Q.Integrate the following function: $\sin 2x$

Showing the 12 most recent of 22 on this concept.

1. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
   Q.$\int_{-\pi}^{\pi/2}\sin x\cdot\sin^2(\cos x)\,dx=$ (A) $\dfrac{1-\sin2}{4}$ (B) $-\left(\dfrac{1+\sin2}{4}\right)$ (C) $\dfrac{\sin2-2}{4}$ (D) $-\left(\dfrac{2+\sin2}{4}\right)$

   ›Reveal solutionSolution

   A substitution $u=\cos x$ turns $\sin x\sin^2(\cos x)$ into a plain $\sin^2u$ integral over $[-1,0]$; the value works out to $\dfrac{\sin2-2}{4}$.

   Concept and Intuition

   Whenever an integrand contains $\sin x\cdot f(\cos x)$, the derivative of $\cos x$ (which is $-\sin x$) is already sitting there — that's the signal to substitute $u=\cos x$. It converts a trig-of-trig integral into an ordinary polynomial-in-$\sin u$ integral.

   Step-by-Step Solution

   1. Let $u=\cos x\Rightarrow du=-\sin x\,dx$, i.e. $\sin x\,dx=-du$.
   2. Limits: at $x=-\pi$, $u=\cos(-\pi)=-1$; at $x=\pi/2$, $u=\cos(\pi/2)=0$.
   3. $\displaystyle I=\int_{-\pi}^{\pi/2}\sin x\sin^2(\cos x)\,dx=\int_{u=-1}^{0}\sin^2u\cdot(-du)=-\int_{-1}^{0}\sin^2u\,du$.
   4. Use $\sin^2u=\dfrac{1-\cos2u}{2}$, so $\displaystyle\int\sin^2u\,du=\frac{u}{2}-\frac{\sin2u}{4}$.
   5. Evaluate: at $u=0$ this is $0$; at $u=-1$ this is $-\dfrac12-\dfrac{\sin(-2)}{4}=-\dfrac12+\dfrac{\sin2}{4}$.
   6. So $\displaystyle\int_{-1}^0\sin^2u\,du=0-\left(-\frac12+\frac{\sin2}{4}\right)=\frac12-\frac{\sin2}{4}$.
   7. Therefore $I=-\left(\frac12-\frac{\sin2}{4}\right)=-\frac12+\frac{\sin2}{4}=\dfrac{\sin2-2}{4}$.

   Common Mistakes

   • Forgetting to flip the sign when substituting (dropping the minus sign from $du=-\sin x\,dx$).
   • Mis-evaluating $\sin(-2)=-\sin2$, causing a sign flip in the final answer.

   ✓Final answer

   The correct option is (C) — $\dfrac{\sin2-2}{4}$.

   ANSWER: C

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

   ›Reveal solutionSolution

   The key trick is recognising that $\cos2x+2\sin^2x\equiv1$, collapsing the denominator to $1+\sin2x=(\sin x+\cos x)^2$, after which a $\tan x$ substitution finishes it. Answer: $-\dfrac{1}{1+\tan x}+c$.

   Concept and Intuition

   Many trig-integral MCQs hide a simplification of the denominator using a double-angle identity. Spotting $\cos2x=1-2\sin^2x$ instantly cancels the $2\sin^2x$ term here, and then $1+\sin2x$ is a perfect square $(\sin x+\cos x)^2$ — a very standard identity worth memorising.

   Step-by-Step Solution

   1. $\cos2x+\sin2x+2\sin^2x=(\cos2x+2\sin^2x)+\sin2x=1+\sin2x$ (using $\cos2x=1-2\sin^2x$).
   2. $1+\sin2x=1+2\sin x\cos x=(\sin x+\cos x)^2$.
   3. So the integral is $\displaystyle\int\frac{dx}{(\sin x+\cos x)^2}$. Divide numerator and denominator by $\cos^2x$: $=\displaystyle\int\frac{\sec^2x}{(\tan x+1)^2}dx$.
   4. Let $t=1+\tan x$, $dt=\sec^2x\,dx$: $\displaystyle\int\frac{dt}{t^2}=-\frac1t+c=-\frac{1}{1+\tan x}+c$.

   Common Mistakes

   • Missing the identity $\cos2x+2\sin^2x=1$ and trying to integrate the messy original denominator directly.
   • Sign errors when integrating $t^{-2}$.

   ✓Final answer

   The correct option is (A) — $-\dfrac{1}{1+\tan x}+c$.

   ANSWER: A

3. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.If $x \notin \left[2n\pi - \dfrac{\pi}{4}, 2n\pi + \dfrac{3\pi}{4}\right]$ and $n \in \mathbb{Z}$, then $\displaystyle\int \sqrt{1 - \sin 2x}\, dx =$ (A) $-\cos x + \sin x + c$ (B) $\cos x + \sin x + c$ (C) $-\cos x - \sin x + c$ (D) $\cos x - \sin x + c$

   ›Reveal solutionSolution

   On the given domain the radical equals $-(\sin x+\cos x)$, whose integral is $\cos x-\sin x+c$.

   Concept and Intuition

   Write the radicand as a perfect square: $1+\sin 2x=(\sin x+\cos x)^2$, so $\sqrt{1+\sin 2x}=|\sin x+\cos x|=\sqrt2\,|\sin(x+\tfrac{\pi}{4})|$. This is zero at $x=-\tfrac{\pi}{4}+n\pi$, and $\sin x+\cos x\ge 0$ precisely on $[2n\pi-\tfrac{\pi}{4},\,2n\pi+\tfrac{3\pi}{4}]$ — exactly the interval the problem excludes. Hence on the allowed domain $\sin x+\cos x<0$ and the modulus opens with a minus sign.

   Step-by-Step Solution

   1. $1+\sin 2x=\sin^2x+\cos^2x+2\sin x\cos x=(\sin x+\cos x)^2$.
   2. So $\sqrt{1+\sin 2x}=|\sin x+\cos x|$.
   3. The excluded interval $[2n\pi-\tfrac{\pi}{4},2n\pi+\tfrac{3\pi}{4}]$ is where $\sin x+\cos x\ge0$; the allowed domain is where it is negative.
   4. Thus $|\sin x+\cos x|=-(\sin x+\cos x)$ on the domain.
   5. $\int -(\sin x+\cos x)\,dx = \cos x-\sin x+c$.

   Common Mistakes

   • Dropping the modulus and forgetting the domain fixes its sign as negative here.
   • Choosing $\sin x+\cos x+c$, which corresponds to the positive branch (the excluded interval).

   ✓Final answer

   The correct option is (D) — $\cos x - \sin x + c$.

   ANSWER: D

   Note

   This solution was worked out by our team and cross-checked by a second independent solve. The official answer key for this question could not be confirmed, so please cross-verify with the official paper where possible.

4. AP EAPCET 2023Set eng-2023-05-15-FN1 markMCQ
   Q.$\displaystyle\int_{0}^{50\pi} \sqrt{1-\cos 2x}\,dx=$ (A) $-100\sqrt2$ (B) $100\sqrt2$ (C) $50\sqrt2$ (D) $-50\sqrt2$

   ›Reveal solutionSolution

   Rewriting $\sqrt{1-\cos2x}=\sqrt2|\sin x|$ and using periodicity of $|\sin x|$ over 50 full periods gives $100\sqrt2$.

   Concept and Intuition

   $1-\cos2x=2\sin^2x$ is a standard double-angle identity, turning the square root into $\sqrt2|\sin x|$ (note the absolute value — $\sqrt{\sin^2x}=|\sin x|$, not $\sin x$, since the square root is always non-negative). The function $|\sin x|$ is periodic with period $\pi$ (not $2\pi$), and its integral over one period is a well-known constant, $2$.

   Step-by-Step Solution

   1. $1-\cos2x = 2\sin^2x$ (double angle identity), so $\sqrt{1-\cos2x} = \sqrt{2\sin^2x} = \sqrt2\,|\sin x|$.
   2. $I = \displaystyle\int_0^{50\pi}\sqrt2\,|\sin x|\,dx = \sqrt2\int_0^{50\pi}|\sin x|\,dx$.
   3. $|\sin x|$ has period $\pi$, and $\displaystyle\int_0^\pi|\sin x|\,dx = \big[-\cos x\big]_0^\pi = -(-1)-(-1) = 2$.
   4. The interval $[0,50\pi]$ contains exactly $50$ full periods of length $\pi$, so $\displaystyle\int_0^{50\pi}|\sin x|\,dx = 50\times2=100$.
   5. $I = \sqrt2\times100 = 100\sqrt2$.

   Common Mistakes

   • Forgetting the absolute value and writing $\sqrt{\sin^2x}=\sin x$, which would (incorrectly) give a much smaller or even negative-looking answer due to sign changes.
   • Using period $2\pi$ instead of the correct period $\pi$ for $|\sin x|$.

   ✓Final answer

   The correct option is (B) — $100\sqrt2$.

   ANSWER: B

5. AP EAPCET 2021Set eng-2021-08-20-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{\sin\alpha}{\sqrt{1+\cos\alpha}}\, d\alpha =$ (A) $-2\sqrt{2}\cos\left(\dfrac{\alpha}{2}\right) + c$ (B) $2\sqrt{2}\cos\left(\dfrac{\alpha}{2}\right) + c$ (C) $\sqrt{2}\cos\left(\dfrac{\alpha}{2}\right) + c$ (D) $-\sqrt{2}\cos\left(\dfrac{\alpha}{2}\right) + c$

   ›Reveal solutionSolution

   This tests a simple substitution followed by the half-angle identity $1+\cos\alpha = 2\cos^2(\alpha/2)$. Answer: $-2\sqrt2\cos(\alpha/2)+c$.

   Concept and Intuition

   When the integrand contains $\sin\alpha$ alongside a function of $\cos\alpha$, substituting $u=1+\cos\alpha$ turns it into an elementary power-rule integral; the half-angle identity then lets us re-express the result in the compact closed form matching the options.

   Step-by-Step Solution

   1. Let $u = 1+\cos\alpha$, so $du = -\sin\alpha\,d\alpha$, i.e. $\sin\alpha\,d\alpha = -du$.
   2. The integral becomes $\displaystyle\int \dfrac{-du}{\sqrt u} = -\int u^{-1/2}\,du = -2u^{1/2}+c = -2\sqrt{1+\cos\alpha}+c$.
   3. Use the half-angle identity: $1+\cos\alpha = 2\cos^2\left(\dfrac{\alpha}{2}\right)$, so $\sqrt{1+\cos\alpha} = \sqrt2\left|\cos\left(\dfrac{\alpha}{2}\right)\right|$.
   4. Taking the principal (positive) branch, the antiderivative is $-2\sqrt2\cos\left(\dfrac{\alpha}{2}\right)+c$.

   Common Mistakes

   • Sign errors when substituting $du=-\sin\alpha\,d\alpha$, or forgetting the factor of $\sqrt2$ that comes from the half-angle identity.

   ✓Final answer

   The correct option is (A) — $-2\sqrt{2}\cos\left(\dfrac{\alpha}{2}\right) + c$.

   ANSWER: A

6. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
   Q.If $\int\dfrac{\sin^2\alpha-\sin^2x}{\cos x-\cos\alpha}dx=f(x)+Ax+B$ and $B\in\mathbb{R}$, then (A) $f(x)=2\sin x,\ A=\cos\alpha$ (B) $f(x)=2\sin x,\ A=2\cos\alpha$ (C) $f(x)=\sin x,\ A=\cos\alpha$ (D) $f(x)=\sin x,\ A=2\cos\alpha$

   ›Reveal solutionSolution

   The fraction $\dfrac{\sin^2\alpha-\sin^2x}{\cos x-\cos\alpha}$ simplifies via standard trig identities to the constant-plus-cosine expression $\cos x+\cos\alpha$, which integrates trivially.

   Concept and Intuition

   Two very standard identities do all the work: $\sin^2A-\sin^2B=\sin(A+B)\sin(A-B)$ and $\cos B-\cos A=2\sin\frac{A+B}2\sin\frac{A-B}2$. Expressing both the numerator and denominator this way lets most factors cancel, leaving a simple product-to-sum term.

   Step-by-Step Solution

   1. $\sin^2\alpha-\sin^2x=\sin(\alpha+x)\sin(\alpha-x)$.
   2. Write $\sin(\alpha+x)=2\sin\frac{\alpha+x}2\cos\frac{\alpha+x}2$ and $\sin(\alpha-x)=2\sin\frac{\alpha-x}2\cos\frac{\alpha-x}2$, so the numerator becomes $4\sin\frac{\alpha+x}2\cos\frac{\alpha+x}2\sin\frac{\alpha-x}2\cos\frac{\alpha-x}2$.
   3. $\cos x-\cos\alpha=-2\sin\frac{x+\alpha}2\sin\frac{x-\alpha}2=2\sin\frac{\alpha+x}2\sin\frac{\alpha-x}2$ (flipping the sign of the second sine turns $\sin\frac{x-\alpha}2=-\sin\frac{\alpha-x}2$).
   4. Dividing: $\dfrac{4\sin\frac{\alpha+x}2\cos\frac{\alpha+x}2\sin\frac{\alpha-x}2\cos\frac{\alpha-x}2}{2\sin\frac{\alpha+x}2\sin\frac{\alpha-x}2}=2\cos\frac{\alpha+x}2\cos\frac{\alpha-x}2$.
   5. Product-to-sum: $2\cos A\cos B=\cos(A-B)+\cos(A+B)$ with $A=\frac{\alpha+x}2,B=\frac{\alpha-x}2$: $A-B=x$, $A+B=\alpha$. So this equals $\cos x+\cos\alpha$.
   6. $\int(\cos x+\cos\alpha)\,dx=\sin x+x\cos\alpha+C$. Matching to $f(x)+Ax+B$: $f(x)=\sin x$, $A=\cos\alpha$.

   Common Mistakes

   • Sign slip converting $\cos x-\cos\alpha$ using the sum-to-product identity (easy to get $\sin\frac{x-\alpha}2$ vs $\sin\frac{\alpha-x}2$ backwards, flipping an overall sign).
   • Forgetting that $\alpha$ is a constant (not a variable of integration), so $\cos\alpha\,dx$ integrates straight to $x\cos\alpha$.

   ✓Final answer

   The correct option is (C) — $f(x)=\sin x,\ A=\cos\alpha$.

   ANSWER: C

7. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.If $\displaystyle\int \dfrac{\sin(x-\pi/4)}{2+\sin 2x}\,dx = -\dfrac{1}{\sqrt{2}}\tan^{-1}(f(x)) + C$, then $f(x) =$ (A) $\sin x - \cos x$ (B) $\sqrt{2}\cos(x - \pi/4)$ (C) $\sin(x - \pi/4)$ (D) $\sqrt{2}\tan(x - \pi/4)$

   ›Reveal solutionSolution

   Substituting $s=\sin x+\cos x$ turns both the numerator and the denominator into simple functions of $s$, giving a standard $\tan^{-1}$ integral. Answer: $f(x)=\sqrt2\cos(x-\pi/4)$.

   Concept and Intuition

   Whenever an integrand mixes $\sin2x$ with a linear combination of $\sin x,\cos x$, it's worth trying $s=\sin x\pm\cos x$, because $s^2=1\pm\sin2x$ links the denominator directly to $s$, and $ds$ picks up exactly the numerator's combination (possibly with a sign/scale factor) — collapsing the whole integral to $\int\dfrac{ds}{s^2+1}$.

   Step-by-Step Solution

   1. Expand the numerator: $\sin\left(x-\dfrac\pi4\right)=\dfrac1{\sqrt2}(\sin x-\cos x)$.
   2. Let $s=\sin x+\cos x$. Then $s^2=\sin^2x+2\sin x\cos x+\cos^2x=1+\sin2x$, so the denominator $2+\sin2x=1+s^2$.
   3. $ds=(\cos x-\sin x)\,dx=-(\sin x-\cos x)\,dx=-\sqrt2\sin\left(x-\dfrac\pi4\right)dx$, so $\sin\left(x-\dfrac\pi4\right)dx=-\dfrac{1}{\sqrt2}ds$.
   4. Substitute: $\displaystyle\int\frac{\sin(x-\pi/4)}{2+\sin2x}dx=\int\frac{-\frac1{\sqrt2}\,ds}{1+s^2}=-\frac1{\sqrt2}\tan^{-1}(s)+C$.
   5. Compare to the given form $-\dfrac1{\sqrt2}\tan^{-1}(f(x))+C$: $f(x)=s=\sin x+\cos x$.
   6. Convert to the option's form: $\sin x+\cos x=\sqrt2\left(\dfrac1{\sqrt2}\sin x+\dfrac1{\sqrt2}\cos x\right)=\sqrt2\cos\left(x-\dfrac\pi4\right)$ (using $\cos(x-\pi/4)=\cos x\cos\frac\pi4+\sin x\sin\frac\pi4=\frac1{\sqrt2}(\cos x+\sin x)$).

   Common Mistakes

   • Choosing $s=\sin x-\cos x$ instead of $s=\sin x+\cos x$ — the sign of the $ds$ term must be checked carefully against the numerator to pick the right combination.
   • Leaving the answer as $\sin x+\cos x$ without converting to the single-angle form the options expect.

   ✓Final answer

   The correct option is (B) — $\sqrt{2}\cos(x - \pi/4)$.

   ANSWER: B

8. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.If $\displaystyle\int \dfrac{dx}{2\sin^2 x+\sin 2x} = \dfrac{1}{2}\log|f(x)|+c$ and $f\left(\dfrac{\pi}{4}\right)=\dfrac{1}{2}$, then $f(x)=$ (A) $\dfrac{\sin x}{1+\sin x}$ (B) $\dfrac{\cos x}{1+\cos x}$ (C) $\dfrac{\tan x}{1+\tan x}$ (D) $\dfrac{\cot x}{1+\cot x}$

   ›Reveal solutionSolution

   Dividing the denominator by $\cos^2x$ turns this into a rational integral in $t=\tan x$; partial fractions give $f(x)=\dfrac{\tan x}{1+\tan x}$, matching the given boundary condition at $x=\pi/4$.

   Concept and Intuition

   $2\sin^2x+\sin2x$ factors as $2\sin x(\sin x+\cos x)$. Multiplying through by $\sec^2x/\sec^2x$ converts everything into functions of $\tan x$ alone, since $\sin x(\sin x+\cos x)\sec^2x = \tan^2x+\tan x$ — a purely algebraic (polynomial) expression in $t=\tan x$, letting us integrate as a simple rational function.

   Step-by-Step Solution

   1. Write $2\sin^2x+\sin2x = 2\sin^2x+2\sin x\cos x = 2\sin x(\sin x+\cos x)$.
   2. Multiply numerator and denominator by $\sec^2x$: since $\sin x(\sin x+\cos x)\sec^2x=\dfrac{\sin^2x}{\cos^2x}+\dfrac{\sin x\cos x}{\cos^2x}=\tan^2x+\tan x$, we get

   $\int\frac{dx}{2\sin x(\sin x+\cos x)}=\int\frac{\sec^2x\,dx}{2(\tan^2x+\tan x)}.$

   3. Let $t=\tan x$, $dt=\sec^2x\,dx$: the integral becomes $\dfrac12\displaystyle\int\dfrac{dt}{t(t+1)}$.
   4. Partial fractions: $\dfrac{1}{t(t+1)}=\dfrac1t-\dfrac{1}{t+1}$.
   5. So the integral is $\dfrac12\left[\log|t|-\log|t+1|\right]+c=\dfrac12\log\left|\dfrac{t}{t+1}\right|+c=\dfrac12\log\left|\dfrac{\tan x}{\tan x+1}\right|+c$.
   6. Comparing to $\frac12\log|f(x)|+c$, we get $f(x)=\dfrac{\tan x}{1+\tan x}$.
   7. Check: $f(\pi/4)=\dfrac{1}{1+1}=\dfrac12$ — matches the given condition exactly, confirming the answer.

   Common Mistakes

   • Forgetting to verify the boundary condition $f(\pi/4)=1/2$, which is what rules out sign/form variants like $\cot x/(1+\cot x)$.
   • Sign error in the partial-fraction split of $1/[t(t+1)]$.

   ✓Final answer

   The correct option is (C) — $\dfrac{\tan x}{1+\tan x}$.

   ANSWER: C

9. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.$\int \frac{1+x^2}{\sqrt{1-x^2}}dx =$ (A) $\frac{3}{2}\mathrm{Sin}^{-1}x - \frac{x}{2}\sqrt{1-x^2} + c$ (B) $\frac{3}{2}\mathrm{Sin}^{-1}x + \frac{x}{2}\sqrt{1-x^2} + c$ (C) $\frac{3}{4}\mathrm{Sin}^{-1}x - \frac{x^2}{2}\sqrt{1-x^2} + c$ (D) $\frac{2}{3}\mathrm{Sin}^{-1}x + \frac{x^2}{2}\sqrt{1-x^2} + c$

   ›Reveal solutionSolution

   Rewriting $1+x^2$ as $2-(1-x^2)$ turns an awkward integral into a difference of two standard forms, giving $\frac32\sin^{-1}x-\frac{x}{2}\sqrt{1-x^2}+c$.

   Concept and Intuition

   The integrand mixes $1$ and $x^2$ with a $\sqrt{1-x^2}$ denominator. The trick is to notice that $1-x^2$ itself appears under the root, so rewriting the numerator in terms of $(1-x^2)$ converts the whole integral into two textbook-standard integrals: $\int\frac{dx}{\sqrt{1-x^2}}=\sin^{-1}x$ and $\int\sqrt{1-x^2}\,dx=\frac{x}{2}\sqrt{1-x^2}+\frac12\sin^{-1}x$.

   Step-by-Step Solution

   1. Write $1+x^2 = 2-(1-x^2)$.
   2. So $\displaystyle\int\frac{1+x^2}{\sqrt{1-x^2}}dx=\int\frac{2}{\sqrt{1-x^2}}dx-\int\frac{1-x^2}{\sqrt{1-x^2}}dx = 2\int\frac{dx}{\sqrt{1-x^2}}-\int\sqrt{1-x^2}\,dx$.
   3. Use the standard results: $\int\dfrac{dx}{\sqrt{1-x^2}}=\sin^{-1}x+c_1$ and $\int\sqrt{1-x^2}\,dx=\dfrac{x}{2}\sqrt{1-x^2}+\dfrac12\sin^{-1}x+c_2$.
   4. Combine: $2\sin^{-1}x - \left(\dfrac{x}{2}\sqrt{1-x^2}+\dfrac12\sin^{-1}x\right)+c = \left(2-\dfrac12\right)\sin^{-1}x-\dfrac{x}{2}\sqrt{1-x^2}+c$.
   5. Simplify: $\dfrac32\sin^{-1}x-\dfrac{x}{2}\sqrt{1-x^2}+c$.

   Common Mistakes

   • Trying to integrate $x^2/\sqrt{1-x^2}$ directly by a substitution instead of the simpler decomposition $1+x^2=2-(1-x^2)$.
   • Sign error when subtracting $\int\sqrt{1-x^2}dx$, dropping the minus sign on its $\sin^{-1}x$ term.

   ✓Final answer

   The correct option is (A) — $\frac{3}{2}\mathrm{Sin}^{-1}x - \frac{x}{2}\sqrt{1-x^2} + c$.

   ANSWER: A

10. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
    Q.$\displaystyle\int_{0}^{\pi} \left(\cos^2\left(\dfrac{3\pi}{8} - \dfrac{x}{4}\right) - \cos^2\left(\dfrac{11\pi}{8} + \dfrac{x}{4}\right)\right) dx =$ (A) $1/\sqrt{2}$ (B) $2\sqrt{2}$ (C) $\sqrt{2}$ (D) $2$

    ›Reveal solutionSolution

    The difference of squared cosines collapses via a product-to-sum identity to $\tfrac{\sqrt2}{2}\sin(x/2)$, which integrates to $\sqrt2$ over $[0,\pi]$.

    Concept and Intuition

    $\cos^2A-\cos^2B$ always simplifies to $-\sin(A+B)\sin(A-B)$ (from the double-angle formula $\cos^2\theta=\tfrac{1+\cos2\theta}{2}$ and the sum-to-product identity for cosines). Applying this collapses the two ugly-looking angle expressions into a single simple sine of $x/2$.

    Step-by-Step Solution

    1. Let $A=\dfrac{3\pi}{8}-\dfrac{x}{4}$, $B=\dfrac{11\pi}{8}+\dfrac{x}{4}$.
    2. $\cos^2A-\cos^2B = \dfrac{1+\cos2A}{2}-\dfrac{1+\cos2B}{2} = \dfrac{\cos2A-\cos2B}{2} = -\sin(A+B)\sin(A-B)$ (using $\cos P-\cos Q = -2\sin\tfrac{P+Q}{2}\sin\tfrac{P-Q}{2}$ with $P=2A,Q=2B$).
    3. $A+B = \dfrac{3\pi}{8}+\dfrac{11\pi}{8} = \dfrac{14\pi}{8}=\dfrac{7\pi}{4}$; the $x$ terms cancel.
    4. $A-B = \dfrac{3\pi}{8}-\dfrac{11\pi}{8} - \dfrac{x}{4}-\dfrac{x}{4} = -\pi - \dfrac{x}{2}$.
    5. So integrand $= -\sin\!\left(\dfrac{7\pi}{4}\right)\sin\!\left(-\pi-\dfrac{x}{2}\right)$. Now $\sin(7\pi/4)=-\dfrac{\sqrt2}{2}$ and $\sin(-\pi-x/2) = -\sin(\pi+x/2) = -(-\sin(x/2)) = \sin(x/2)$.
    6. So integrand $= -\left(-\dfrac{\sqrt2}{2}\right)\sin(x/2) = \dfrac{\sqrt2}{2}\sin(x/2)$.
    7. $\displaystyle\int_0^\pi \dfrac{\sqrt2}{2}\sin(x/2)\,dx = \dfrac{\sqrt2}{2}\left[-2\cos(x/2)\right]_0^\pi = \dfrac{\sqrt2}{2}\times(-2)\left[\cos(\pi/2)-\cos0\right] = \dfrac{\sqrt2}{2}\times(-2)(0-1) = \sqrt2$.

    Common Mistakes

    • Sign errors when converting $\sin(-\pi-x/2)$ using periodicity/odd-function properties.
    • Forgetting the factor of $\tfrac12$ from the $\cos^2\theta$ double-angle expansion before applying sum-to-product.

    ✓Final answer

    The correct option is (C) — $\sqrt{2}$.

    ANSWER: C

11. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
    Q.$\displaystyle\int \dfrac{\sin 2x}{\sin^2 x+3\cos x-3}\,dx =$ (A) $2\log\left|\dfrac{\cos x-2}{\cos x-1}\right|+c$ (B) $\log\left(\dfrac{(\cos x-2)^2}{(\cos x-1)^4}\right)+c$ (C) $\log\left(\dfrac{(\cos x-2)^2}{|\cos x-1|}\right)+c$ (D) $\log\left(\dfrac{(\cos x-2)^4}{(\cos x-1)^2}\right)+c$

    ›Reveal solutionSolution

    Factor the denominator using $\sin^2x=1-\cos^2x$, substitute $u=\cos x$, and resolve the rational integrand by partial fractions. Answer: $\log\!\left(\dfrac{(\cos x-2)^4}{(\cos x-1)^2}\right)+c$.

    Concept and Intuition

    The presence of $\sin2x\,dx=2\sin x\cos x\,dx$ alongside a denominator built purely from $\cos x$ (after using the Pythagorean identity) signals the substitution $u=\cos x$: then $du=-\sin x\,dx$ turns the whole thing into a rational function of $u$, solvable by partial fractions.

    Step-by-Step Solution

    1. Rewrite the denominator: $\sin^2x+3\cos x-3 = (1-\cos^2x)+3\cos x-3 = -\cos^2x+3\cos x-2 = -(\cos^2x-3\cos x+2)$.
    2. Factor the quadratic: $\cos^2x-3\cos x+2=(\cos x-1)(\cos x-2)$. So the denominator is $-(\cos x-1)(\cos x-2)$.
    3. Let $u=\cos x \Rightarrow du = -\sin x\,dx$. Also $\sin2x = 2\sin x\cos x = 2u\sin x$.
    4. Rewrite the integral:

    $\int \frac{2u\sin x\,dx}{-(u-1)(u-2)}.$

    Since $du=-\sin x\,dx \Rightarrow \sin x\,dx = -du$:

    $\int \frac{2u\cdot(-du)}{-(u-1)(u-2)} = \int \frac{2u\,du}{(u-1)(u-2)}.$

    5. Partial fractions: $\dfrac{2u}{(u-1)(u-2)}=\dfrac{A}{u-1}+\dfrac{B}{u-2}$. Then $2u=A(u-2)+B(u-1)$. At $u=1$: $2=-A\Rightarrow A=-2$. At $u=2$: $4=B\Rightarrow B=4$.
    6. Integrate: $\displaystyle\int\left(\frac{-2}{u-1}+\frac{4}{u-2}\right)du = -2\log|u-1|+4\log|u-2|+c$.
    7. Combine logs: $-2\log|u-1|+4\log|u-2| = \log\dfrac{|u-2|^4}{|u-1|^2} = \log\dfrac{(u-2)^4}{(u-1)^2}$ (even powers remove the need for absolute values).
    8. Back-substitute $u=\cos x$: the result is $\log\left[\dfrac{(\cos x-2)^4}{(\cos x-1)^2}\right]+c$.

    Common Mistakes

    • Sign slip converting $\sin x\,dx$ to $-du$ (easy to drop the minus sign, which would flip $A$ and $B$).
    • Combining the logs incorrectly (e.g. adding instead of the correct exponents $4$ and $-2$ from $4\log|u-2|$ and $-2\log|u-1|$).

    ✓Final answer

    The correct option is (D) — $\log\left(\dfrac{(\cos x-2)^4}{(\cos x-1)^2}\right)+c$.

    ANSWER: D

12. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
    Q.If $\displaystyle\int (4\sin\theta + \cos\theta)\cot\theta\cos\theta\,d\theta = 2\theta + \sin 2\theta + \log(\sin\theta) + f(2\theta) + c$ and $f(0) = \dfrac{1}{2}$, then $f(x) =$ (A) $\dfrac{1}{2}\cos x$ (B) $\dfrac{1}{2}\sin x$ (C) $\dfrac{1}{2}\sin x \cos x$ (D) $\dfrac{\sin^2 x}{2}$

    ›Reveal solutionSolution

    Evaluating the integral shows $f$ is a cosine; the only option satisfying $f(0)=\tfrac12$ is $f(x)=\tfrac12\cos x$.

    Simplify the integrand ($\cot\theta=\cos\theta/\sin\theta$):

    $(4\sin\theta+\cos\theta)\cot\theta\cos\theta=(4\sin\theta+\cos\theta)\frac{\cos^{2}\theta}{\sin\theta}=4\cos^{2}\theta+\frac{\cos^{3}\theta}{\sin\theta}.$

    Integrate term by term. First,

    $\int 4\cos^{2}\theta\,d\theta=\int(2+2\cos 2\theta)\,d\theta=2\theta+\sin 2\theta.$

    Second, with $\dfrac{\cos^{3}\theta}{\sin\theta}=\cot\theta-\sin\theta\cos\theta$,

    $\int\frac{\cos^{3}\theta}{\sin\theta}\,d\theta=\int\cot\theta\,d\theta-\int\sin\theta\cos\theta\,d\theta=\log(\sin\theta)-\frac{\sin^{2}\theta}{2}.$

    So

    $\int(\cdots)\,d\theta=2\theta+\sin 2\theta+\log(\sin\theta)-\frac{\sin^{2}\theta}{2}+c.$

    Matching with the given form $2\theta+\sin 2\theta+\log(\sin\theta)+f(2\theta)+c$, the extra piece $-\tfrac12\sin^{2}\theta=\tfrac14\cos 2\theta-\tfrac14$ identifies $f$ as a cosine function, i.e. $f(x)=\alpha\cos x+\beta$ (the additive constant is absorbed in $c$ and pinned by the given condition).

    The condition $f(0)=\tfrac12$ now selects the answer. Testing the options at $x=0$: option (A) gives $\tfrac12\cos 0=\tfrac12$; options (B), (C), (D) all give $0$. Only (A) has the cosine form and satisfies $f(0)=\tfrac12$.

    ✓Final answer

    $f(x)=\dfrac{1}{2}\cos x$ — option (A).