Q.Evaluate the definite integral: $\int_0^{\pi/4} \sin 2x \ dx$

Showing the 12 most recent of 22 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.$\displaystyle\int_0^{\pi/2} \dfrac{\sin 6x}{\sin x}\,dx =$ (A) $\dfrac{26}{15}$ (B) $\dfrac{256}{15}$ (C) $\dfrac{64}{15}$ (D) $\dfrac{17}{15}$

   ›Reveal solutionSolution

   Expanding $\sin6x/\sin x$ as a sum of cosines (a Chebyshev-type identity) and integrating term by term gives $26/15$.

   Concept and Intuition

   Ratios like $\sin(n\theta)/\sin\theta$ are polynomials in $\cos\theta$ (Chebyshev polynomials of the second kind), and can always be rewritten as a finite sum of cosines of multiples of $\theta$. This makes the integral trivial term by term, instead of trying to integrate the ratio directly.

   Step-by-Step Solution

   1. Claim: $\sin6\theta=\sin\theta\,[2\cos5\theta+2\cos3\theta+2\cos\theta]$. Verify using the product-to-sum identity $2\sin\theta\cos k\theta=\sin(k+1)\theta-\sin(k-1)\theta$: taking $k=5,3,1$ gives $(\sin6\theta-\sin4\theta)+(\sin4\theta-\sin2\theta)+(\sin2\theta-\sin0)=\sin6\theta$ — the sum telescopes perfectly.
   2. So $\dfrac{\sin6x}{\sin x}=2\cos5x+2\cos3x+2\cos x$.
   3. Integrate: $\displaystyle\int_0^{\pi/2}2\cos5x\,dx=\left[\tfrac{2}{5}\sin5x\right]_0^{\pi/2}=\tfrac25\sin\tfrac{5\pi}{2}=\tfrac25(1)=\tfrac25$. …

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

   ›Reveal solutionSolution

   Splitting $\cos2\theta/(\sin2\theta+\cos2\theta)$ into a constant plus an exact-derivative piece gives $\dfrac12\theta+\dfrac14\log(\sin2\theta+\cos2\theta)+c$, which rewrites as option (C).

   Concept and Intuition

   Integrals of the form $\int\dfrac{1}{1+\tan k\theta}d\theta$ are handled by converting to sines/cosines and splitting the numerator into a piece that's a constant multiple of the denominator (integrates to $\theta$) plus a piece that's proportional to the derivative of the denominator (integrates to a log). Recognising $\cos2\theta = \tfrac12[(\sin2\theta+\cos2\theta)+(\cos2\theta-\sin2\theta)]$ is the standard split.

   Step-by-Step Solution

   1. $\dfrac{1}{1+\tan2\theta} = \dfrac{\cos2\theta}{\cos2\theta+\sin2\theta}$.
   2. Write $\cos2\theta = \dfrac12(\sin2\theta+\cos2\theta) + \dfrac12(\cos2\theta-\sin2\theta)$.
   3. So the integrand $= \dfrac12 + \dfrac12\cdot\dfrac{\cos2\theta-\sin2\theta}{\sin2\theta+\cos2\theta}$.
   4. Note $\dfrac{d}{d\theta}(\sin2\theta+\cos2\theta) = 2\cos2\theta-2\sin2\theta = 2(\cos2\theta-\sin2\theta)$, so the second term integrates to $\dfrac12\cdot\dfrac12\log|\sin2\theta+\cos2\theta| = \dfrac14\log(\sin2\theta+\cos2\theta)$.
   5. Total: $\int = \dfrac12\theta + \dfrac14\log(\sin2\theta+\cos2\theta)+c$. …

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3. 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

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

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4. 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: $\dfrac{1}{t(t+1)}=\dfrac1t-\dfrac{1}{t+1}$. …

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5. 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 $\int\sqrt{1-x^2}\,dx=\dfrac{x}{2}\sqrt{1-x^2}+\dfrac12\sin^{-1}x+c_2$. …

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6. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.If $\int \frac{\cos 8x + 1}{\cot 2x - \tan 2x}dx = A\cos 8x + c$, then $A =$ (A) $-\frac{1}{16}$ (B) $\frac{1}{16}$ (C) $-\frac{1}{8}$ (D) $\frac{1}{8}$

   ›Reveal solutionSolution

   Simplifying the trigonometric expression using double-angle identities collapses the integrand to $\tfrac12\sin8x$, whose antiderivative directly gives $A=-\tfrac1{16}$.

   Concept and Intuition

   This problem looks intimidating because of the $\cot2x-\tan2x$ combination, but that combination has a clean double-angle identity: $\cot\theta-\tan\theta=2\cot2\theta$. Combined with $1+\cos8x=2\cos^24x$, nearly everything cancels, leaving a single simple sine term to integrate.

   Step-by-Step Solution

   1. Simplify the denominator: $\cot2x-\tan2x=\dfrac{\cos2x}{\sin2x}-\dfrac{\sin2x}{\cos2x}=\dfrac{\cos^22x-\sin^22x}{\sin2x\cos2x}=\dfrac{\cos4x}{\tfrac12\sin4x}=2\cot4x$.
   2. Simplify the numerator: $\cos8x+1=2\cos^24x$ (using $\cos2\theta=2\cos^2\theta-1$ with $\theta=4x$).
   3. So the integrand is $\dfrac{2\cos^24x}{2\cot4x}=\cos^24x\cdot\tan4x=\cos4x\sin4x$.
   4. Use $\sin\theta\cos\theta=\tfrac12\sin2\theta$: $\cos4x\sin4x=\tfrac12\sin8x$. …

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7. 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. $1+\sin2x=1+2\sin x\cos x=(\sin x+\cos x)^2$. …

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8. 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: $\displaystyle\int\left(\frac{-2}{u-1}+\frac{4}{u-2}\right)du = -2\log|u-1|+4\log|u-2|+c$. …

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9. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.$\displaystyle\int_{\pi/4}^{\pi/3} \dfrac{\cos x-\sin x}{\sin 2x}\,dx =$ (A) $\dfrac{1}{2}\log\left[\dfrac{(3+2\sqrt{2})(2-\sqrt{3})}{\sqrt{3}}\right]$ (B) $\dfrac{1}{2}\log\left[\dfrac{(3-2\sqrt{2})(2+\sqrt{3})}{\sqrt{3}}\right]$ (C) $\log\left[\dfrac{(3-2\sqrt{2})(2-\sqrt{3})}{\sqrt{3}}\right]$ (D) $\log\left[\dfrac{(3+2\sqrt{2})(2-\sqrt{3})}{\sqrt{3}}\right]$

   ›Reveal solutionSolution

   This tests splitting a trig rational integrand using $\sin 2x=2\sin x\cos x$ into standard $\csc$/$\sec$ integrals; the answer is (A).

   Concept and Intuition

   Whenever you see $\sin 2x$ in a denominator paired with $\cos x - \sin x$ in the numerator, it is almost always meant to be split via the double-angle identity into two elementary reciprocal-trig integrals, each of which has a well-known logarithmic antiderivative.

   Step-by-Step Solution

   1. Write $\sin 2x = 2\sin x \cos x$, so

   $\frac{\cos x-\sin x}{\sin 2x}=\frac{\cos x-\sin x}{2\sin x\cos x}=\frac12\left(\frac{1}{\sin x}-\frac{1}{\cos x}\right)=\frac12(\csc x-\sec x).$

   2. Use the standard antiderivatives $\int \csc x\,dx=-\log|\csc x+\cot x|$ and $\int\sec x\,dx=\log|\sec x+\tan x|$. So

   $\int\frac12(\csc x-\sec x)\,dx=-\frac12\log\big[(\csc x+\cot x)(\sec x+\tan x)\big]+C.$

   3. At $x=\pi/3$: $\csc x+\cot x=\frac{2}{\sqrt3}+\frac1{\sqrt3}=\sqrt3$, and $\sec x+\tan x=2+\sqrt3$. Product $=\sqrt3(2+\sqrt3)=2\sqrt3+3$.
   4. At $x=\pi/4$: $\csc x+\cot x=\sqrt2+1$, and $\sec x+\tan x=\sqrt2+1$. Product $=(\sqrt2+1)^2=3+2\sqrt2$.
   5. So the definite integral $=-\frac12\left[\log(2\sqrt3+3)-\log(3+2\sqrt2)\right]=\frac12\log\dfrac{3+2\sqrt2}{2\sqrt3+3}$. …

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10. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
    Q.$\displaystyle\int_0^{400\pi} \sqrt{1 - \cos 2x}\,dx =$ (A) $100\sqrt{2}$ (B) $200\sqrt{2}$ (C) $400\sqrt{2}$ (D) $800\sqrt{2}$

    ›Reveal solutionSolution

    This tests the identity $1-\cos2x=2\sin^2x$ combined with periodicity of $|\sin x|$ over a large interval. The integral equals $800\sqrt2$, option (D).

    Concept and Intuition

    $\sqrt{1-\cos2x}$ simplifies via the double-angle identity to $\sqrt2|\sin x|$ — the absolute value is essential because a square root is always non-negative, even though $\sin x$ itself oscillates in sign. Since $|\sin x|$ is periodic with period $\pi$ (not $2\pi$), a huge interval like $[0,400\pi]$ is just many identical copies of one period, so we only need the integral over one period and multiply by the number of periods.

    Step-by-Step Solution

    1. $1-\cos2x=2\sin^2x \Rightarrow \sqrt{1-\cos2x}=\sqrt{2\sin^2x}=\sqrt2\,|\sin x|$.
    2. $|\sin x|$ has period $\pi$ since $|\sin(x+\pi)|=|-\sin x|=|\sin x|$.
    3. On $[0,\pi]$, $\sin x\ge0$, so $\int_0^\pi|\sin x|\,dx=\int_0^\pi\sin x\,dx=[-\cos x]_0^\pi=2$.
    4. The interval $[0,400\pi]$ contains exactly $400$ such periods, so $\int_0^{400\pi}|\sin x|\,dx=400\times2=800$. …

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11. 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 …

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12. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
    Q.$\displaystyle\int\sin^4x\cos^4x\,dx=$ (A) $\dfrac{1}{128}(-2\sin^3x\cos x-3\sin x\cos x+3)+c$ (B) $\dfrac{1}{256}(-2\sin^3 2x\cos2x-3\sin2x\cos2x+6x)+c$ (C) $\dfrac{1}{128}(2\sin^3x\cos x-3\sin x\cos x+3x)+c$ (D) $\dfrac{1}{256}(3\sin^3x\cos x-2\sin x\cos x+2)+c$

    ›Reveal solutionSolution

    This tests reduction of $\sin^4x\cos^4x$ via the double-angle substitution $\sin x\cos x=\frac12\sin2x$; the answer is (B).

    Concept and Intuition

    Products of even powers of $\sin x,\cos x$ are attacked by writing $\sin x\cos x=\tfrac12\sin2x$ first, which collapses the problem to a single power of $\sin(2x)$, and then a standard power-reduction formula finishes it.

    Step-by-Step Solution

    1. $\sin^4x\cos^4x=(\sin x\cos x)^4=\left(\frac{\sin2x}{2}\right)^4=\frac{1}{16}\sin^4(2x)$.
    2. Use $\sin^4u=\dfrac{3}{8}-\dfrac12\cos2u+\dfrac18\cos4u$ with $u=2x$: $\sin^4(2x)=\dfrac38-\dfrac12\cos4x+\dfrac18\cos8x$.
    3. So the integrand is $\dfrac{1}{16}\left(\dfrac38-\dfrac12\cos4x+\dfrac18\cos8x\right)$. Integrating term by term:

    $\int\sin^4x\cos^4x\,dx=\frac{1}{16}\left(\frac{3x}{8}-\frac{\sin4x}{8}+\frac{\sin8x}{64}\right)+c=\frac{3x}{128}-\frac{\sin4x}{128}+\frac{\sin8x}{1024}+c.$

    4. Convert to functions of $2x$: $\sin4x=2\sin2x\cos2x$, and $\sin8x=2\sin4x\cos4x=4\sin2x\cos2x(1-2\sin^2 2x)=4\sin2x\cos2x-8\sin^3(2x)\cos2x$. …

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