Q.Find $\int \cos 6x \sqrt{1 + \sin 6x}\, dx$

Showing the 12 most recent of 51 on this concept.

1. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
   Q.$\int \cos\sqrt{x}\, dx =$ (A) $2\sqrt{x}\sin\sqrt{x} + 2\cos\sqrt{x} + c$ (B) $2\sqrt{x}\sin\sqrt{x} + 2\sin\sqrt{x} + c$ (C) $2\sqrt{x}\sin\sqrt{x} - 2\cos\sqrt{x} + c$ (D) $\sqrt{x}\cos\sqrt{x} - 2\sin\sqrt{x} + c$

   ›Reveal solutionSolution

   Substitute $t=\sqrt x$ to turn the integral into a standard integration-by-parts problem. Answer: $2\sqrt x\sin\sqrt x+2\cos\sqrt x+c$.

   Concept and Intuition

   Whenever you see $\sqrt x$ trapped inside a trig or exponential function, substituting $t=\sqrt x$ converts it into a polynomial-times-trig integral solvable by parts.

   Step-by-Step Solution

   1. Let $t=\sqrt x \Rightarrow x=t^2,\ dx=2t\,dt$.
   2. $\int\cos\sqrt x\,dx = \int\cos t\cdot 2t\,dt = 2\int t\cos t\,dt$.
   3. Integrate by parts: $\int t\cos t\,dt = t\sin t - \int\sin t\,dt = t\sin t+\cos t$.
   4. So the integral $=2(t\sin t+\cos t)+c = 2t\sin t+2\cos t+c$.
   5. Substitute back $t=\sqrt x$: $=2\sqrt x\sin\sqrt x+2\cos\sqrt x+c$.

   Common Mistakes

   • Forgetting the factor of $2t$ from $dx=2t\,dt$ when substituting.

   ✓Final answer

   The correct option is (A) — $2\sqrt x\sin\sqrt x+2\cos\sqrt x+c$.

   ANSWER: A

2. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
   Q.$\int(x^{3m}+x^{2m}+x^m)(2x^{2m}+3x^m+6)^{\frac{1}{m}}dx=$ (A) $\dfrac{1}{6(m+1)}\left(2x^{3m}+3x^{2m}+6x^m\right)^{\frac{m+1}{m}}+C$ (B) $\dfrac{1}{6(m+1)}\left(2x^{3m}+3x^{2m}+6x^m\right)^{\frac{m-1}{m}}+C$ (C) $\dfrac{1}{6(m+1)}\left(2x^{3m}+3x^{2m}+6\right)^{\frac{m+1}{m}}+C$ (D) $\dfrac{1}{6(m-1)}\left(2x^{3m}+mx^{2m}+6x^m\right)^{\frac{m-1}{m}}+C$

   ›Reveal solutionSolution

   Recognising that $2x^{3m}+3x^{2m}+6x^m$ equals $x^m$ times the bracket under the $1/m$-power root lets the whole integrand be rewritten as (a constant times) $g^{1/m}g'$ for $g=2x^{3m}+3x^{2m}+6x^m$ — a pure "power rule" integral.

   Concept and Intuition

   Whenever an integrand looks like $g(x)^{1/m}\cdot g'(x)$ (up to a constant factor), the antiderivative is immediately $\dfrac{g^{1/m+1}}{1/m+1}$. The main work here is algebraic: spotting that the "outside" factor $(x^{3m}+x^{2m}+x^m)$ is secretly related to the derivative of $g=2x^{3m}+3x^{2m}+6x^m$, and that the "inside" bracket $(2x^{2m}+3x^m+6)$ is just $g/x^m$.

   Step-by-Step Solution

   1. Let $g=2x^{3m}+3x^{2m}+6x^m$. Factor: $g=x^m(2x^{2m}+3x^m+6)$, so $2x^{2m}+3x^m+6=g/x^m$, and $(2x^{2m}+3x^m+6)^{1/m}=g^{1/m}/x$ (since $(g/x^m)^{1/m}=g^{1/m}/x$).
   2. Differentiate $g$: $g'=6mx^{3m-1}+6mx^{2m-1}+6mx^{m-1}=6mx^{m-1}(x^{2m}+x^m+1)$.
   3. Rewrite the integrand: $(x^{3m}+x^{2m}+x^m)(2x^{2m}+3x^m+6)^{1/m}=x^m(x^{2m}+x^m+1)\cdot\dfrac{g^{1/m}}x=x^{m-1}(x^{2m}+x^m+1)\,g^{1/m}$.
   4. From step 2, $x^{m-1}(x^{2m}+x^m+1)=\dfrac{g'}{6m}$. So the integrand equals $\dfrac{g'}{6m}g^{1/m}$.
   5. $\int\dfrac{g'}{6m}g^{1/m}\,dx=\dfrac1{6m}\int g^{1/m}\,dg=\dfrac1{6m}\cdot\dfrac{g^{1/m+1}}{1/m+1}+C=\dfrac1{6(m+1)}g^{(m+1)/m}+C$.
   6. Substituting back: $\dfrac1{6(m+1)}\left(2x^{3m}+3x^{2m}+6x^m\right)^{(m+1)/m}+C$.

   Common Mistakes

   • Mistaking the bracket raised to $1/m$ (which is $2x^{2m}+3x^m+6$, not $2x^{3m}+3x^{2m}+6x^m$) for $g$ itself — the two are related by a factor of $x^m$, which is exactly the subtlety that makes the problem work out.
   • Losing track of the exponent arithmetic $1/m+1=(m+1)/m$.

   ✓Final answer

   The correct option is (A) — $\dfrac1{6(m+1)}\left(2x^{3m}+3x^{2m}+6x^m\right)^{\frac{m+1}m}+C$.

   ANSWER: A

3. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If $\displaystyle\int e^{\sin^3 x}(\sin^8 x + 2\sin^5 x)\cos x\, dx = \dfrac{1}{3} e^{\sin^3 x} f(x) + c$, then $f(x) =$ (A) $\sin^8 x$ (B) $\sin^6 x$ (C) $\cos^8 x$ (D) $\cos^6 x$

   ›Reveal solutionSolution

   Substituting $t=\sin x$ and testing $f(t)=t^6$ against the target derivative confirms $f(x)=\sin^6x$.

   Concept and Intuition

   When an integral has the form $e^{g(x)}\cdot(\text{stuff})\cdot g'(x)$, substituting $u=g(x)$ turns it into $\int e^u\cdot h(u)\,du$ for a polynomial $h$. Matching the answer's assumed shape $\tfrac13e^{u}f$ and differentiating (product rule, since both $e^u$ and $f$ depend on $u$) lets us solve for $f$ by comparing polynomial coefficients — much safer than guessing an antiderivative by inspection alone.

   Step-by-Step Solution

   1. Let $t=\sin x$, so $dt=\cos x\,dx$. The integral becomes $\displaystyle\int e^{t^3}(t^8+2t^5)\,dt$.
   2. We're told this equals $\dfrac13 e^{t^3}f(t)+c$. Differentiate the RHS w.r.t. $t$: $\dfrac{d}{dt}\left[\dfrac13e^{t^3}f(t)\right] = \dfrac13\left[3t^2e^{t^3}f(t) + e^{t^3}f'(t)\right] = e^{t^3}\left[t^2f(t)+\dfrac13f'(t)\right]$.
   3. This must equal $e^{t^3}(t^8+2t^5)$, so $t^2f(t) + \dfrac13f'(t) = t^8+2t^5$.
   4. Try $f(t)=t^6$: $f'(t)=6t^5$. Then $t^2\cdot t^6 + \dfrac13\cdot 6t^5 = t^8+2t^5$ — matches exactly.
   5. So $f(t)=t^6$, i.e. $f(x)=\sin^6x$.

   Common Mistakes

   • Forgetting the product rule when differentiating $e^{t^3}f(t)$ (both factors depend on $t$).
   • Guessing $f(t)=t^8$ directly from the leading term without checking the $2t^5$ term also matches.

   ✓Final answer

   The correct option is (B) — $\sin^6 x$.

   ANSWER: B

4. 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: $\dfrac12\left[\dfrac23t^{3/2}-4t^{-1/2}\right]+C=\dfrac13t^{3/2}-2t^{-1/2}+C$.
   6. Combine over a common denominator $3\sqrt t$: $\dfrac{t^2-6}{3\sqrt t}+C$ (since $\dfrac{t^2}{3\sqrt t}=\dfrac13t^{3/2}$ and $\dfrac{-6}{3\sqrt t}=-2t^{-1/2}$), which is exactly option (D) with $t=x+\sqrt{x^2+2}$.

   Common Mistakes

   • Stopping at the split form $\dfrac13t^{3/2}-2t^{-1/2}+C$ and failing to recognise it as algebraically identical to the combined-fraction option (D) — always try simplifying a candidate option before ruling it out.
   • Sign or algebra slips solving for $x$ in terms of $t$.

   ✓Final answer

   The correct option is (D) — $\dfrac{\left(x+\sqrt{x^2+2}\right)^2 - 6}{3\sqrt{x+\sqrt{x^2+2}}} + C$.

   ANSWER: D

5. 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 $x^{-5}dx = -\dfrac{dt}{4}$.
   4. The integral becomes $\displaystyle\int t^{-3/4}\left(-\dfrac{dt}{4}\right) = -\dfrac14\cdot\dfrac{t^{1/4}}{1/4}+c = -t^{1/4}+c$.
   5. Substituting back: $-\left(1+\dfrac{1}{x^4}\right)^{1/4}+c$.

   Common Mistakes

   • Forgetting the negative sign that comes from $dt=-4x^{-5}dx$.
   • Not factoring $x^4$ out correctly before substituting, leading to a mismatched power.

   ✓Final answer

   The correct option is (D) — $-\left(1+\dfrac{1}{x^4}\right)^{1/4}+c$.

   ANSWER: D

6. 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

   • Missing that one $\sec^2x$ factor must be reserved for $du$, leaving $\tan^6x$ (6, not 8, powers of secant used up).
   • Writing $\tan 7x$ (a different function) instead of $\tan^7x$ (a power).

   ✓Final answer

   The correct option is (B) — $\dfrac{\tan^7x}{7}+c$.

   ANSWER: B

7. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{(x^5+1)^{6/5}} =$ (A) $\dfrac{1}{\sqrt[5]{x^5+1}} + c$ (B) $\dfrac{\sqrt[5]{x^5+1}}{x} + c$ (C) $\dfrac{x}{\sqrt[5]{x^5+1}} + c$ (D) $\sqrt[5]{x^5+1} + c$

   ›Reveal solutionSolution

   Recognizing the integrand as the derivative of $\dfrac{x}{\sqrt[5]{x^5+1}}$ (verified by direct differentiation) gives the antiderivative immediately.

   Concept and Intuition

   For integrals of the form $\int\frac{dx}{(x^n+1)^{(n+1)/n}}$, a useful trick is to guess that the antiderivative looks like $\frac{x}{(x^n+1)^{1/n}}$ (a ratio designed to make the product-rule differentiation collapse nicely), and then verify by differentiating it — if it reproduces the integrand exactly, we're done. This is often faster than a substitution for this particular family.

   Step-by-Step Solution

   1. Guess the antiderivative $g(x) = \dfrac{x}{(x^5+1)^{1/5}} = x(x^5+1)^{-1/5}$.
   2. Differentiate using the product rule: $g'(x) = (x^5+1)^{-1/5} + x\cdot\left(-\dfrac15\right)(x^5+1)^{-6/5}\cdot5x^4$.
   3. Simplify the second term: $x\cdot\left(-\dfrac15\right)(5x^4)(x^5+1)^{-6/5} = -x^5(x^5+1)^{-6/5}$.
   4. So $g'(x) = (x^5+1)^{-1/5} - x^5(x^5+1)^{-6/5}$.
   5. Factor out $(x^5+1)^{-6/5}$: $g'(x) = (x^5+1)^{-6/5}\left[(x^5+1) - x^5\right] = (x^5+1)^{-6/5}\times 1 = (x^5+1)^{-6/5}$.
   6. This matches the integrand $\dfrac{1}{(x^5+1)^{6/5}}$ exactly.
   7. So $\displaystyle\int\dfrac{dx}{(x^5+1)^{6/5}} = \dfrac{x}{(x^5+1)^{1/5}}+c = \dfrac{x}{\sqrt[5]{x^5+1}}+c$.

   Common Mistakes

   • Forgetting the chain rule factor of $5x^4$ when differentiating $(x^5+1)^{-1/5}$-type expressions.
   • Choosing the wrong candidate antiderivative form (e.g. without the $x$ in the numerator), which would not reproduce the integrand upon differentiation.

   ✓Final answer

   The correct option is (C) — $\dfrac{x}{\sqrt[5]{x^5+1}} + c$.

   ANSWER: C

8. 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 $\dfrac13\displaystyle\int\dfrac{du}{1+u^2}$.
   5. This is the standard arctangent integral: $\dfrac13\mathrm{Tan}^{-1}(u) + c$.
   6. Substitute back $u=t^3=\tan^3x$: the answer is $\dfrac13\mathrm{Tan}^{-1}(\tan^3x)+c$.

   Common Mistakes

   • Mis-simplifying $\dfrac{\sin^2x\cos^2x}{\cos^6x}$ (arithmetic slip in exponents), which changes the power of $\tan x$ obtained.
   • Forgetting the second substitution ($u=t^3$) and trying to directly integrate $\int\frac{t^2}{1+t^6}dt$ as though it were already a standard arctan form.

   ✓Final answer

   The correct option is (C) — $\dfrac{1}{3}\mathrm{Tan}^{-1}(\tan^3 x) + c$.

   ANSWER: C

9. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.If $\displaystyle\int (x^6 + x^4 + x^2)\sqrt{2x^4 + 3x^2 + 6}\,dx = f(x) + c$, then $f(3) =$ (A) $\dfrac{3}{2}(95)^{3/2}$ (B) $\dfrac{3}{2}(195)^{3/2}$ (C) $\dfrac{3}{2}(265)^{3/2}$ (D) $\dfrac{3}{2}(175)^{3/2}$

   ›Reveal solutionSolution

   Guess the antiderivative in the form $(ax^3+bx)(2x^4+3x^2+6)^{3/2}$, match

   coefficients by differentiating, solve for $a,b$, then evaluate at $x=3$ to get

   $f(3)=\dfrac32(195)^{3/2}$.

   Concept and Intuition

   When an integrand is a polynomial times the square root of another polynomial,

   and the polynomial factor's degree matches what you'd get by differentiating a

   $(\text{poly})\cdot(\text{radicand})^{3/2}$ ansatz, the fastest route is the

   "reverse chain rule with an undetermined polynomial coefficient" method: guess

   the shape of the antiderivative with unknown coefficients, differentiate, and

   match coefficients of like powers of $x$ to solve for them.

   Step-by-Step Solution

   1. Guess $f(x)=(ax^3+bx)\,(2x^4+3x^2+6)^{3/2}$.
   2. Differentiate (product + chain rule), then factor out $(2x^4+3x^2+6)^{1/2}$: $f'(x)=(2x^4+3x^2+6)^{1/2}\Big[(3ax^2+b)(2x^4+3x^2+6)+(ax^3+bx)(12x^3+9x)\Big]$.
   3. Expand the bracket: it collects to $18a\,x^6+(18a+14b)x^4+(18a+12b)x^2+6b$.
   4. Match this to the target $x^6+x^4+x^2$ (constant term $0$): $6b=0\Rightarrow b=0$; then $18a=1\Rightarrow a=\dfrac1{18}$.
   5. Check remaining coefficients with $a=1/18,\ b=0$: $18a+14b=1$ ✓ (matches $x^4$ coefficient $1$), $18a+12b=1$ ✓ (matches $x^2$ coefficient $1$). All consistent.
   6. So $f(x)=\dfrac{x^3}{18}(2x^4+3x^2+6)^{3/2}$.
   7. At $x=3$: $2(3)^4+3(3)^2+6=162+27+6=195$, and $x^3=27$.
   8. $f(3)=\dfrac{27}{18}(195)^{3/2}=\dfrac32(195)^{3/2}$.

   Common Mistakes

   • Guessing the wrong power on the ansatz's radical factor (e.g. using $(\cdot)^{1/2}$ instead of $(\cdot)^{3/2}$) — the chain rule from a $3/2$ power is exactly what produces both an $x^6$ and lower-power terms needed to match all three terms in the integrand.
   • Arithmetic slip evaluating $2\cdot3^4=162$ (not $2\cdot81=182$ or similar) — worth recomputing $3^4=81$ carefully.

   ✓Final answer

   The correct option is (B) — $\dfrac32(195)^{3/2}$.

   ANSWER: B

10. 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. $=\dfrac{t^6}{6}-\dfrac{t^4}{4}+\dfrac{t^2}{2}+c=\dfrac{\tan^6x}{6}-\dfrac{\tan^4x}{4}+\dfrac{\tan^2x}{2}+c$.
    4. Factor out $\dfrac{\tan^2x}{12}$: $\dfrac{\tan^2x}{12}\left(2\tan^4x-3\tan^2x+6\right)+c$ (check: $\dfrac{2}{12}=\dfrac16$, $\dfrac{3}{12}=\dfrac14$, $\dfrac{6}{12}=\dfrac12$ — all match).

    Common Mistakes

    • Missing the factorization $t^6+1=(t^2+1)(t^4-t^2+1)$ and instead trying to expand $\tan^7x$ directly, which is much harder.
    • Sign error in the middle term when factoring $t^4-t^2+1$ vs $t^4+t^2+1$.

    ✓Final answer

    The correct option is (A) — $\dfrac{\tan^2 x}{12}(2\tan^4 x - 3\tan^2 x+6)+c$.

    ANSWER: A

11. 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.
    5. To match the option's shape, try $F(x)=A\dfrac{\cos^4x}{(1+\sin x)^n}$ and differentiate: using $\cos^2x=(1-\sin x)(1+\sin x)$, one finds $F'(x)=\cos^3x/(1+\sin x)^4$ exactly when $n=4$ and $A=-\dfrac14$ — i.e. $F(x)=-\dfrac{\cos^4x}{4(1+\sin x)^4}$ is also a valid antiderivative (differs from step 4's form only by the constant $\tfrac14$, confirmed by evaluating both at, say, $x=0$ and $x=\pi/2$).

    Common Mistakes

    • Sign error: option (C) has the same magnitude but wrong sign — differentiating (C) gives $+\cos^3x/(1+\sin x)^4$, not matching.
    • Using the wrong power $n=5$ (options A/B) instead of the correct $n=4$.

    ✓Final answer

    The correct option is (D) — $-\dfrac{\cos^4 x}{4(1 + \sin x)^4} + c$.

    ANSWER: D

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 $s=t^2-1$, $ds=2t\,dt$: $-\dfrac12\int \sqrt{s}\cdot\dfrac{ds}{2} = -\dfrac14\cdot\dfrac{2}{3}s^{3/2} = -\dfrac16 (t^2-1)^{3/2}+c$.
    7. Convert back: $t^2-1 = x^{-4}-1 = \dfrac{1-x^4}{x^4}$, so $(t^2-1)^{3/2} = \dfrac{(1-x^4)^{3/2}}{x^6}$.
    8. Integral $= -\dfrac{1}{6x^6}(1-x^4)^{3/2}+c = -\dfrac{1}{6x^6}\left\{\sqrt{1-x^4}\right\}^3 + c$.
    9. Matching to $f(x)\{\sqrt{1-x^4}\}^n+c$: $f(x)=-\dfrac{1}{6x^6}$, $n=3$.
    10. $(f(x))^n = \left(-\dfrac{1}{6x^6}\right)^3 = -\dfrac{1}{216x^{18}}$.

    Common Mistakes

    • Losing the sign when cubing a negative fraction.
    • Misreading $(f(x))^n$ as $f(x^n)$ or $n\cdot f(x)$.

    ✓Final answer

    The correct option is (B) — $\dfrac{-1}{216x^{18}}$.

    ANSWER: B