Q.Evaluate $\int_{-1}^1 \sin^5 x \cos^4 x\, dx$

Showing the 12 most recent of 48 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.If $\displaystyle\int_0^{2\pi} \sin^m x \cos^n x\,dx = k \int_0^{\pi/2} \sin^m x \cos^n x\,dx$, then (A) $k$ divides $(mn)$ (B) $k$ divides $(m+n)$ (C) When $(mn)$ is divided by $k$, it leaves the reminder 1 (D) When $(m+n)$ is divided by $k$, it leaves the reminder 3

   ›Reveal solutionSolution

   The constant $k$ relating the full-period and quarter-period integrals is $4$ only when $m,n$ are both even (else $0$); in that case $k=4$ always divides $mn$, proving option (A).

   Concept and Intuition

   $\sin^mx\cos^nx$ has extra symmetry over $[0,2\pi]$ that lets us relate the integral over the whole period to the integral over just the first quadrant $[0,\pi/2]$. The key is checking how the integrand transforms under the reflections $x\to\pi-t,\ \pi+t,\ 2\pi-t$.

   Step-by-Step Solution

   1. Split $[0,2\pi]$ into $[0,\pi/2],[\pi/2,\pi],[\pi,3\pi/2],[3\pi/2,2\pi]$ and substitute $x=\pi-t,\ \pi+t,\ 2\pi-t$ respectively (each $t\in[0,\pi/2]$).
   2. On $[\pi/2,\pi]$: $\sin x=\sin t,\ \cos x=-\cos t\Rightarrow$ integrand $=(-1)^n\sin^mt\cos^nt$.
   3. On $[\pi,3\pi/2]$: $\sin x=-\sin t,\ \cos x=-\cos t\Rightarrow$ integrand $=(-1)^{m+n}\sin^mt\cos^nt$.
   4. On $[3\pi/2,2\pi]$: $\sin x=-\sin t,\ \cos x=\cos t\Rightarrow$ integrand $=(-1)^m\sin^mt\cos^nt$.
   5. Adding all four pieces: $\displaystyle\int_0^{2\pi}=\big[1+(-1)^n+(-1)^{m+n}+(-1)^m\big]\int_0^{\pi/2}\sin^mx\cos^nx\,dx$, so $k=1+(-1)^n+(-1)^{m+n}+(-1)^m$. …

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2. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.$\displaystyle\int_0^5 x(5 - x)^{20}\, dx =$ (A) $\dfrac{5^{20}}{441}$ (B) $\dfrac{5^{21}}{441}$ (C) $\dfrac{5^{21}}{462}$ (D) $\dfrac{5^{22}}{462}$

   ›Reveal solutionSolution

   The substitution $u=5-x$ turns the integral into two elementary power integrals, giving $\dfrac{5^{22}}{462}$.

   Concept and Intuition

   Whenever an integral has a linear factor times a high power of $(a-x)$, the substitution $u=a-x$ swaps the roles so the high power becomes the simple variable, and the linear factor becomes $(a-u)$ — turning the whole thing into two standard power-rule integrals.

   Step-by-Step Solution

   1. Let $u = 5-x$, so $x=5-u$ and $dx=-du$. Limits: $x=0\Rightarrow u=5$; $x=5\Rightarrow u=0$.
   2. $\displaystyle\int_0^5 x(5-x)^{20}dx = \int_5^0 (5-u)u^{20}(-du) = \int_0^5(5-u)u^{20}\,du$.
   3. Split: $= 5\displaystyle\int_0^5 u^{20}\,du - \int_0^5 u^{21}\,du = 5\left[\dfrac{u^{21}}{21}\right]_0^5 - \left[\dfrac{u^{22}}{22}\right]_0^5 = \dfrac{5\cdot5^{21}}{21} - \dfrac{5^{22}}{22} = \dfrac{5^{22}}{21}-\dfrac{5^{22}}{22}$. …

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3. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.$\displaystyle\int_0^{\pi/2} \dfrac{200\sin x + 100\cos x}{\sin x + \cos x}\, dx =$ (A) $50\pi$ (B) $25\pi$ (C) $75\pi$ (D) $150\pi$

   ›Reveal solutionSolution

   Using the King's rule $x\to a-x$ on $[0,\pi/2]$ and adding the two equal-value forms of the integral gives $I=75\pi$.

   Concept and Intuition

   For $\int_0^{a} f(x)dx$, the substitution $x\to a-x$ leaves the integral's value unchanged but can transform the integrand into a different-looking (but equal-valued) expression. Adding the original and transformed integrands often makes the sines and cosines combine into a constant, collapsing the whole problem to $\int$ of a constant.

   Step-by-Step Solution

   1. Let $I = \displaystyle\int_0^{\pi/2}\dfrac{200\sin x+100\cos x}{\sin x+\cos x}dx$.
   2. Substitute $x\to \dfrac{\pi}{2}-x$: $\sin x\to\cos x$, $\cos x\to\sin x$, and the limits/interval are unchanged, so $I = \displaystyle\int_0^{\pi/2}\dfrac{200\cos x+100\sin x}{\cos x+\sin x}dx$ as well.
   3. Add the two expressions for $I$: $2I = \displaystyle\int_0^{\pi/2}\dfrac{(200\sin x+100\cos x)+(200\cos x+100\sin x)}{\sin x+\cos x}dx = \int_0^{\pi/2}\dfrac{300\sin x+300\cos x}{\sin x+\cos x}dx$. …

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4. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.$\displaystyle\int_0^{\pi/2} \dfrac{\sin^3 x}{\sin x + \cos x}\, dx =$ (A) $\dfrac{\pi + 1}{4}$ (B) $\dfrac{\pi - 2}{2}$ (C) $\dfrac{\pi - 2}{4}$ (D) $\dfrac{\pi - 1}{4}$

   ›Reveal solutionSolution

   By symmetry $I=J$ (the $\cos^3$ analogue) and $I+J=\pi/2-1/2$ via a sum-of-cubes factorisation, giving $I=(\pi-1)/4$.

   Concept and Intuition

   Define $J$ as the same integral with $\sin^3x$ replaced by $\cos^3x$. The substitution $x\to\pi/2-x$ shows $I=J$ exactly (not just numerically), because it swaps sine and cosine while leaving $\sin x+\cos x$ and the interval unchanged. Then the sum-of-cubes identity $\sin^3x+\cos^3x = (\sin x+\cos x)(1-\sin x\cos x)$ lets $I+J$ collapse to an elementary integral, and since $I=J$, each equals half of that sum.

   Step-by-Step Solution

   1. Let $I=\displaystyle\int_0^{\pi/2}\dfrac{\sin^3x}{\sin x+\cos x}dx$ and $J=\displaystyle\int_0^{\pi/2}\dfrac{\cos^3x}{\sin x+\cos x}dx$.
   2. Substituting $u=\pi/2-x$ in $I$: $\sin x\to\cos u$, $\cos x\to\sin u$, limits unchanged, giving $I=\displaystyle\int_0^{\pi/2}\dfrac{\cos^3u}{\cos u+\sin u}du = J$. So $I=J$.
   3. $I+J = \displaystyle\int_0^{\pi/2}\dfrac{\sin^3x+\cos^3x}{\sin x+\cos x}dx$. Using $\sin^3x+\cos^3x=(\sin x+\cos x)(\sin^2x-\sin x\cos x+\cos^2x)=(\sin x+\cos x)(1-\sin x\cos x)$, this simplifies to $\displaystyle\int_0^{\pi/2}(1-\sin x\cos x)\,dx$. …

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5. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.$\displaystyle\int_{\frac{\pi}{2}}^{\frac{4051\pi}{2}} \dfrac{\cos^2 2x}{1+\sin 2x}\,dx =$ (A) $2026\pi$ (B) $2047\pi$ (C) $2027\pi$ (D) $2025\pi$

   ›Reveal solutionSolution

   The Pythagorean identity collapses $\cos^2 2x/(1+\sin2x)$ to $1-\sin2x$; integrating this over the given huge range gives $2025\pi$ after the cosine boundary terms cancel.

   Concept and Intuition

   A fraction like $\dfrac{\cos^2\theta}{1+\sin\theta}$ almost always simplifies using $\cos^2\theta = 1-\sin^2\theta = (1-\sin\theta)(1+\sin\theta)$, cancelling the $(1+\sin\theta)$ factor. This turns a seemingly hard rational-trig integral into a trivial polynomial-in-trig integral. The huge integration range is a distractor meant to make direct integration look unpleasant — the simplification removes that entirely.

   Step-by-Step Solution

   1. Simplify the integrand:

   $\frac{\cos^2 2x}{1+\sin 2x} = \frac{1-\sin^2 2x}{1+\sin2x} = \frac{(1-\sin2x)(1+\sin2x)}{1+\sin2x} = 1-\sin2x$

   (this holds wherever $1+\sin2x\neq 0$, which is almost everywhere, so it doesn't affect the definite integral).

   2. Integrate: $\displaystyle\int (1-\sin2x)\,dx = x + \frac{1}{2}\cos2x + C$.

   3. Evaluate at the upper limit $x = \dfrac{4051\pi}{2}$: here $2x = 4051\pi$. Since $4051$ is odd, $\cos(4051\pi) = -1$.

   So the antiderivative's value is $\dfrac{4051\pi}{2} - \dfrac12$.

   4. Evaluate at the lower limit $x=\dfrac{\pi}{2}$: here $2x=\pi$, so $\cos\pi=-1$. …

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6. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.$\displaystyle\int_{-2\pi}^{2\pi} (1+\cos x)^3(1-\cos x)^4\,dx =$ (A) $0$ (B) $5\pi$ (C) $\dfrac{5\pi}{2}$ (D) $\dfrac{5\pi}{4}$

   ›Reveal solutionSolution

   Factoring $(1+\cos x)^3(1-\cos x)^4$ into $\sin^6x(1-\cos x)$ and splitting the integral shows the $\cos x\sin^6x$ part vanishes by symmetry, leaving $4\int_0^\pi\sin^6x\,dx=\frac{5\pi}{4}$.

   Concept and Intuition

   Products of $(1\pm\cos x)$ raised to powers usually simplify via $(1+\cos x)(1-\cos x)=1-\cos^2x=\sin^2x$. Pairing three factors from each gives $\sin^6x$, leaving one leftover $(1-\cos x)$ factor. Splitting that leftover separates the integral into an even, periodic $\sin^6x$ piece and an odd-derivative $\sin^6x\cos x$ piece that integrates to zero over a range where $\sin x$ returns to the same value at both ends.

   Step-by-Step Solution

   1. Rewrite the product:

   $(1+\cos x)^3(1-\cos x)^4 = \big[(1+\cos x)(1-\cos x)\big]^3\cdot(1-\cos x) = (\sin^2x)^3(1-\cos x) = \sin^6x - \sin^6x\cos x$

   2. Split the integral:

   $\int_{-2\pi}^{2\pi}\sin^6x\,dx \;-\; \int_{-2\pi}^{2\pi}\sin^6x\cos x\,dx$

   3. Second integral: $\sin^6x\cos x = \frac{d}{dx}\left(\frac{\sin^7x}{7}\right)$, so

   $\int_{-2\pi}^{2\pi}\sin^6x\cos x\,dx = \left[\frac{\sin^7x}{7}\right]_{-2\pi}^{2\pi} = \frac{0-0}{7} = 0$

   since $\sin(2\pi)=\sin(-2\pi)=0$.

   4. First integral: $\sin^6x$ has period $\pi$ (because $\sin(x+\pi)=-\sin x \Rightarrow \sin^6(x+\pi)=\sin^6x$). The interval $[-2\pi,2\pi]$ has length $4\pi$, i.e. exactly $4$ periods of length $\pi$:

   $\int_{-2\pi}^{2\pi}\sin^6x\,dx = 4\int_0^\pi \sin^6x\,dx$ …

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7. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.$\displaystyle\int_{-2}^{0} \left(x^3+3x^2+3x+3+(x+1)\cos(x+1)\right)\,dx =$ (A) $4$ (B) $3$ (C) $2$ (D) $1$

   ›Reveal solutionSolution

   Shifting by $u=x+1$ centers the integral on $[-1,1]$ and reveals that most of the integrand is odd (vanishes by symmetry), leaving just a constant term to integrate — giving $4$.

   Concept and Intuition

   Integrals over symmetric intervals $[-a,a]$ simplify enormously once you spot odd vs even structure: an odd function integrates to zero over such an interval, so isolating the even (symmetric) part of the integrand is often the fastest route, avoiding any need to actually antidifferentiate terms like $u\cos u$.

   Step-by-Step Solution

   1. Substitute $u=x+1$ (so $x=u-1$, $dx=du$); when $x=-2$, $u=-1$; when $x=0$, $u=1$.
   2. Rewrite $x^3+3x^2+3x+3$ in terms of $u$: since $(x+1)^3=x^3+3x^2+3x+1$, we have $x^3+3x^2+3x+3=(x+1)^3+2=u^3+2$.
   3. The full integrand becomes $u^3+2+u\cos u$, so the integral is $\displaystyle\int_{-1}^{1}\left(u^3+2+u\cos u\right)du$.
   4. $u^3$ is odd, so $\int_{-1}^1 u^3\,du=0$. …

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8. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.$\displaystyle\int_{-3\pi/2}^{-\pi/2} \left((x+\pi)^3 + \cos^2(x+3\pi)\right) dx =$ (A) $\dfrac{\pi}{8}$ (B) $\dfrac{\pi}{2}$ (C) $\dfrac{\pi}{4}-1$ (D) $\dfrac{\pi^4}{32}$

   ›Reveal solutionSolution

   Shifting the variable to $u=x+\pi$ turns the limits into a symmetric interval about zero; the odd cubic term vanishes and only the even $\cos^2u$ term survives, giving $\pi/2$.

   Concept and Intuition

   Whenever an integral's limits and integrand both have a shift-symmetry, substituting to center the interval at 0 lets us exploit odd/even function properties: odd functions integrate to zero over $[-a,a]$, and even functions can be doubled over $[0,a]$.

   Step-by-Step Solution

   1. Let $u = x+\pi$, so $du=dx$. When $x=-3\pi/2$, $u=-\pi/2$; when $x=-\pi/2$, $u=\pi/2$.
   2. $\cos^2(x+3\pi) = \cos^2((x+\pi)+2\pi) = \cos^2(u+2\pi) = \cos^2u$ (cosine has period $2\pi$).
   3. The integral becomes $\displaystyle\int_{-\pi/2}^{\pi/2}\left(u^3+\cos^2u\right)du$.
   4. $u^3$ is an odd function, so $\displaystyle\int_{-\pi/2}^{\pi/2}u^3\,du = 0$.
   5. $\cos^2u$ is even, so $\displaystyle\int_{-\pi/2}^{\pi/2}\cos^2u\,du = 2\int_0^{\pi/2}\cos^2u\,du$. …

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9. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.$\int_{-\pi}^{\pi} \frac{\cos^2 x}{1+a^x}dx$, ($a > 0$) $=$ (A) $a\pi$ (B) $\frac{\pi}{a}$ (C) $\frac{\pi}{2}$ (D) $2\pi$

   ›Reveal solutionSolution

   This is the classic "King's rule" property $\int_{-a}^{a}\frac{f(x)}{1+b^x}dx=\frac12\int_{-a}^{a}f(x)dx$ for even $f$; here it directly gives $I=\pi/2$, independent of $a$.

   Concept and Intuition

   Whenever an even function $f(x)$ is divided by $1+b^x$ and integrated over a symmetric interval $[-a,a]$, replacing $x\to-x$ swaps $1/(1+b^x)$ with $b^x/(1+b^x)$ — and since these two fractions add to exactly $1$, averaging the original and transformed integral removes the exponential entirely, leaving half of $\int_{-a}^a f(x)dx$. This makes the $a^x$ term a red herring: the answer depends only on $\cos^2x$.

   Step-by-Step Solution

   1. Let $I=\displaystyle\int_{-\pi}^{\pi}\frac{\cos^2x}{1+a^x}dx$.
   2. Substitute $x\to -x$ (valid since limits are symmetric): $I=\displaystyle\int_{-\pi}^{\pi}\frac{\cos^2(-x)}{1+a^{-x}}dx=\int_{-\pi}^{\pi}\frac{\cos^2x}{1+a^{-x}}dx$ (using $\cos^2(-x)=\cos^2x$).
   3. Simplify $\dfrac{1}{1+a^{-x}} = \dfrac{a^x}{a^x+1}$, so $I=\displaystyle\int_{-\pi}^\pi\frac{\cos^2x\cdot a^x}{1+a^x}dx$.
   4. Add the two expressions for $I$: $2I=\displaystyle\int_{-\pi}^\pi\cos^2x\left[\frac{1}{1+a^x}+\frac{a^x}{1+a^x}\right]dx=\int_{-\pi}^\pi\cos^2x\,dx$. …

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10. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
    Q.$\displaystyle\int_0^{\pi/2} \dfrac{x\sin 2x}{1 + 4\cos^2 2x}\, dx =$ (A) $\dfrac{\pi}{8}\left(\mathrm{Tan}^{-1} 2\right)$ (B) $\dfrac{\pi}{4}\left(\mathrm{Tan}^{-1} 2\right)$ (C) $\dfrac{\pi}{2}\left(\mathrm{Tan}^{-1} 2\right)$ (D) $\dfrac{\pi}{8}\left(\mathrm{Tan}^{-1} 4\right)$

    ›Reveal solutionSolution

    Use the King's-rule symmetry $\int_0^a f(x)dx=\int_0^a f(a-x)dx$ to eliminate the explicit $x$, then substitute; the result is $\dfrac\pi8\mathrm{Tan}^{-1}2$.

    Concept and Intuition

    Whenever an integrand is $x$ times a function that is itself symmetric (or anti-symmetric) under $x\to a-x$, replacing $x$ by $a-x$ and adding to the original kills the explicit $x$-dependence, leaving a much simpler integral to evaluate.

    Step-by-Step Solution

    1. Let $I=\displaystyle\int_0^{\pi/2}\frac{x\sin2x}{1+4\cos^22x}\,dx$.
    2. Replace $x\to\frac\pi2-x$: since $\sin(\pi-2x)=\sin2x$ and $\cos(\pi-2x)=-\cos2x$ (so $\cos^2$ is unchanged),

    $I=\int_0^{\pi/2}\frac{\left(\frac\pi2-x\right)\sin2x}{1+4\cos^22x}\,dx=\frac\pi2\int_0^{\pi/2}\frac{\sin2x}{1+4\cos^22x}\,dx - I.$

    3. So $2I=\dfrac\pi2 J$ where $J=\displaystyle\int_0^{\pi/2}\frac{\sin2x}{1+4\cos^22x}\,dx$.
    4. Let $w=\cos2x$, $dw=-2\sin2x\,dx$. Limits: $x=0\Rightarrow w=1$; $x=\pi/2\Rightarrow w=-1$.

    $J=\int_{1}^{-1}\frac{-dw/2}{1+4w^2}=\int_{-1}^{1}\frac{dw/2}{1+4w^2}=\int_0^1\frac{dw}{1+4w^2}$

    (using the evenness of the integrand over the symmetric interval). …

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11. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
    Q.$\displaystyle\int_0^{\pi} \dfrac{x\sin x}{1+\cos^2 x}\,dx =$ (A) $\dfrac{\pi^2}{4}$ (B) $\dfrac{\pi}{2}$ (C) $\dfrac{\pi^2}{2}$ (D) $\dfrac{\pi}{4}$

    ›Reveal solutionSolution

    This is the classic $\int_0^\pi x f(\sin x,\cos x)\,dx$ trick using the substitution $x\to \pi-x$; the answer is (A).

    Concept and Intuition

    For integrals of the form $\int_0^\pi x\,g(\cos x)\sin x\,dx$, substituting $x\to\pi-x$ (using $\sin(\pi-x)=\sin x$ and $\cos(\pi-x)=-\cos x$, but here $\cos^2$ is unaffected) converts the $x$ factor into $\pi - x$, letting you solve for the integral algebraically in terms of a simpler integral without the $x$ weight.

    Step-by-Step Solution

    1. Let $I=\displaystyle\int_0^\pi\frac{x\sin x}{1+\cos^2 x}dx$.
    2. Substitute $x\to\pi-x$: $\sin(\pi-x)=\sin x$, $\cos^2(\pi-x)=\cos^2 x$, so

    $I=\int_0^\pi\frac{(\pi-x)\sin x}{1+\cos^2x}dx=\pi\int_0^\pi\frac{\sin x}{1+\cos^2x}dx-I.$

    3. So $2I=\pi\displaystyle\int_0^\pi\frac{\sin x}{1+\cos^2 x}dx$. …

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12. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
    Q.$\displaystyle\int_{0}^{\pi/2} \log|\tan x + \cot x|\,dx =$ (A) $\pi \log 2$ (B) $-\pi \log 2$ (C) $\dfrac{\pi}{2}\log 2$ (D) $2\pi \log 2$

    ›Reveal solutionSolution

    Simplify $\tan x+\cot x=2/\sin2x$, split the log, and use the classical result $\int_0^\pi \log\sin u\,du=-\pi\log2$. Answer: $\pi\log2$.

    Concept and Intuition

    $\tan x + \cot x$ always simplifies to $\dfrac{2}{\sin 2x}$ via the Pythagorean identity — a very common simplification in definite-integral problems. This converts the problem into the well-known integral of $\log \sin$, whose value over a full "half-period" ($0$ to $\pi$) is a standard memorized result, $-\pi\log 2$.

    Step-by-Step Solution

    1. $\tan x + \cot x = \dfrac{\sin x}{\cos x}+\dfrac{\cos x}{\sin x} = \dfrac{\sin^2x+\cos^2x}{\sin x\cos x} = \dfrac{1}{\sin x\cos x} = \dfrac{2}{\sin 2x}$.
    2. So $\log|\tan x+\cot x| = \log 2 - \log(\sin 2x)$ (since $\sin2x>0$ on $(0,\pi/2)$).
    3. $\displaystyle\int_0^{\pi/2}\log|\tan x+\cot x|\,dx = \frac{\pi}{2}\log 2 - \int_0^{\pi/2}\log(\sin2x)\,dx$.
    4. Let $I=\displaystyle\int_0^{\pi/2}\log(\sin2x)\,dx$. Substitute $u=2x,\,du=2dx$: $I=\dfrac12\displaystyle\int_0^{\pi}\log(\sin u)\,du$. …

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