Q.Evaluate $\int_{\pi/6}^{\pi/3} \dfrac{dx}{1 + \sqrt{\tan x}}$

Showing the 12 most recent of 22 on this concept.

1. 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: …

   🔒 Start your 14-day free trial to see the full solution →
2. 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:

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

   🔒 Start your 14-day free trial to see the full solution →
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. …

   🔒 Start your 14-day free trial to see the full solution →
4. 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 $\pi/2$: $\int_{-\pi/2}^{\pi/2} \sin^2 x \cos^3 x \, dx = 2 \int_{0}^{\pi/2} \sin^2 x \cos^3 x \, dx.$ …

   🔒 Start your 14-day free trial to see the full solution →
5. 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 $\int_0^{\pi/2} \cos^{2n} x \, dx = \frac{(2n-1)!!}{(2n)!!} \cdot \frac{\pi}{2}$. …

   🔒 Start your 14-day free trial to see the full solution →
6. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
   Q.If $f$ is defined on $\mathbb{R}$ such that $f(x)f(-x)=9$, then $\int_{-23}^{23} \frac{dx}{3+f(x)}=$ (A) $\frac{51}{3}$ (B) $\frac{49}{3}$ (C) $\frac{46}{3}$ (D) $\frac{46}{6}$

   ›Reveal solutionSolution

   Using the symmetry $f(x)f(-x)=9$ together with the substitution $x\to -x$, the integrand collapses to a constant. The paired integral gives $2I=\tfrac{46}{3}$, so $I=\tfrac{46}{6}$.

   Setting up the symmetry. Let

   $I=\int_{-23}^{23}\frac{dx}{3+f(x)}.$

   Applying the substitution $x\to -x$ (which leaves the limits $-23$ to $23$ unchanged) gives an equal value:

   $I=\int_{-23}^{23}\frac{dx}{3+f(-x)}.$

   Using the given relation. Since $f(x)f(-x)=9$, we have $f(-x)=\dfrac{9}{f(x)}$. Substituting,

   $I=\int_{-23}^{23}\frac{dx}{3+\dfrac{9}{f(x)}}=\int_{-23}^{23}\frac{f(x)\,dx}{3f(x)+9}=\int_{-23}^{23}\frac{f(x)}{3\,[\,3+f(x)\,]}\,dx.$

   Adding the two forms of $I$. …

   🔒 Start your 14-day free trial to see the full solution →
7. 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: $\int_{-1}^{0} -x^2 \, dx = -\left[ \frac{x^3}{3} \right]_{-1}^{0} = -\left(0 - \frac{(-1)^3}{3}\right) = -\left(0 + \frac{1}{3}\right) = -\frac{1}{3}$. …

   🔒 Start your 14-day free trial to see the full solution →
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 $\tfrac{\pi}{2}$:

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

   🔒 Start your 14-day free trial to see the full solution →
9. 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 $0\to\pi/2$: …

   🔒 Start your 14-day free trial to see the full solution →
10. 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

    $$$$

    🔒 Start your 14-day free trial to see the full solution →
11. 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 $$2I = \int_0^1 1\,dx = \bigl[x\bigr]_{0}^{1} = 1. …$$

    🔒 Start your 14-day free trial to see the full solution →
12. 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. …

    🔒 Start your 14-day free trial to see the full solution →