Q.Evaluate the integral using substitution $\int_{0}^{\pi/2}\sqrt{\sin\phi}\,\cos^5\phi\,d\phi$

Showing the 12 most recent of 51 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.$\displaystyle\int_{\log 2}^{\log 3} \dfrac{e^{3x} - 3e^x}{e^{2x} - 1}\,dx =$ (A) $\log\left(\dfrac{2e}{3}\right)$ (B) $\log\left(\dfrac{4e}{3}\right)$ (C) $\dfrac{2e}{3}$ (D) $\dfrac{4e}{3}$

   ›Reveal solutionSolution

   A substitution $t=e^x$ turns this exponential integral into a rational one; the value works out to $\log(2e/3)$.

   Concept and Intuition

   Whenever an integral is built entirely from powers of $e^x$, substituting $t=e^x$ converts it into an algebraic (rational function) integral, which is usually far easier to handle with partial fractions.

   Step-by-Step Solution

   1. Let $t=e^x$, so $dt=e^xdx=t\,dx$, i.e. $dx=dt/t$. When $x=\log2$, $t=2$; when $x=\log3$, $t=3$.
   2. Rewrite the integrand: $\dfrac{e^{3x}-3e^x}{e^{2x}-1}=\dfrac{t^3-3t}{t^2-1}$. Multiplying by $dx=dt/t$ gives $\dfrac{t^3-3t}{t(t^2-1)}dt=\dfrac{t^2-3}{t^2-1}dt$.
   3. So the integral becomes $\displaystyle\int_2^3\dfrac{t^2-3}{t^2-1}dt$.
   4. Split: $\dfrac{t^2-3}{t^2-1}=\dfrac{(t^2-1)-2}{t^2-1}=1-\dfrac{2}{t^2-1}$.
   5. $\displaystyle\int_2^3 1\,dt=1$. And $\displaystyle\int_2^3\dfrac{dt}{t^2-1}=\left[\tfrac12\log\left|\tfrac{t-1}{t+1}\right|\right]_2^3=\tfrac12\left[\log\tfrac24-\log\tfrac13\right]=\tfrac12\log\tfrac{2/4}{1/3}=\tfrac12\log\tfrac32$.
   6. So the whole integral $=1-2\cdot\tfrac12\log\tfrac32=1-\log\tfrac32=\log e-\ln3+\ln2=\log\left(\dfrac{2e}{3}\right)$.

   Common Mistakes

   • Forgetting to convert $dx$ into $dt/t$ after substitution (dropping the extra factor of $t$).
   • Sign slip when combining $1-\log(3/2)$ into a single logarithm.

   ✓Final answer

   The correct option is (A) — $\log\left(\dfrac{2e}{3}\right)$.

   ANSWER: A

2. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.If $\displaystyle\int \dfrac{\sin 2\theta}{(1 + \sin\theta)(3 - \cos 2\theta)}\, d\theta = \dfrac{1}{2}\tan^{-1}(\sin\theta) + \dfrac{1}{4}\log(f(\theta)) + c$ then $f\left(\dfrac{\pi}{2}\right) - f(0) =$ (A) $\dfrac{1}{2}$ (B) $-\dfrac{1}{2}$ (C) $0$ (D) $-\dfrac{3}{4}$

   ›Reveal solutionSolution

   Reducing the integral via $t=\sin\theta$ and partial fractions identifies $f(\theta)=\dfrac{1+\sin^2\theta}{(1+\sin\theta)^2}$, giving $f(\pi/2)-f(0)=-1/2$.

   Concept and Intuition

   Double-angle identities collapse $\sin2\theta$ and $3-\cos2\theta$ into expressions purely in $\sin\theta$ and $\cos\theta$; then $t=\sin\theta$ (since $\cos\theta\,d\theta = dt$ appears naturally) turns the whole thing into a rational-function integral solvable by partial fractions — a very standard pattern for trig integrals with even powers/mixed degree-2 denominators.

   Step-by-Step Solution

   1. $\sin2\theta = 2\sin\theta\cos\theta$; $\cos2\theta = 1-2\sin^2\theta \Rightarrow 3-\cos2\theta = 2+2\sin^2\theta = 2(1+\sin^2\theta)$.
   2. Integrand becomes $\dfrac{2\sin\theta\cos\theta}{(1+\sin\theta)\cdot 2(1+\sin^2\theta)} = \dfrac{\sin\theta\cos\theta}{(1+\sin\theta)(1+\sin^2\theta)}$.
   3. Substitute $t=\sin\theta$, $dt=\cos\theta\,d\theta$: integral $= \displaystyle\int \dfrac{t}{(1+t)(1+t^2)}dt$.
   4. Partial fractions: $\dfrac{t}{(1+t)(1+t^2)} = \dfrac{-1/2}{1+t} + \dfrac{(1/2)t+1/2}{1+t^2}$ (solve $t=A(1+t^2)+(Bt+C)(1+t)$, giving $A=-1/2,\ B=1/2,\ C=1/2$).
   5. Integrate: $-\dfrac12\log(1+t) + \dfrac14\log(1+t^2) + \dfrac12\tan^{-1}t + c$.
   6. Compare to the given form $\dfrac12\tan^{-1}(\sin\theta)+\dfrac14\log f(\theta)+c$: the $\tan^{-1}$ terms match; the log terms give $\dfrac14\log f(\theta) = \dfrac14\log(1+t^2)-\dfrac12\log(1+t) = \dfrac14\log\!\left[\dfrac{1+t^2}{(1+t)^2}\right]$, so $f(\theta) = \dfrac{1+\sin^2\theta}{(1+\sin\theta)^2}$.
   7. $f(\pi/2) = \dfrac{1+1}{(1+1)^2} = \dfrac{2}{4}=\dfrac12$; $f(0) = \dfrac{1+0}{(1+0)^2}=1$.
   8. $f(\pi/2)-f(0) = \dfrac12-1 = -\dfrac12$.

   Common Mistakes

   • Losing the factor of 2 when converting $-\dfrac12\log(1+t)$ into the $\dfrac14\log f(\theta)$ form (it must become $\log(1+t)^{-2}$ inside the single combined log).
   • Sign errors in the partial-fraction constants $A,B,C$.

   ✓Final answer

   The correct option is (B) — $-\dfrac{1}{2}$.

   ANSWER: B

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 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

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 $\dfrac13\displaystyle\int\dfrac{du}{1+u^2}$.
   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

6. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.If $\displaystyle\int (x+5)\sqrt{x-5}\,dx = \dfrac{2\sqrt{x-5}}{15} f(x) + c$, then $f(6) =$ (A) $5$ (B) $20$ (C) $100$ (D) $53$

   ›Reveal solutionSolution

   Substituting $u=x-5$ turns the integral into a simple power-rule computation; matching it to the given form and evaluating $f$ at $x=6$ gives $53$.

   Concept and Intuition

   An integrand with $\sqrt{x-a}$ multiplying a linear expression in $x$ is best handled by substituting $u=x-a$, which converts everything into pure powers of $u$ that integrate directly via the power rule. After integrating, we factor the result to match the given answer template and read off $f(x)$.

   Step-by-Step Solution

   1. Let $u = x-5$, so $x = u+5$, and $x+5 = u+10$. Also $dx = du$.
   2. The integral becomes $\displaystyle\int (u+10)\sqrt u\, du = \displaystyle\int \left(u^{3/2}+10u^{1/2}\right)du$.
   3. Integrate term by term: $\displaystyle\int u^{3/2}du = \dfrac{2}{5}u^{5/2}$, and $\displaystyle\int 10u^{1/2}du = 10\times\dfrac23 u^{3/2} = \dfrac{20}{3}u^{3/2}$.
   4. So the integral $= \dfrac25 u^{5/2} + \dfrac{20}{3}u^{3/2} + c$.
   5. Factor out $\dfrac{2}{15}u^{3/2}$ (the common factor, chosen to match the given form's $\frac{2}{15}$): $\dfrac25 u^{5/2} = \dfrac{2}{15}u^{3/2}(3u)$ and $\dfrac{20}{3}u^{3/2} = \dfrac{2}{15}u^{3/2}(50)$. So the integral $= \dfrac{2}{15}u^{3/2}(3u+50)+c = \dfrac{2}{15}u^{1/2}\cdot u(3u+50) + c$.
   6. Rewrite as $\dfrac{2\sqrt u}{15}\left[u(3u+50)\right]+c = \dfrac{2\sqrt{x-5}}{15}\left[3(x-5)^2+50(x-5)\right]+c$.
   7. Matching to the given form $\dfrac{2\sqrt{x-5}}{15}f(x)+c$, we get $f(x) = 3(x-5)^2+50(x-5)$.
   8. Evaluate at $x=6$: $u=x-5=1$, so $f(6) = 3(1)^2+50(1) = 3+50=53$.

   Common Mistakes

   • Forgetting to substitute back $x+5=u+10$ correctly (losing the constant shift), which changes the coefficient of the linear term inside the integral.
   • Errors in the factoring step when trying to force the result into the $\dfrac{2}{15}u^{3/2}(\ldots)$ template.

   ✓Final answer

   The correct option is (D) — $53$.

   ANSWER: D

7. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.If $\displaystyle\int \dfrac{1}{\sin^3 x+\cos^3 x}\,dx = A\log\left|\dfrac{\sqrt{2}+t}{\sqrt{2}-t}\right|+B\,\mathrm{Tan}^{-1}(t)+c$, then $\left(\dfrac{B}{A},t\right)=$ (A) $(2\sqrt{2},\sin x+\cos x)$ (B) $\left(\dfrac{\sqrt{2}}{9},\sin x+\cos x\right)$ (C) $\left(\dfrac{\sqrt{2}}{9},\sin x-\cos x\right)$ (D) $(2\sqrt{2},\sin x-\cos x)$

   ›Reveal solutionSolution

   Factoring the sum of cubes and substituting $t=\sin x-\cos x$ turns the trigonometric integral into a clean rational-function integral in $t$, giving $B/A=2\sqrt2$ with $t=\sin x-\cos x$.

   Concept and Intuition

   $\sin^3x+\cos^3x$ factors as a sum of cubes: $(\sin x+\cos x)(1-\sin x\cos x)$. Both $\sin x+\cos x$ and $1-\sin x\cos x$ can be written purely in terms of $u=\sin x-\cos x$, because $(\sin x+\cos x)^2+(\sin x-\cos x)^2=2$ and $1-\sin x\cos x=\tfrac{1+(\sin x-\cos x)^2}{2}$. Crucially, $\dfrac{du}{dx}=\cos x+\sin x$, which is exactly the factor left over after using the second identity — so the whole integral collapses into a rational function of $u$ alone.

   Step-by-Step Solution

   1. Factor: $\sin^3x+\cos^3x=(\sin x+\cos x)(1-\sin x\cos x)$.
   2. Let $u=\sin x-\cos x$. Then $u^2=1-2\sin x\cos x$, so $1-\sin x\cos x=\dfrac{1+u^2}{2}$.
   3. Also $(\sin x+\cos x)^2=1+2\sin x\cos x=2-u^2$, so $\sin x+\cos x=\sqrt{2-u^2}$ (taking the appropriate branch), and $\dfrac{du}{dx}=\cos x+\sin x=\sqrt{2-u^2}$.
   4. So $\sin^3x+\cos^3x=\sqrt{2-u^2}\cdot\dfrac{1+u^2}{2}$, and

   $I=\int\frac{dx}{\sin^3x+\cos^3x}=\int\frac{2\,dx}{\sqrt{2-u^2}(1+u^2)}=\int\frac{2}{\sqrt{2-u^2}(1+u^2)}\cdot\frac{du}{\sqrt{2-u^2}}=\int\frac{2\,du}{(2-u^2)(1+u^2)}.$

   5. Partial fractions (by symmetry, only even terms survive): $\dfrac{2}{(2-u^2)(1+u^2)}=\dfrac{2/3}{2-u^2}+\dfrac{2/3}{1+u^2}$.
   6. Integrate: $\displaystyle\int\frac{du}{2-u^2}=\frac{1}{2\sqrt2}\log\left|\frac{\sqrt2+u}{\sqrt2-u}\right|$ and $\int\frac{du}{1+u^2}=\mathrm{Tan}^{-1}u$.
   7. So $I=\dfrac{1}{3\sqrt2}\log\left|\dfrac{\sqrt2+u}{\sqrt2-u}\right|+\dfrac{2}{3}\mathrm{Tan}^{-1}u+c$, matching the given form with $A=\dfrac{1}{3\sqrt2}$, $B=\dfrac{2}{3}$, $t=u=\sin x-\cos x$.
   8. Hence $\dfrac{B}{A}=\dfrac{2/3}{1/(3\sqrt2)}=2\sqrt2$.

   Common Mistakes

   • Trying $t=\sin x+\cos x$ directly — it doesn't decouple the $\sqrt{2-t^2}$ factor from the rest as cleanly as $t=\sin x-\cos x$ does here.
   • Sign slip when computing $(\sin x+\cos x)^2$ vs $(\sin x-\cos x)^2$ and which one equals $1\pm2\sin x\cos x$.

   ✓Final answer

   The correct option is (D) — $(2\sqrt{2},\sin x-\cos x)$.

   ANSWER: D

8. 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

9. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.If $\int x^5 e^{-4x^3}dx = \frac{1}{48}e^{-4x^3}f(x) + c$, then $f(x) =$ (A) $-2x^3 - 1$ (B) $-4x^3 - 1$ (C) $-2x^2 + 1$ (D) $4x^3 + 1$

   ›Reveal solutionSolution

   Substituting $u=x^3$ converts the integral into a simple integration-by-parts problem $\int ue^{-4u}du$; matching the result to the given form yields $f(x)=-4x^3-1$.

   Concept and Intuition

   The presence of $x^5$ alongside $e^{-4x^3}$ is a strong hint to substitute $u=x^3$, since then $x^2\,dx$ (part of $du$) combines with the remaining $x^3=u$ to leave a clean polynomial-times-exponential integral, solvable by the standard integration-by-parts reduction formula for $\int u\,e^{ku}du$.

   Step-by-Step Solution

   1. Let $u=x^3$, so $du=3x^2\,dx \Rightarrow x^2dx=\dfrac{du}{3}$.
   2. Rewrite $x^5e^{-4x^3}dx = x^3\cdot x^2 e^{-4x^3}dx = u\,e^{-4u}\cdot\dfrac{du}{3}$.
   3. So the integral becomes $\dfrac13\displaystyle\int u\,e^{-4u}\,du$.
   4. Integrate by parts with first function $u$, second $e^{-4u}$: $\int u e^{-4u}du = u\cdot\left(-\dfrac{e^{-4u}}{4}\right) - \int\left(-\dfrac{e^{-4u}}{4}\right)du = -\dfrac{u}{4}e^{-4u} - \dfrac{1}{16}e^{-4u}$.
   5. So the full integral is $\dfrac13\left[-\dfrac{u}{4}e^{-4u}-\dfrac1{16}e^{-4u}\right] = -\dfrac{e^{-4u}}{48}(4u+1)$.
   6. Substitute back $u=x^3$: $-\dfrac{e^{-4x^3}}{48}(4x^3+1) = \dfrac{1}{48}e^{-4x^3}\cdot\left[-(4x^3+1)\right]$.
   7. Comparing with $\dfrac{1}{48}e^{-4x^3}f(x)+c$, we read off $f(x)=-4x^3-1$.

   Common Mistakes

   • Forgetting the $\tfrac13$ factor that comes from $x^2\,dx = du/3$.
   • Sign errors in the by-parts step, flipping the final sign of $f(x)$.

   ✓Final answer

   The correct option is (B) — $-4x^3 - 1$.

   ANSWER: B

10. 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

11. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
    Q.$\displaystyle\int \dfrac{\cos x - \sin x}{10 + \sin 2x}\, dx =$ (A) $\dfrac{1}{2}\log(10 + \sin 2x) + c$ (B) $\dfrac{1}{3}\log(10 + \sin 2x) + c$ (C) $\dfrac{1}{3}\, \mathrm{Tan}^{-1}\left(\dfrac{\sin x + \cos x}{3}\right) + c$ (D) $\dfrac{1}{3}\, \mathrm{Tan}^{-1}(10 + \sin 2x) + c$

    ›Reveal solutionSolution

    The numerator $\cos x-\sin x$ is exactly $d(\sin x+\cos x)$, and the denominator rewrites in terms of $u=\sin x+\cos x$ via $\sin2x=u^2-1$. That collapses the integral to a standard $\int du/(a^2+u^2)$ arctangent form. Answer: $\dfrac13\mathrm{Tan}^{-1}\!\left(\dfrac{\sin x+\cos x}{3}\right)+c$.

    Concept and Intuition

    Whenever an integrand contains both $\sin x-\cos x$ (or $\cos x-\sin x$) and $\sin2x$, it is worth trying $u=\sin x+\cos x$ (or $\sin x-\cos x$) as the substitution, because $u^2=1\pm\sin2x$ links the two.

    Step-by-Step Solution

    1. Let $u=\sin x+\cos x$. Then $du=(\cos x-\sin x)\,dx$ — this is exactly the numerator times $dx$.
    2. Also $u^2=\sin^2x+\cos^2x+2\sin x\cos x=1+\sin2x$, so $\sin2x=u^2-1$.
    3. Denominator: $10+\sin2x=10+u^2-1=9+u^2$.
    4. The integral becomes $\displaystyle\int\frac{du}{9+u^2}=\frac13\mathrm{Tan}^{-1}\!\left(\frac u3\right)+c$.
    5. Substitute back: $=\dfrac13\mathrm{Tan}^{-1}\!\left(\dfrac{\sin x+\cos x}{3}\right)+c$.

    Common Mistakes

    • Trying $u=\sin x-\cos x$ instead, which does not match the numerator's sign here.
    • Forgetting the $\frac13$ scaling factor from $\int du/(a^2+u^2)=\frac1a\tan^{-1}(u/a)$ with $a=3$.

    ✓Final answer

    The correct option is (C) — $\dfrac{1}{3}\, \mathrm{Tan}^{-1}\left(\dfrac{\sin x + \cos x}{3}\right) + c$.

    ANSWER: C

12. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
    Q.If $\displaystyle\int \dfrac{dx}{\sin^3 x+\cos^3 x} = A\log\left|\dfrac{\sqrt{2}+t}{\sqrt{2}-t}\right|+B\mathrm{Tan}^{-1}(t)+c$, then $\left(\dfrac{B}{A},t\right) =$ (A) $(3\sqrt{2},\sin x-\cos x)$ (B) $(2\sqrt{2},\sin x-\cos x)$ (C) $\left(\dfrac{\sqrt{2}}{3},\sin x-\cos x\right)$ (D) $\left(\dfrac{3}{\sqrt{2}},\sin x+\cos x\right)$

    ›Reveal solutionSolution

    The standard $\sin^3x+\cos^3x$ integral, solved via the substitution $t=\sin x-\cos x$; matching to the given $A\log|\cdots|+B\tan^{-1}t$ form gives $B/A=2\sqrt2$ with $t=\sin x-\cos x$.

    Concept and Intuition

    $\sin^3x+\cos^3x$ factors as a sum of cubes, $(\sin x+\cos x)(\sin^2x-\sin x\cos x+\cos^2x)=(\sin x+\cos x)(1-\sin x\cos x)$. Both remaining factors can be expressed in terms of $t=\sin x-\cos x$ (since $t^2=1-2\sin x\cos x$ links $\sin x\cos x$ to $t$, and $dt=(\sin x+\cos x)\,dx$ conveniently cancels the other factor).

    Step-by-Step Solution

    1. Factor: $\sin^3x+\cos^3x = (\sin x+\cos x)(1-\sin x\cos x)$.
    2. Let $t=\sin x-\cos x \Rightarrow dt = (\cos x+\sin x)\,dx$, and $t^2 = 1-2\sin x\cos x \Rightarrow \sin x\cos x = \dfrac{1-t^2}{2}$.
    3. So $1-\sin x\cos x = 1-\dfrac{1-t^2}{2} = \dfrac{1+t^2}{2}$.
    4. Also $(\sin x+\cos x)^2 = 1+2\sin x\cos x = 1+(1-t^2) = 2-t^2$, so $\sin x+\cos x=\sqrt{2-t^2}$.
    5. Rewrite the integral:

    $\int\frac{dx}{(\sin x+\cos x)(1-\sin x\cos x)} = \int\frac{dx}{\sqrt{2-t^2}\cdot\frac{1+t^2}{2}}.$

    Since $dx = \dfrac{dt}{\sin x+\cos x} = \dfrac{dt}{\sqrt{2-t^2}}$:

    $= \int \frac{1}{\sqrt{2-t^2}\cdot\frac{1+t^2}{2}}\cdot\frac{dt}{\sqrt{2-t^2}} = \int\frac{2\,dt}{(2-t^2)(1+t^2)}.$

    6. Split using $1=\dfrac{(2-t^2)+(1+t^2)}{3}$:

    $\frac{2}{(2-t^2)(1+t^2)} = \frac23\cdot\frac{(2-t^2)+(1+t^2)}{(2-t^2)(1+t^2)} = \frac23\left[\frac{1}{1+t^2}+\frac{1}{2-t^2}\right].$

    7. Integrate each piece: $\displaystyle\int\frac{dt}{1+t^2}=\tan^{-1}t$; $\displaystyle\int\frac{dt}{2-t^2}=\frac{1}{2\sqrt2}\log\left|\frac{\sqrt2+t}{\sqrt2-t}\right|$ (standard form $\int\frac{dx}{a^2-x^2}=\frac{1}{2a}\log\left|\frac{a+x}{a-x}\right|$ with $a=\sqrt2$).
    8. So the integral is

    $\frac23\tan^{-1}t + \frac23\cdot\frac{1}{2\sqrt2}\log\left|\frac{\sqrt2+t}{\sqrt2-t}\right|+c = \frac{1}{3\sqrt2}\log\left|\frac{\sqrt2+t}{\sqrt2-t}\right| + \frac23\tan^{-1}t + c.$

    9. Matching to $A\log\left|\frac{\sqrt2+t}{\sqrt2-t}\right|+B\tan^{-1}t+c$: $A=\dfrac{1}{3\sqrt2}$, $B=\dfrac23$.
    10. $\dfrac{B}{A} = \dfrac{2/3}{1/(3\sqrt2)} = \dfrac23\times3\sqrt2 = 2\sqrt2$, and $t=\sin x-\cos x$.

    Common Mistakes

    • Using $t=\sin x+\cos x$ instead of $\sin x-\cos x$ (the sign matters — it's the difference that makes $dt$ match the other factor cleanly here).
    • Errors in the standard log-form constant $\dfrac{1}{2a}$ with $a=\sqrt2$, or forgetting the extra $\frac23$ scaling from the partial-fraction split.

    ✓Final answer

    The correct option is (B) — $(2\sqrt2,\ \sin x-\cos x)$.

    ANSWER: B