Q.Prove that $\int_{-1}^{1}x^{17}\cos^4 x\,dx=0$

Showing the 12 most recent of 22 on this concept.

1. 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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2. 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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3. 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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4. 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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5. 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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6. 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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7. 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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8. 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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9. 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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10. 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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11. TG EAPCET 2023Set eng-2023-05-14-FN1 markMCQ
    Q.$\int_{-1}^{1} x |x| dx =$ (A) $1$ (B) $\dfrac{1}{2}$ (C) $0$ (D) $\dfrac{2}{3}$

    ›Reveal solutionSolution

    The integral of an odd function over a symmetric interval is zero. Since $x|x|$ is odd, the definite integral from $-1$ to $1$ is $0$.

    The key idea here is symmetry. When you integrate an odd function over an interval symmetric about zero, the positive and negative contributions cancel perfectly. The function $f(x) = x|x|$ is odd because $f(-x) = (-x)|-x| = -x|x| = -f(x)$. So instead of doing any messy piecewise integration, we can immediately see the result.

    Let’s verify this step by step to be thorough.

    1. Understand the function

       The absolute value makes the function piecewise:

       • For $x \ge 0$, $|x| = x$, so $x|x| = x \cdot x = x^2$.
       • For $x < 0$, $|x| = -x$, so $x|x| = x \cdot (-x) = -x^2$. So $f(x) = \begin{cases} -x^2, & x < 0 \\ x^2, & x \ge 0 \end{cases}$. This confirms it’s odd: the graph for negative $x$ is the mirror image (with opposite sign) of the graph for positive $x$.

    2. Split the integral at the symmetry point

       Since the function changes definition at $x=0$, we write:

    $\int_{-1}^{1} x|x| \, dx = \int_{-1}^{0} (-x^2) \, dx + \int_{0}^{1} x^2 \, dx.$

    3. Evaluate each piece
       • For the left part: $\int_{-1}^{0} -x^2 \, dx = -\left[ \frac{x^3}{3} \right]_{-1}^{0} = -\left(0 - \frac{(-1)^3}{3}\right) = -\left(0 + \frac{1}{3}\right) = -\frac{1}{3}$. …

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12. TG EAPCET 2022Set eng-2022-07-20-AN1 markMCQ
    Q.If $\int_{-a}^{a} f(x) dx = \int_{0}^{a} f(x) dx + \int_{0}^{a} g(x) dx$ then $g(x) =$ (A) $-f(x)$ (B) $f(x)$ (C) $f(-x)$ (D) $f(x) + f(-x)$

    ›Reveal solutionSolution

    The property of definite integrals over symmetric limits splits into two parts: one from $0$ to $a$ of $f(x)$, and another from $0$ to $a$ of $f(-x)$. So $g(x) = f(-x)$, which is option (C).

    The key idea here is how an integral over a symmetric interval $[-a, a]$ can be broken into two integrals from $0$ to $a$ — one using the original function and the other using a reflected version. This is a standard trick in definite integration, especially when dealing with even and odd functions, but it works for any function.

    Let’s see why this works.

    1. Split the original integral at zero. For any function $f$, we can write:

    $\int_{-a}^{a} f(x) \, dx = \int_{-a}^{0} f(x) \, dx + \int_{0}^{a} f(x) \, dx$

    2. Change variable in the first integral. Let $x = -t$. Then $dx = -dt$, and when $x = -a$, $t = a$; when $x = 0$, $t = 0$. So:

    $\int_{-a}^{0} f(x) \, dx = \int_{a}^{0} f(-t) \, (-dt) = \int_{0}^{a} f(-t) \, dt$

    The limits flip and the minus signs cancel. Since the variable of integration is dummy, we can rename $t$ back to $x$:

    $\int_{-a}^{0} f(x) \, dx = \int_{0}^{a} f(-x) \, dx$

    3. Combine the two pieces. Substituting back:

    $\int_{-a}^{a} f(x) \, dx = \int_{0}^{a} f(-x) \, dx + \int_{0}^{a} f(x) \, dx$

    4. Compare with the given form. The problem states: $\int_{-a}^{a} f(x) \, dx = \int_{0}^{a} f(x) \, dx + \int_{0}^{a} g(x) \, dx$ …

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