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

Showing the 12 most recent of 22 on this concept.

1. 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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2. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.If $\int_{e^{\cos x}}^{e^{-\cos x}} \frac{\sin x}{e^{\cos x} + e^{-\cos x}} dx = -\frac{1}{2} \left[ f(x) + \log \left( e^{\cos x} + e^{-\cos x} \right) \right] + c$ and $f\left(\frac{\pi}{3}\right) = \frac{1}{2}$, then the number of points at which $f(x)$ attains the maximum value in $[-2\pi, 2\pi]$ is (A) 4 (B) 5 (C) 3 (D) 6

   ›Reveal solutionSolution

   Solving the integral gives $f(x)=\cos x$; it attains its maximum ($=1$) at $x=-2\pi,\,0,\,2\pi$ — 3 points.

   The integrand simplifies as (writing $t=\cos x$, so $\sin x\,dx=-dt$):

   $\frac{e^{\cos x}\sin x}{e^{\cos x}+e^{-\cos x}}=\frac{\sin x}{2}+\frac12\cdot\frac{\sin x\,(e^{\cos x}-e^{-\cos x})}{e^{\cos x}+e^{-\cos x}}.$

   The second term integrates to $-\tfrac12\log\!\left(e^{\cos x}+e^{-\cos x}\right)$ (its derivative is exactly that fraction, up to sign). Matching with the given form

   $-\tfrac12\Big[f(x)+\log\!\left(e^{\cos x}+e^{-\cos x}\right)\Big]+c,$

   the first term forces $-\tfrac12 f'(x)=\tfrac{\sin x}{2}$, i.e. $f'(x)=-\sin x$, so

   $f(x)=\cos x + C.$ …

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3. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.Let $f:\mathbb{R} \to \mathbb{R}$ be an odd function and $\int_{-1}^{1} x^3 f''(x) dx = \frac{8}{5}$. If $g(x) = x f(x)$, $g'(1) = \frac{8}{3}$ and $\int_{0}^{1} g(x) dx = \frac{2}{15}$, then $f(1) =$ (A) $\frac{1}{2}$ (B) $\frac{2}{3}$ (C) $\frac{3}{4}$ (D) $\frac{4}{5}$

   ›Reveal solutionSolution

   Using integration by parts and the given oddness of $f$, we reduce the integral condition to an equation involving $f(1)$ and $g'(1)$, then solve to find $f(1) = \frac{2}{3}$.

   We are told $f$ is odd: $f(-x) = -f(x)$. This implies $f(0)=0$ and $f'$ is even, $f''$ is odd. The function $g(x) = x f(x)$ is even (product of odd and odd), so $g$ is even. That symmetry will simplify integrals.

   We have three pieces of data:

   1. $\int_{-1}^{1} x^3 f''(x) \, dx = \frac{8}{5}$
   2. $g'(1) = \frac{8}{3}$
   3. $\int_{0}^{1} g(x) \, dx = \frac{2}{15}$

   We want $f(1)$.

   ---

   1. Use integration by parts on the first integral. Since $f''$ is odd and $x^3$ is odd, their product is even, so

   $\int_{-1}^{1} x^3 f''(x) \, dx = 2 \int_{0}^{1} x^3 f''(x) \, dx = \frac{8}{5}.$

   Hence

   $\int_{0}^{1} x^3 f''(x) \, dx = \frac{4}{5}.$

   Now integrate by parts: let $u = x^3$, $dv = f''(x) dx$, so $du = 3x^2 dx$, $v = f'(x)$. Then

   $\int_{0}^{1} x^3 f''(x) dx = \left[ x^3 f'(x) \right]_{0}^{1} - \int_{0}^{1} 3x^2 f'(x) dx.$

   At $x=0$, the term vanishes; at $x=1$, we get $1^3 f'(1) = f'(1)$. So

   $\frac{4}{5} = f'(1) - 3 \int_{0}^{1} x^2 f'(x) dx. \qquad(1)$

   2. Relate $g'(1)$ to $f$ and $f'$. Since $g(x) = x f(x)$, differentiate:

   $g'(x) = f(x) + x f'(x).$

   At $x=1$,

   $g'(1) = f(1) + f'(1) = \frac{8}{3}. \qquad(2)$

   3. Use the integral of $g$.

   $\int_{0}^{1} g(x) dx = \int_{0}^{1} x f(x) dx = \frac{2}{15}. \qquad(3)$

   4. Connect the integral in (1) to the known integral (3). Integrate $\int x^2 f'(x) dx$ by parts: let $u = x^2$, $dv = f'(x) dx$, so $du = 2x dx$, $v = f(x)$. Then

   $\int_{0}^{1} x^2 f'(x) dx = \left[ x^2 f(x) \right]_{0}^{1} - \int_{0}^{1} 2x f(x) dx.$

   At $x=0$, term is $0$; at $x=1$, we get $1^2 f(1) = f(1)$. So

   $\int_{0}^{1} x^2 f'(x) dx = f(1) - 2 \int_{0}^{1} x f(x) dx.$

   Using (3), $\int_{0}^{1} x f(x) dx = \frac{2}{15}$, so

   $\int_{0}^{1} x^2 f'(x) dx = f(1) - 2 \cdot \frac{2}{15} = f(1) - \frac{4}{15}. \qquad(4)$

   5. Substitute (4) into (1). From (1): $$$$

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4. 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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5. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.Let $m, n, p, q$ be four positive integers. If $\int_{0}^{2\pi} \sin^m x \cos^n x dx = 4 \int_{0}^{\frac{\pi}{2}} \sin^m x \cos^n x dx$, $\int_{0}^{2\pi} \sin^p x \cos^q x dx = 0$, $\int_{0}^{\pi} \sin^r x \cos^q x dx = 0$, $a = m + n + p$ and $b = m + n + q$, then (A) a is even number and b is odd number (B) a is odd number and b is even number (C) Both a and b are even numbers (D) Both a and b are odd numbers

   ›Reveal solutionSolution

   Both $a$ and $b$ are odd numbers — option (D).

   Analyse each condition by symmetry.

   • $\displaystyle\int_{0}^{2\pi}\sin^{m}x\cos^{n}x\,dx=4\int_{0}^{\pi/2}\sin^{m}x\cos^{n}x\,dx$ requires the integrand to be non-negative with quarter-period symmetry, i.e. $m$ and $n$ are both even.
   • $\displaystyle\int_{0}^{2\pi}\sin^{p}x\cos^{q}x\,dx=0$: under $x\to 2\pi-x$ the integral picks up a factor $(-1)^{p}$, so it vanishes only if $p$ is odd. …

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6. 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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7. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.$\int_{-2}^{2} x^4 (4-x^2)^7 dx =$ (A) $4\pi$ (B) $\dfrac{\pi}{16}$ (C) $28\pi$ (D) $\dfrac{3\pi}{128}$

   ›Reveal solutionSolution

   With the even integrand and the substitution $x=2\sin\theta$, the integral reduces to a Beta function. For the value to be a multiple of $\pi$ the intended power is $(4-x^2)^{7/2}$; evaluating gives $28\pi$. Answer: (C).

   Concept

   An integral of the form $\int_{-2}^{2}x^{4}(4-x^2)^{m}\,dx$ with $m$ a half-integer becomes a Beta/Gamma expression under $x=2\sin\theta$, producing a rational multiple of $\pi$. The choices here ($4\pi,\ \pi/16,\ 28\pi,\ 3\pi/128$) are all multiples of $\pi$, which is only possible when the exponent is $\tfrac{7}{2}$ (a half-integer), so the integrand is read as $x^{4}(4-x^2)^{7/2}$.

   Note

   A strict polynomial power $(4-x^2)^{7}$ would integrate to a rational number with no $\pi$, inconsistent with every option. The exponent is therefore taken as $\tfrac{7}{2}$, which the printed options require; the solution below uses that reading and lands on the exam key.

   Solution

   1. Use symmetry. The integrand is even, so

   $I=\int_{-2}^{2}x^{4}(4-x^2)^{7/2}\,dx=2\int_{0}^{2}x^{4}(4-x^2)^{7/2}\,dx.$

   2. Substitute $x=2\sin\theta$, $dx=2\cos\theta\,d\theta$, with $4-x^2=4\cos^2\theta$:

   $x^{4}=16\sin^4\theta,\qquad (4-x^2)^{7/2}=(4\cos^2\theta)^{7/2}=128\cos^7\theta.$

   Therefore

   $I=2\int_{0}^{\pi/2}16\sin^4\theta\cdot128\cos^7\theta\cdot2\cos\theta\,d\theta=8192\int_{0}^{\pi/2}\sin^4\theta\cos^8\theta\,d\theta.$

   3. Beta function. Using $\int_{0}^{\pi/2}\sin^{2p-1}\theta\cos^{2q-1}\theta\,d\theta=\tfrac12 B(p,q)$ with $p=\tfrac52,\ q=\tfrac92$: …

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8. 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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9. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
   Q.If $m,\, l,\, r,\, s,\, n$ are integers such that, $9 > m > l > s > n > r > 2$ and

   [!FORMULA] $\int_{-\pi}^{\pi} \sin^r x \cos^s x\, dx = 4 \left| \int_{-\pi}^{\pi} \sin^l x \cos^r x\, dx \right|, \quad \int_{0}^{\frac{7\pi}{2}} \sin^r x \cos^s x\, dx = 4 \left| \int_{0}^{\pi} \sin^l x \cos^r x\, dx \right| \text{ and}$

   [!FORMULA] $\int_{-\frac{\pi}{2}}^{\frac{7\pi}{2}} \sin^l x \cos^m x\, dx = 0,$

   then (A) $(s-2)(l-2) = mr$ (B) $(s-2)(l+2) = rm + 5$ (C) $(s-2)(s+2) = ln - 3$ (D) $(l-2)(l+2) = ms - 5$

   ›Reveal solutionSolution

   The chain of inequalities leaves only six integer sets, and the symmetry conditions on the trigonometric integrals fix the parities. Checking the four printed relations, only $(s-2)(s+2)=log-3$ survives — option (C).

   The concept first

   Two symmetry facts do all the work.

   1. An odd power of sine kills a symmetric integral. $\sin x$ is odd, $\cos x$ is even, so $\sin^{p}x\cos^{q}x$ is an odd function when $p$ is odd, and therefore $\int_{-\pi}^{\pi}\sin^{p}x\cos^{q}x\,dx=0$.
   2. An odd power of cosine kills the $[0,\pi]$ integral. Under $x\mapsto\pi-x$, $\sin x$ is unchanged but $\cos x\mapsto-\cos x$; so $\int_{0}^{\pi}\sin^{p}x\cos^{q}x\,dx=0$ when $q$ is odd.

   Combining: on $[-\pi,\pi]$ the integral is non-zero only if both exponents are even, and then

   $\int_{-\pi}^{\pi}\sin^{p}x\cos^{q}x\,dx=2\int_{0}^{\pi}\sin^{p}x\cos^{q}x\,dx=4\int_{0}^{\pi/2}\sin^{p}x\cos^{q}x\,dx .$

   That is exactly the shape of the given equations — the factor $4$ is a symmetry statement, not an accident. The condition that the third integral (taken over a whole number of periods) is $0$ forces one of $l,m$ to be odd.

   Step 1 — List the candidate sets

   All five integers lie strictly between $2$ and $9$, i.e. in $\{3,4,5,6,7,8\}$, and are strictly increasing in the order $r<n<s<l<m$. So we just drop one member of that six-element set:

   $(r,n,s,l,m)\in\{(4,5,6,7,8),(3,5,6,7,8),(3,4,6,7,8),(3,4,5,7,8),(3,4,5,6,8),(3,4,5,6,7)\}.$

   Step 2 — Apply the parity conditions …

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10. 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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11. 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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12. 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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