Q.Evaluate $\int_0^{\pi/2} \log \sin x\, dx$

Showing the 12 most recent of 22 on this concept.

1. TG EAPCET 2021Set eng-2021-08-06-FN1 markMCQ
   Q.

   [!FORMULA] $\int_{0}^{\pi} x f(\sin x) dx =$

   (A) $2\pi \int_{0}^{\pi/4} f(\sin x) dx$ (B) $\pi \int_{0}^{\pi/4} f(\sin x) dx$ (C) $2\pi \int_{0}^{\pi/2} f(\sin x) dx$ (D) $\pi \int_{0}^{\pi/2} f(\sin x) dx$

   ›Reveal solutionSolution

   Use the property $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$ with $a=\pi$ to rewrite the integral, then add the two forms. The result is $\pi \int_0^{\pi} f(\sin x)\,dx$, which simplifies to $\pi \int_0^{\pi/2} f(\sin x)\,dx$ because $f(\sin x)$ is symmetric about $\pi/2$. The correct option is (D).

   The key idea here is a classic trick for integrals of the form $\int_0^a x\,g(x)\,dx$: replace $x$ by $a-x$ and add the two expressions. This often cancels the $x$ factor and leaves a simpler integral.

   Let $I = \int_{0}^{\pi} x \, f(\sin x) \, dx$. The function $f(\sin x)$ depends on $x$ only through $\sin x$, which has the property $\sin(\pi - x) = \sin x$. That symmetry is what we will exploit.

   1. Apply the substitution $x \to \pi - x$. Let $t = \pi - x$. Then $dx = -dt$, and when $x=0$, $t=\pi$; when $x=\pi$, $t=0$. So

   $I = \int_{0}^{\pi} x \, f(\sin x) \, dx = \int_{\pi}^{0} (\pi - t) \, f(\sin(\pi - t)) \, (-dt) = \int_{0}^{\pi} (\pi - t) \, f(\sin t) \, dt.$

   Since the dummy variable doesn’t matter, rename $t$ back to $x$:

   $I = \int_{0}^{\pi} (\pi - x) \, f(\sin x) \, dx.$

   2. Add the two expressions for $I$. We now have two forms:

   $I = \int_{0}^{\pi} x \, f(\sin x) \, dx \quad \text{and} \quad I = \int_{0}^{\pi} (\pi - x) \, f(\sin x) \, dx.$

   Adding them:

   $2I = \int_{0}^{\pi} \bigl[ x + (\pi - x) \bigr] f(\sin x) \, dx = \int_{0}^{\pi} \pi \, f(\sin x) \, dx.$

   Hence

   $I = \frac{\pi}{2} \int_{0}^{\pi} f(\sin x) \, dx.$

   3. Simplify the limits using symmetry. …

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2. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.$\int_{-2\pi}^{2\pi} \sin^4 x \cos^6 x \, dx =$ (A) $\dfrac{3\pi}{128}$ (B) $\dfrac{9\pi}{32}$ (C) $\dfrac{9\pi}{64}$ (D) $\dfrac{3\pi}{64}$

   ›Reveal solutionSolution

   The integral of an even power of sine and cosine over a full period can be reduced using symmetry and the Beta function; the value is $\frac{3\pi}{64}$, which corresponds to option (D).

   The key insight: the integrand $\sin^4 x \cos^6 x$ is an even function (since both sine and cosine are raised to even powers, the product is symmetric about $x=0$). Also, over $[-2\pi, 2\pi]$, the function repeats its pattern four times (period $\pi$ for the product of even powers). So we can simplify the integral to a multiple of an integral over $[0, \pi/2]$, where the classic Beta-function reduction applies.

   1. Use symmetry and periodicity. The function $f(x) = \sin^4 x \cos^6 x$ has period $\pi$ (because $\sin^2 x$ and $\cos^2 x$ have period $\pi$, and even powers preserve that). Over $[-2\pi, 2\pi]$, which is 4 periods of length $\pi$, we have

   $\int_{-2\pi}^{2\pi} f(x)\,dx = 4 \int_{0}^{\pi} f(x)\,dx.$

   Also, $f(x)$ is even, so $\int_{0}^{\pi} f(x)\,dx = 2\int_{0}^{\pi/2} f(x)\,dx$. Thus

   $\int_{-2\pi}^{2\pi} f(x)\,dx = 8 \int_{0}^{\pi/2} \sin^4 x \cos^6 x \, dx.$

   2. Reduce to a Beta integral. Recall the Beta function identity:

   $\int_{0}^{\pi/2} \sin^{2a-1} x \cos^{2b-1} x \, dx = \frac{1}{2} B(a,b) = \frac{\Gamma(a)\Gamma(b)}{2\,\Gamma(a+b)}.$

   Here we have $\sin^4 x = \sin^{2\cdot 2.5 - 1} x$? Let's match carefully: we want exponents $2a-1 = 4$ and $2b-1 = 6$, so

   $2a-1 = 4 \implies a = \frac{5}{2}, \qquad 2b-1 = 6 \implies b = \frac{7}{2}.$

   Then

   $\int_{0}^{\pi/2} \sin^4 x \cos^6 x \, dx = \frac{1}{2} B\!\left(\frac{5}{2}, \frac{7}{2}\right).$

   3. Evaluate the Beta function using Gamma values.

   $B\!\left(\frac{5}{2}, \frac{7}{2}\right) = \frac{\Gamma(5/2)\,\Gamma(7/2)}{\Gamma(6)}.$

   Recall $\Gamma(n) = (n-1)!$ for integers, and $\Gamma(1/2) = \sqrt{\pi}$.

   $\Gamma(5/2) = \frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi} = \frac{3\sqrt{\pi}}{4},$

   $\Gamma(7/2) = \frac{5}{2} \cdot \frac{3}{2} \cdot \frac{1}{2} \cdot \sqrt{\pi} = \frac{15\sqrt{\pi}}{8},$

   $\Gamma(6) = 5! = 120.$

   So

   $$$$

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3. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
   Q.$\int_{-\pi/15}^{\pi/15} \frac{\cos 5x}{1 + e^{5x}} \, dx =$ (A) $\frac{1}{5}$ (B) $\frac{\sqrt{3}}{10}$ (C) $\frac{1}{15}$ (D) $\frac{1}{10}$

   ›Reveal solutionSolution

   The symmetry trick $\int_{-a}^{a}\frac{f(x)}{1+e^{cx}}dx=\int_0^a f(x)\,dx$ (for even $f$) reduces this to $\int_0^{\pi/15}\cos 5x\,dx=\frac{\sqrt3}{10}$.

   Use the king-property symmetry. Let

   $I=\int_{-\pi/15}^{\pi/15}\frac{\cos 5x}{1+e^{5x}}\,dx.$

   Replacing $x\to -x$ (limits symmetric) and using $\cos(-5x)=\cos 5x$:

   $I=\int_{-\pi/15}^{\pi/15}\frac{\cos 5x\;e^{5x}}{1+e^{5x}}\,dx.$

   Adding the two forms, since $\dfrac{1}{1+e^{5x}}+\dfrac{e^{5x}}{1+e^{5x}}=1$:

   $2I=\int_{-\pi/15}^{\pi/15}\cos 5x\,dx=2\int_{0}^{\pi/15}\cos 5x\,dx.$

   Evaluate. …

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4. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.$\lim_{n \to \infty} \frac{(2n(2n-1)\ldots(n+2)(n+1))^{1/n}}{n} =$ (A) $\int_{0}^{1} \log x \, dx$ (B) $\int_{0}^{1} (x+1)\log(x+1) \, dx$ (C) $\int_{0}^{1} \log(1+x) \, dx$ (D) $\int_{0}^{1} x \log x \, dx$

   ›Reveal solutionSolution

   The product is $\prod_{k=1}^{n}(n+k)$; taking $\tfrac1n$-th power and dividing by $n$ turns the log into a Riemann sum for $\displaystyle\int_0^1\log(1+x)\,dx$. Answer (C).

   Rewrite the product

   $2n(2n-1)\cdots(n+2)(n+1)=\prod_{k=1}^{n}(n+k).$

   So the limit is

   $L=\lim_{n\to\infty}\frac{1}{n}\left(\prod_{k=1}^{n}(n+k)\right)^{1/n} =\lim_{n\to\infty}\left(\prod_{k=1}^{n}\frac{n+k}{n}\right)^{1/n} =\lim_{n\to\infty}\left(\prod_{k=1}^{n}\Bigl(1+\frac{k}{n}\Bigr)\right)^{1/n}.$

   Take logarithms …

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5. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
   Q.$\int_{0}^{3} \left[ \sin\left(\frac{\pi}{3}x\right) - \cos\left(\frac{\pi}{3}x\right) \right] dx =$ (A) $\dfrac{-6}{\pi}$ (B) $0$ (C) $\dfrac{-3}{\pi}$ (D) $\dfrac{6}{\pi}$

   ›Reveal solutionSolution

   Integrating over one full period-related span $[0,3]$: the $\sin$ term contributes $\tfrac6\pi$ and the $\cos$ term contributes $0$, so the integral is $\tfrac6\pi$ (option D).

   $\int_0^3\left[\sin\!\left(\tfrac{\pi}{3}x\right)-\cos\!\left(\tfrac{\pi}{3}x\right)\right]dx.$

   Sine part:

   $\int_0^3\sin\!\left(\tfrac{\pi}{3}x\right)dx=\left[-\dfrac{3}{\pi}\cos\!\left(\tfrac{\pi}{3}x\right)\right]_0^3=-\dfrac{3}{\pi}\big(\cos\pi-\cos0\big)=-\dfrac3\pi(-1-1)=\dfrac6\pi.$

   Cosine part: …

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6. TG EAPCET 2021Set eng-2021-08-04-FN1 markMCQ
   Q.

   [!FORMULA] $\int_{-1}^{1} \frac{\log(1+x)}{1+x^2} dx = \int_{0}^{1} \frac{\log(1+x)}{1+x^2} dx + \int_{0}^{1} f(x) dx \text{ then } f(x) =$

   (A) $\frac{\log(1+x)}{1+x^2}$ (B) $-\frac{\log(1+x)}{1+x^2}$ (C) $\frac{\log(1-x)}{1+x^2}$ (D) $0$

   ›Reveal solutionSolution

   The key idea is to split the integral at $0$ and then use the substitution $x \to -x$ on the negative half to rewrite it as an integral from $0$ to $1$; the function $f(x)$ turns out to be $\frac{\log(1-x)}{1+x^2}$, which is option (C).

   The problem gives you a split of the original integral from $-1$ to $1$ into two parts: one from $-1$ to $0$ and one from $0$ to $1$. The second part is already written as $\int_0^1 \frac{\log(1+x)}{1+x^2} dx$. The first part, $\int_{-1}^0 \frac{\log(1+x)}{1+x^2} dx$, is what needs to be transformed into $\int_0^1 f(x) dx$. So we need to find $f(x)$ such that

   $\int_{-1}^0 \frac{\log(1+x)}{1+x^2} dx = \int_0^1 f(x) dx.$

   The natural way to convert an integral over a negative interval to one over a positive interval is a change of variable that flips the limits. Let’s work through it.

   1. Set up the substitution. On the interval $[-1, 0]$, let $x = -t$. Then when $x = -1$, $t = 1$; when $x = 0$, $t = 0$. Also $dx = -dt$. The integral becomes

   $\int_{-1}^0 \frac{\log(1+x)}{1+x^2} dx = \int_{1}^{0} \frac{\log(1 - t)}{1 + t^2} (-dt) = \int_{0}^{1} \frac{\log(1 - t)}{1 + t^2} dt.$

   The minus sign from $dx = -dt$ flips the limits back to $0$ to $1$, and $x^2 = t^2$ so the denominator is unchanged.

   2. Identify $f(x)$. The variable of integration is a dummy, so rename $t$ back to $x$. We have

   $\int_{-1}^0 \frac{\log(1+x)}{1+x^2} dx = \int_0^1 \frac{\log(1 - x)}{1 + x^2} dx.$

   Therefore, the function $f(x)$ that makes the original equation hold is

   $f(x) = \frac{\log(1 - x)}{1 + x^2}.$ …

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7. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.$\int_{-\pi/8}^{\pi/8} \frac{\sin^4(4x)}{1+e^{4x}} \, dx =$ (A) $\frac{3\pi}{128}$ (B) $\frac{3\pi}{256}$ (C) $\frac{3\pi}{64}$ (D) $\frac{3\pi}{32}$

   ›Reveal solutionSolution

   The $\dfrac{1}{1+e^{4x}}$ symmetry trick reduces the integral to $\displaystyle\int_0^{\pi/8}\sin^4(4x)\,dx=\dfrac{3\pi}{64}$.

   Apply the symmetric-interval identity. For an even function $g$,

   $\int_{-a}^{a}\frac{g(x)}{1+e^{4x}}\,dx=\int_0^{a}g(x)\,dx.$

   This follows from adding $I$ to its $x\to-x$ image: $\dfrac{1}{1+e^{4x}}+\dfrac{1}{1+e^{-4x}}=1$. Here $g(x)=\sin^4(4x)$ is even and $a=\dfrac{\pi}{8}$, so

   $I=\int_0^{\pi/8}\sin^4(4x)\,dx.$

   Evaluate. Substitute $u=4x$, $du=4\,dx$; limits $0\to\pi/2$: …

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8. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
   Q.$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \cos^2 x (\sin x + \cos x) dx =$ (A) $\dfrac{2}{3}$ (B) $\dfrac{3}{10}$ (C) $\dfrac{4}{15}$ (D) $\dfrac{5}{18}$

   ›Reveal solutionSolution

   The integrand is an odd function over a symmetric interval, so the integral evaluates to zero; none of the given positive options match, but the correct answer is 0, which is not listed — the intended answer is (C) only if the problem had a misprint, but strictly the integral is zero.

   The key insight is symmetry. When integrating over $[-\pi/2, \pi/2]$, check if the function is odd or even. An odd function integrated over a symmetric interval always gives zero. Here, $\sin^2 x \cos^2 x$ is even, but $(\sin x + \cos x)$ is a sum of an odd and an even part. The product of an even function with an odd function is odd, and that part integrates to zero. The even part (from $\cos x$) also integrates to zero because of the specific powers? Let’s check carefully.

   1. Separate the integrand:

   $\sin^2 x \cos^2 x (\sin x + \cos x) = \sin^2 x \cos^2 x \sin x + \sin^2 x \cos^2 x \cos x.$

   2. Analyze parity:

      • $\sin^2 x \cos^2 x$ is even because $\sin^2 x$ and $\cos^2 x$ are both even.
      • $\sin x$ is odd, so $\sin^2 x \cos^2 x \cdot \sin x$ is odd.
      • $\cos x$ is even, so $\sin^2 x \cos^2 x \cdot \cos x$ is even.

   3. Integrate the odd part:

      For any odd function $f(x)$, $\int_{-a}^{a} f(x)\,dx = 0$.

      Thus,

   $\int_{-\pi/2}^{\pi/2} \sin^2 x \cos^2 x \sin x \, dx = 0.$

   4. Integrate the even part: The even part is $\sin^2 x \cos^3 x$. Over a symmetric interval, we can double the integral from $0$ to $\pi/2$: $\int_{-\pi/2}^{\pi/2} \sin^2 x \cos^3 x \, dx = 2 \int_{0}^{\pi/2} \sin^2 x \cos^3 x \, dx.$ …

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9. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.$\int\limits_{0}^{\pi} (\sin^3 x + \cos^2 x)^2 dx =$ (A) $\dfrac{15\pi}{16} + \dfrac{8}{15}$ (B) $\dfrac{11\pi}{16} + \dfrac{8}{15}$ (C) $\dfrac{15\pi}{16} + \dfrac{4}{15}$ (D) $\dfrac{11\pi}{16} + \dfrac{4}{15}$

   ›Reveal solutionSolution

   The integral simplifies by expanding the square, using symmetry and reduction formulas; the final value is $\frac{11\pi}{16} + \frac{8}{15}$, which corresponds to option (B).

   We start with

   $I = \int_0^\pi (\sin^3 x + \cos^2 x)^2 \, dx.$

   The key idea: expand the square, then use symmetry and standard trigonometric integrals. The presence of both odd and even powers suggests splitting the interval or using reduction formulas.

   1. Expand the square

   $(\sin^3 x + \cos^2 x)^2 = \sin^6 x + 2\sin^3 x \cos^2 x + \cos^4 x.$

   So

   $I = \int_0^\pi \sin^6 x \, dx + 2\int_0^\pi \sin^3 x \cos^2 x \, dx + \int_0^\pi \cos^4 x \, dx.$

   2. Use symmetry for the middle integral The function $\sin^3 x \cos^2 x$ is odd about $x = \pi/2$? Check: $\sin(\pi - x) = \sin x$, $\cos(\pi - x) = -\cos x$, so $\sin^3(\pi-x)\cos^2(\pi-x) = \sin^3 x \cos^2 x$. Actually it's symmetric, not odd. But better: note that $\sin^3 x \cos^2 x$ is an odd function with respect to $x = \pi/2$? Let's test: replace $x$ by $\pi - x$, the product is unchanged. So it's symmetric. However, we can compute directly using substitution $u = \cos x$:

   $\int_0^\pi \sin^3 x \cos^2 x \, dx = \int_0^\pi \sin x (1-\cos^2 x) \cos^2 x \, dx.$

   Let $u = \cos x$, $du = -\sin x \, dx$, limits: $x=0 \to u=1$, $x=\pi \to u=-1$. Then

   $\int_0^\pi \sin^3 x \cos^2 x \, dx = \int_1^{-1} (1-u^2) u^2 (-du) = \int_{-1}^1 (u^2 - u^4) \, du.$

   The integrand is even, so

   $= 2\int_0^1 (u^2 - u^4) \, du = 2\left[\frac{u^3}{3} - \frac{u^5}{5}\right]_0^1 = 2\left(\frac13 - \frac15\right) = 2\cdot\frac{2}{15} = \frac{4}{15}.$

   Thus

   $2\int_0^\pi \sin^3 x \cos^2 x \, dx = \frac{8}{15}.$

   3. Compute $\int_0^\pi \sin^6 x \, dx$ Use the reduction formula or known result:

   $\int_0^\pi \sin^{2n} x \, dx = \pi \cdot \frac{(2n-1)!!}{(2n)!!}.$

   For $n=3$, $\sin^6 x$ has even power, so

   $\int_0^\pi \sin^6 x \, dx = \pi \cdot \frac{5!!}{6!!} = \pi \cdot \frac{5 \cdot 3 \cdot 1}{6 \cdot 4 \cdot 2} = \pi \cdot \frac{15}{48} = \frac{5\pi}{16}.$

   (Check: $5!! = 15$, $6!! = 48$, yes.)

   4. Compute $\int_0^\pi \cos^4 x \, dx$ Since $\cos^4 x$ is symmetric about $\pi/2$, we can also use $\int_0^\pi \cos^4 x \, dx = 2\int_0^{\pi/2} \cos^4 x \, dx$. Use the reduction formula for $\int_0^{\pi/2} \cos^{2n} x \, dx = \frac{(2n-1)!!}{(2n)!!} \cdot \frac{\pi}{2}$. …

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10. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
    Q.Let $f:[0,1] \to \mathbb{R}$ be a function defined as $f(x) + f(1-x) = 1$. Then $\int_{0}^{1} f(x) dx =$ (A) $0$ (B) $1$ (C) $\dfrac{1}{2}$ (D) $\dfrac{1}{4}$

    ›Reveal solutionSolution

    The functional equation $f(x)+f(1-x)=1$ forces the average value of $f$ over $[0,1]$ to be $\frac12$, so the integral is $\frac12$. The correct option is (C).

    Concept & Intuition

    The given condition $f(x)+f(1-x)=1$ is a symmetry relation: the value at $x$ and the value at its mirror point $1-x$ always sum to $1$. This means the graph of $f$ is symmetric about the point $\left(\frac12,\frac12\right)$. If you average $f$ over the whole interval, the contributions from $x$ and $1-x$ together always give $1$, so the overall average must be $\frac12$. The integral is just the average value times the length of the interval.

    ---

    1. Set up the integral and use the substitution $x \to 1-x$. Let $I = \int_0^1 f(x)\,dx$. Substitute $u = 1-x$, so $du = -dx$ and when $x=0$, $u=1$; when $x=1$, $u=0$. Then

    $I = \int_0^1 f(x)\,dx = \int_1^0 f(1-u)\,(-du) = \int_0^1 f(1-u)\,du.$

    Renaming the dummy variable back to $x$, we have

    $I = \int_0^1 f(1-x)\,dx.$

    2. Add the two expressions for $I$. We now have two representations:

    $I = \int_0^1 f(x)\,dx \quad\text{and}\quad I = \int_0^1 f(1-x)\,dx.$

    Adding them gives

    $2I = \int_0^1 \bigl[f(x) + f(1-x)\bigr]\,dx.$

    3. Use the given functional equation. The condition $f(x)+f(1-x)=1$ holds for every $x\in[0,1]$. Therefore $$2I = \int_0^1 1\,dx = \bigl[x\bigr]_{0}^{1} = 1. …$$

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11. TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
    Q.$\int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x}+\sqrt{x}} \, dx =$ (A) $\frac{1}{2}$ (B) $\frac{3}{2}$ (C) $2$ (D) $1$

    ›Reveal solutionSolution

    This integral is a classic symmetric-invariance problem: using the substitution $x \to 9 - x$ shows the integrand and its complement sum to 1, so the integral over $[3,6]$ is half the interval length, giving $\frac{3}{2}$.

    Concept & Intuition

    When you see an integral of the form $\int_a^b \frac{f(x)}{f(x) + f(a+b-x)} \, dx$, there’s a beautiful trick: the integrand and its “mirror image” add to 1. Here $a=3$, $b=6$, so $a+b=9$. The denominator is $\sqrt{9-x} + \sqrt{x}$, and the numerator is $\sqrt{x}$. If we replace $x$ by $9-x$, the numerator becomes $\sqrt{9-x}$ and the denominator stays the same (just swapped order). So the original integrand $I(x)$ and $I(9-x)$ sum to 1. Integrating over a symmetric interval around the midpoint $x=4.5$ then gives half the length of the interval.

    Step-by-step solution

    1. Define the integral Let

    $I = \int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x} + \sqrt{x}} \, dx.$

    2. Apply the substitution $x \to 9 - u$ Set $u = 9 - x$. Then $dx = -du$, and when $x=3$, $u=6$; when $x=6$, $u=3$. So

    $I = \int_{6}^{3} \frac{\sqrt{9-u}}{\sqrt{u} + \sqrt{9-u}} \, (-du) = \int_{3}^{6} \frac{\sqrt{9-u}}{\sqrt{u} + \sqrt{9-u}} \, du.$

    Since $u$ is a dummy variable, rename it $x$:

    $I = \int_{3}^{6} \frac{\sqrt{9-x}}{\sqrt{x} + \sqrt{9-x}} \, dx.$

    3. Add the two expressions for $I$ We now have two representations:

    $I = \int_{3}^{6} \frac{\sqrt{x}}{\sqrt{9-x} + \sqrt{x}} \, dx \quad \text{and} \quad I = \int_{3}^{6} \frac{\sqrt{9-x}}{\sqrt{x} + \sqrt{9-x}} \, dx.$

    Adding them:

    $$2I = \int_{3}^{6} \left( \frac{\sqrt{x}}{\sqrt{9-x} + \sqrt{x}} + \frac{\sqrt{9-x}}{\sqrt{x} + \sqrt{9-x}} \right) dx. …$$

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12. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
    Q.$\frac{3}{25} \int_{0}^{25\pi} \sqrt{|\cos x - \cos^3 x|} \, dx =$ (A) $8$ (B) $4$ (C) $1$ (D) $0$

    ›Reveal solutionSolution

    $\sqrt{|\cos x - \cos^3 x|} = \sqrt{|\cos x|}\,|\sin x|$, whose integral over one period $\pi$ is $\tfrac{4}{3}$; over $25\pi$ it is $\tfrac{100}{3}$, and $\tfrac{3}{25}\cdot\tfrac{100}{3} = 4$.

    Simplifying the integrand.

    $\cos x - \cos^3 x = \cos x\,(1-\cos^2 x) = \cos x\,\sin^2 x.$

    Since $\sin^2 x \ge 0$,

    $\sqrt{|\cos x - \cos^3 x|} = \sqrt{|\cos x|\,\sin^2 x} = \sqrt{|\cos x|}\,|\sin x|.$

    Periodicity.

    Both $|\cos x|$ and $|\sin x|$ have period $\pi$, so the integrand has period $\pi$. The interval $[0,25\pi]$ contains exactly $25$ periods.

    Integral over one period $[0,\pi]$.

    On $[0,\pi]$, $\sin x \ge 0$. Split where $\cos x$ changes sign at $\tfrac{\pi}{2}$:

    $\int_0^{\pi}\sqrt{|\cos x|}\,\sin x\,dx = \int_0^{\pi/2}\sqrt{\cos x}\,\sin x\,dx + \int_{\pi/2}^{\pi}\sqrt{-\cos x}\,\sin x\,dx.$ …

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