Q.Evaluate: $\int_0^{\pi/2} [\log (\sin x) - \log (2 \cos x)] dx$

Showing the 12 most recent of 45 on this concept.

1. CBSE 2026Set 65/1/11 markMCQ
   Q.The value of $\displaystyle\int_{-1}^{1} \frac{x^3}{x^2 + 2|x| + 1}\,dx$ is (A) $0$ (B) $\log 2$ (C) $2\log 2$ (D) $\dfrac{1}{2}\log 2$

   ›Reveal solutionSolution

   The integrand is an odd function (satisfies $f(-x) = -f(x)$), so its integral over the symmetric interval $[-1, 1]$ vanishes. The value is $0$.

   When you see an integral over a symmetric interval like $[-1, 1]$, the first instinct should be to check whether the integrand has any special symmetry. Two types matter:

   • Even function: $f(-x) = f(x)$ for all $x$. Then $\int_{-a}^{a} f(x)\,dx = 2\int_{0}^{a} f(x)\,dx$.
   • Odd function: $f(-x) = -f(x)$ for all $x$. Then $\int_{-a}^{a} f(x)\,dx = 0$.

   The second property is powerful: if you can show the integrand is odd, the integral is zero immediately, no computation needed. The geometric reason is that the area under the curve from $-a$ to $0$ exactly cancels the area from $0$ to $a$.

   Let's check our integrand $f(x) = \frac{x^3}{x^2 + 2|x| + 1}$.

   Step-by-step verification

   1. Compute $f(-x)$.

      Substitute $-x$ for $x$:

   $f(-x) = \frac{(-x)^3}{(-x)^2 + 2|-x| + 1} = \frac{-x^3}{x^2 + 2|x| + 1}$

   Notice that $(-x)^3 = -x^3$, $(-x)^2 = x^2$, and crucially $|-x| = |x|$ (the absolute value kills the sign).

   2. Compare with $f(x)$.

      We have

   $f(-x) = \frac{-x^3}{x^2 + 2|x| + 1} = -\frac{x^3}{x^2 + 2|x| + 1} = -f(x)$

   This is exactly the definition of an odd function.

   3. Apply the symmetry property.

      Since $f(x)$ is odd and the interval $[-1, 1]$ is symmetric about the origin, …

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2. CBSE 2026Set 65/3/11 markMCQ
   Q.$\int_{-1}^{1} (1-|x|)\, dx$ is equal to: (A) $2 \int_{0}^{1} (1+x)\, dx$ (B) $2 \int_{-1}^{0} (1+x)\, dx$ (C) $0$ (D) $2 \int_{-1}^{0} (1-x)\, dx$

   ›Reveal solutionSolution

   The integral $\int_{-1}^{1} (1-|x|)\, dx$ evaluates to $1$. Using symmetry, the integrand is even, so the integral equals $2\int_{0}^{1} (1-x)\, dx$, which matches option (B) after a variable substitution.

   The key here is the absolute value function $|x|$. It makes the integrand $1-|x|$ an even function — symmetric about the $y$-axis. For any even function $f(x)$, we have the property:

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

   This is because the area from $-a$ to $0$ is a mirror image of the area from $0$ to $a$. So instead of dealing with the absolute value directly, we can exploit this symmetry to simplify the integral.

   Let’s work through it step by step.

   1. Identify the symmetry. The function $f(x) = 1 - |x|$ is even because $|x|$ is even, and subtracting an even function from a constant keeps it even. So:

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

   But for $x \ge 0$, $|x| = x$. Therefore:

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

   2. Evaluate the integral directly (to know the target value). Compute:

   $2 \int_{0}^{1} (1-x)\, dx = 2 \left[ x - \frac{x^2}{2} \right]_{0}^{1} = 2 \left( 1 - \frac{1}{2} \right) = 2 \cdot \frac{1}{2} = 1$

   So the integral equals $1$. Now we check which option also gives $1$.

   3. Examine each option.
      • (A) $2 \int_{0}^{1} (1+x)\, dx = 2 \left[ x + \frac{x^2}{2} \right]_{0}^{1} = 2 \left( 1 + \frac{1}{2} \right) = 3$ — not equal to $1$.
      • (B) $2 \int_{-1}^{0} (1+x)\, dx$ — let’s evaluate this carefully. For $x$ from $-1$ to $0$, $1+x$ is positive? Actually at $x=-1$, $1+(-1)=0$; at $x=0$, $1+0=1$. So:

   $2 \int_{-1}^{0} (1+x)\, dx = 2 \left[ x + \frac{x^2}{2} \right]_{-1}^{0} = 2 \left( 0 - \left( -1 + \frac{1}{2} \right) \right) = 2 \left( 0 - \left( -\frac{1}{2} \right) \right) = 2 \cdot \frac{1}{2} = 1$

    This matches!
   

   • (C) $0$ — clearly not $1$.
   • (D) $2 \int_{-1}^{0} (1-x)\, dx$ — for $x$ from $-1$ to $0$, $1-x$ ranges from $2$ to $1$, so: …

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3. CBSE 2026Set V11 markMCQ
   Q.The value of $\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^7 x\, dx$

   (a) $1$

   (b) $0$

   (c) $-1$

   (d) $7$

   ›Reveal solutionSolution

   An odd function integrated over symmetric limits gives $0$; answer (b).

   Let $g(x)=\sin^7 x$. Then

   $g(-x)=\sin^7(-x)=(-\sin x)^7=-\sin^7 x=-g(x),$

   so $g$ is odd. For any odd function, …

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4. CBSE 2026Set A1 markMCQ
   Q.$\int_{-\pi}^{\pi}\sin^5 x\,dx =$

   (a) $\frac{3\pi}{4}$

   (b) $2\pi$

   (c) $\frac{5\pi}{6}$

   (d) $0$

   ›Reveal solutionSolution

   Odd function over a symmetric interval integrates to $0$.

   …

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5. CBSE 2026Set ANNUAL1 markMCQ
   Q.Write the value of $\int_{-a}^{a}\{f(x)-f(-x)\}\,dx$.

   (a) 0

   (b) $2f(a)$

   (c) $2f(-a)$

   (d) $2f(0)$

   ›Reveal solutionSolution

   $g(x)=f(x)-f(-x)$ is an odd function, so $\displaystyle\int_{-a}^{a}g(x)\,dx=0$.

   Let $g(x)=f(x)-f(-x)$. Check whether $g$ is odd or even by evaluating $g(-x)$:

   $g(-x)=f(-x)-f(-(-x))=f(-x)-f(x)=-\big[f(x)-f(-x)\big]=-g(x)$

   Since $g(-x)=-g(x)$, $g$ is an odd function.

   …

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6. CBSE 2026Set ANNUAL1 mark
   Q.Evaluate $\displaystyle\int_{-1}^1 \sin^5 x \cos^4 x\, dx$.

   ›Reveal solutionSolution

   Recognise that the integrand is an odd function; the integral of any odd function over a symmetric interval $[-a,a]$ is always zero.

   Let $f(x) = \sin^5x\cos^4x$.

   Test for odd/even symmetry:

   $f(-x) = \sin^5(-x)\cos^4(-x) = (-\sin x)^5(\cos x)^4 = -\sin^5x\cos^4x = -f(x)$

   (using $\sin(-x)=-\sin x$ so $\sin^5(-x)=-\sin^5x$, and $\cos(-x)=\cos x$ so $\cos^4(-x)=\cos^4x$.)

   …

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7. CBSE 2025Set 65/2/11 markMCQ
   Q.If $f(2a - x) = f(x)$, then $\int_{0}^{2a} f(x)\,dx$ is: (A) $\int_{0}^{2a} f\left(\dfrac{x}{2}\right) dx$ (B) $\int_{0}^{a} f(x)\,dx$ (C) $2\int_{a}^{0} f(x)\,dx$ (D) $2\int_{0}^{a} f(x)\,dx$

   ›Reveal solutionSolution

   The condition $f(2a-x) = f(x)$ implies that the function $f(x)$ is symmetric about the line $x=a$, which simplifies the definite integral $\int_{0}^{2a} f(x)\,dx$ to $2\int_{0}^{a} f(x)\,dx$.

   This problem tests your understanding of a fundamental property of definite integrals related to symmetry. The condition $f(2a-x) = f(x)$ is key here. It tells us something profound about the function's behaviour over the interval $[0, 2a]$.

   Concept and Intuition: Symmetry in Definite Integrals

   Imagine the interval $[0, 2a]$ on the x-axis. The midpoint of this interval is $x=a$.

   The condition $f(2a-x) = f(x)$ means that the value of the function at any point $x$ is the same as its value at the point $2a-x$.

   Let's pick a point $x_1$ in the interval $[0, a]$. Its symmetric counterpart with respect to $x=a$ is $x_2 = 2a - x_1$.

   For example, if $a=5$ and $x_1=2$, then $2a-x_1 = 10-2=8$. The condition $f(2)=f(8)$ means the function has the same height at $x=2$ and $x=8$. Both points are $3$ units away from $x=5$.

   This implies that the graph of $f(x)$ is symmetric about the vertical line $x=a$.

   When a function is symmetric about $x=a$ over the interval $[0, 2a]$, the area under the curve from $0$ to $a$ must be exactly equal to the area under the curve from $a$ to $2a$.

   Therefore, the total area from $0$ to $2a$ is simply twice the area from $0$ to $a$. This is the intuition behind the property we are about to derive.

   If $f(2a-x) = f(x)$, then $\int_{0}^{2a} f(x)\,dx = 2\int_{0}^{a} f(x)\,dx$.

   If $f(2a-x) = -f(x)$, then $\int_{0}^{2a} f(x)\,dx = 0$.

   Let's prove this property step-by-step.

   1. Split the integral: We can split the given integral into two parts at the midpoint $a$:

   $\int_{0}^{2a} f(x)\,dx = \int_{0}^{a} f(x)\,dx + \int_{a}^{2a} f(x)\,dx$

   Let's call the second integral $I_2 = \int_{a}^{2a} f(x)\,dx$.
   

   2. Apply substitution to the second integral:

   To make use of the given condition $f(2a-x) = f(x)$, we perform a substitution in $I_2$.

   Let $t = 2a - x$.

   Then, differentiating with respect to $x$, we get $dt = -dx$.

   We also need to change the limits of integration:

   * When $x = a$, $t = 2a - a = a$.

   * When $x = 2a$, $t = 2a - 2a = 0$.

   Substituting these into $I_2$:
   

   $I_2 = \int_{a}^{0} f(2a-t)(-dt)$

   3. Simplify the substituted integral: Using the property $\int_{b}^{a} g(t)\,dt = -\int_{a}^{b} g(t)\,dt$, we can reverse the limits and remove the negative sign:

   $I_2 = -\int_{a}^{0} f(2a-t)\,dt = \int_{0}^{a} f(2a-t)\,dt$

   4. Apply the given condition: We are given that $f(2a-x) = f(x)$. Since $t$ is just a dummy variable, this also means $f(2a-t) = f(t)$. …

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8. CBSE 2025Set E1 markMCQ
   Q.$\int_{-1}^{1} \sin^7 x \cos^{13} x\,dx =$

   (a) $0$

   (b) $1$

   (c) $20$

   (d) $6$

   ›Reveal solutionSolution

   An odd function integrated over a symmetric interval gives $0$.

   Let $f(x)=\sin^7x\cos^{13}x$. Since $\sin^7(-x)=-\sin^7x$ (odd power of an odd function) and $\cos^{13}(-x)=\cos^{13}x$ (even function),

   $f(-x)=-\sin^7x\cos^{13}x=-f(x),$ …

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9. CBSE 2025Set E1 markMCQ
   Q.$\int_{\alpha}^{\beta} \phi(x)\,dx + \int_{\beta}^{\alpha} \phi(x)\,dx =$

   (a) $2$

   (b) $1$

   (c) $0$

   (d) $2\int_{\alpha}^{\beta} \phi(x)\,dx$

   ›Reveal solutionSolution

   By the property $\int_b^a\phi=-\int_a^b\phi$, the two integrals cancel to $0$.

   A basic property of definite integrals is $\int_{\beta}^{\alpha}\phi(x)\,dx=-\int_{\alpha}^{\beta}\phi(x)\,dx$. Therefore …

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10. CBSE 2025Set E1 markMCQ
    Q.$\int_{-1}^{1} \sin^{13}x \cos^{12}x\,dx =$

    (a) $0$

    (b) $1$

    (c) $\frac{1}{2}$

    (d) $2$

    ›Reveal solutionSolution

    $\sin^{13}x\cos^{12}x$ is odd, so its integral over $[-1,1]$ is $0$.

    Let $f(x) = \sin^{13}x\cos^{12}x$. Since $\sin(-x) = -\sin x$ and $\cos(-x) = \cos x$:

    $f(-x) = (-\sin x)^{13}(\cos x)^{12} = -\sin^{13}x\cos^{12}x = -f(x),$ …

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11. CBSE 2025Set E1 markMCQ
    Q.$\int_{-1}^{1} \sin x \cdot \cos^3 x\,dx =$

    (a) $2$

    (b) $1$

    (c) $0$

    (d) $-1$

    ›Reveal solutionSolution

    $\sin x\cos^3 x$ is odd (odd $\times$ even), so its integral over $[-1,1]$ is $0$.

    Check parity: $\sin x$ is odd and $\cos^3 x$ is even, so their product $\sin x\cos^3 x$ is odd. The integral of an odd function over a sym …

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12. CBSE 2025Set ANNUAL1 markMCQ
    Q.$\displaystyle\int_{-a}^{a} f(x)\, dx = 0$ if

    (a) $f(x)$ is an even function

    (b) $f(x)$ is an odd function

    (c) $f(2a-x) = f(x)$

    (d) $f(2a-x) = -f(x)$

    ›Reveal solutionSolution

    This is the standard symmetry property of definite integrals over [-a, a].

    Property: $\displaystyle\int_{-a}^{a} f(x)\,dx = \begin{cases} 2\int_0^a f(x)\,dx & \text{if } f \text{ is even} \\ 0 & \text{if } f \text{ is odd} \end{cases}$

    …

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