}dx = x^{-7}\cdot\left(-\dfrac{x^3}{2}\right)dt = -\dfrac{x^{-4}}{2}dt = -\dfrac{t^2}{2}dt$.
    4. $1-x^4 = 1-\dfrac{1}{t^2} = \dfrac{t^2-1}{t^2}$, so $\sqrt{1-x^4}=\dfrac{\sqrt{t^2-1}}{t}$ (for $t>0$).
    5. The integral becomes $\displaystyle\int \dfrac{\sqrt{t^2-1}}{t}\cdot\left(-\dfrac{t^2}{2}\right)dt = -\dfrac12\int t\sqrt{t^2-1}\,dt$.
    6. Let $s=t^2-1$, $ds=2t\,dt$: $-\dfrac12\int \sqrt{s}\cdot\dfrac{ds}{2} = -\dfrac14\cdot\dfrac{2}{3}s^{3/2} = -\dfrac16 (t^2-1)^{3/2}+c$. …

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