Q.By using the properties of definite integrals, evaluate the integral $\int_{0}^{\pi/2}\cos^2 x\,dx$

Showing the 12 most recent of 45 on this concept.

1. 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 $f(2a-t) = f(t)$. Substituting this into our expression for $I_2$:

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

   Since the variable of integration is a dummy variable, we can replace $t$ with $x$:
   

   $I_2 = \int_{0}^{a} f(x)\,dx$

   This confirms our intuition that the area from $a$ to $2a$ is the same as the area from $0$ to $a$.
   

   5. Combine the results:

   Now, substitute this back into our original split integral from Step 1:

   $\int_{0}^{2a} f(x)\,dx = \int_{0}^{a} f(x)\,dx + I_2$

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

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

   This matches option (D).

   Tip

   This property is often used in competitive exams. Recognizing the form $f(2a-x) = f(x)$ or $f(a-x) = f(x)$ (for an integral from $0$ to $a$) can save significant time. Always look for symmetry in the integrand and limits.

   Watch out

   Do not confuse $f(2a-x) = f(x)$ with $f(-x) = f(x)$ (an even function) or $f(x+T) = f(x)$ (a periodic function). Each property has specific implications for definite integrals over different intervals.

   ---

   ✓Final answer

   Given $f(2a - x) = f(x)$, the value of $\int_{0}^{2a} f(x)\,dx$ is $\boxed{2\int_{0}^{a} f(x)\,dx}$.

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:

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

    Not equal to $1$.
   

   Watch out

   A common mistake is to forget that $|x|$ changes definition at $0$. If you try to integrate $1-|x|$ from $-1$ to $1$ without splitting, you’ll get the wrong sign. Always split at the point where the absolute value expression changes — here at $x=0$.

   Tip

   Notice that option (B) is actually the same as $2\int_{0}^{1} (1-x)\, dx$ after substituting $u = -x$ in the integral. Because if $x$ runs from $-1$ to $0$, then $u = -x$ runs from $1$ to $0$, and $1+x = 1 - u$. So:

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

   which is exactly our symmetric form. So (B) is just a disguised version of the correct expression.

   ✓Final answer

   The correct option is (B).

3. CBSE 2024Set 65/1/11 markMCQ
   Q.$\displaystyle\int_a^b f(x)\, dx$ is equal to: (A) $\displaystyle\int_a^b f(a-x)\, dx$ (B) $\displaystyle\int_a^b f(a+b-x)\, dx$ (C) $\displaystyle\int_a^b f(x-(a+b))\, dx$ (D) $\displaystyle\int_a^b f((a-x)+(b-x))\, dx$

   ›Reveal solutionSolution

   The definite integral $\int_a^b f(x)\, dx$ remains unchanged if we replace $x$ with $a+b-x$. This is a fundamental property of definite integrals, making option (B) the correct choice.

   The question asks us to identify an equivalent expression for the definite integral $\displaystyle\int_a^b f(x)\, dx$. This involves understanding a key property of definite integrals related to symmetry.

   Concept and Intuition: The King Property of Definite Integrals

   One of the most useful properties of definite integrals is that the value of the integral remains the same if we replace the variable of integration, $x$, with $a+b-x$ within the limits $[a, b]$. This is often called the "King Property" or Property 4 in many textbooks.

   Why does this work?

   Imagine the interval of integration is $[a, b]$. The transformation $x \to a+b-x$ effectively reflects the variable $x$ about the midpoint of the interval, which is $\frac{a+b}{2}$.

   • If $x=a$ (the lower limit), then $a+b-x = a+b-a = b$ (the upper limit).
   • If $x=b$ (the upper limit), then $a+b-x = a+b-b = a$ (the lower limit).
   • If $x = \frac{a+b}{2}$ (the midpoint), then $a+b-x = a+b-\frac{a+b}{2} = \frac{a+b}{2}$ (the midpoint itself).

   This reflection means that the "shape" of the function being integrated over the interval, when viewed from $x$ or from $a+b-x$, is essentially the same, just traversed in the opposite direction, which is accounted for by the change in the differential $dx$.

   Let's prove this property using a substitution.

   1. Define the integral:

      Let $I = \displaystyle\int_a^b f(x)\, dx$.

   2. Introduce a substitution:

      We will use the substitution $t = a+b-x$. This is the core idea behind this property.

   3. Change the limits of integration:

      When $x=a$ (the lower limit), $t = a+b-a = b$.

      When $x=b$ (the upper limit), $t = a+b-b = a$.

   4. Change the differential:

      Differentiate the substitution $t = a+b-x$ with respect to $x$:

      $\frac{dt}{dx} = \frac{d}{dx}(a+b-x) = 0+0-1 = -1$.

      So, $dt = -dx$, which means $dx = -dt$.

   5. Substitute into the integral:

      Now, replace $x$ with $a+b-t$, $dx$ with $-dt$, and change the limits from $a$ to $b$ to $b$ to $a$:

      $I = \displaystyle\int_b^a f(a+b-t)\, (-dt)$.

   6. Simplify using properties of definite integrals:

      We know that $\displaystyle\int_p^q -g(t)\, dt = -\displaystyle\int_p^q g(t)\, dt$.

      So, $I = -\displaystyle\int_b^a f(a+b-t)\, dt$.

      Also, we know that $\displaystyle\int_p^q g(t)\, dt = -\displaystyle\int_q^p g(t)\, dt$.

      Applying this, $\displaystyle\int_b^a f(a+b-t)\, dt = -\displaystyle\int_a^b f(a+b-t)\, dt$.

      Substituting this back into the expression for $I$:

      $I = - \left( -\displaystyle\int_a^b f(a+b-t)\, dt \right) = \displaystyle\int_a^b f(a+b-t)\, dt$.

   7. Replace the dummy variable:

      The value of a definite integral does not depend on the variable of integration. We can replace $t$ with $x$ without changing the value:

      $I = \displaystyle\int_a^b f(a+b-x)\, dx$.

      The fundamental property used here is:

      $\int_a^b f(x)\, dx = \int_a^b f(a+b-x)\, dx$

   Comparing this result with the given options:

   (A) $\displaystyle\int_a^b f(a-x)\, dx$

   (B) $\displaystyle\int_a^b f(a+b-x)\, dx$

   (C) $\displaystyle\int_a^b f(x-(a+b))\, dx$

   (D) $\displaystyle\int_a^b f((a-x)+(b-x))\, dx$

   Our derived expression matches option (B).

   ✓Final answer

   The integral $\displaystyle\int_a^b f(x)\, dx$ is equal to $\boxed{\displaystyle\int_a^b f(a+b-x)\, dx}$.

4. 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,

   $\int_{-1}^{1} \frac{x^3}{x^2 + 2|x| + 1}\,dx = 0$

   Tip

   Whenever you see $x^3$ (or any odd power of $x$) in the numerator and even powers or absolute values in the denominator over a symmetric interval, check for odd symmetry first. It saves enormous effort.

   Watch out

   Don't be distracted by the absolute value $|x|$ in the denominator. The key is that $|x|$ is an even function, so the entire denominator $x^2 + 2|x| + 1$ is even. An odd function (numerator) divided by an even function (denominator) is odd.

   ✓Final answer

   The value is $\boxed{0}$, so the correct option is (A).

5. CBSE 2024Set 65/3/11 markMCQ
   Q.$\int_{-a}^{a} f(x)\, dx = 0$, if: (A) $f(-x) = f(x)$ (B) $f(-x) = -f(x)$ (C) $f(a-x) = f(x)$ (D) $f(a-x) = -f(x)$

   ›Reveal solutionSolution

   The integral of a function over a symmetric interval $[-a, a]$ vanishes if and only if the function is odd, i.e., $f(-x) = -f(x)$. The answer is (B).

   Understanding Definite Integral Symmetry

   When we integrate a function over an interval symmetric about the origin, $[-a, a]$, the geometry of the function determines whether contributions from the left and right halves reinforce or cancel each other.

   The key insight is to split the integral at the center of symmetry and examine what happens when we change variables. For any function $f(x)$, we can write:

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

   Now let's see what each option implies.

   Step-by-step analysis

   1. Split and substitute in the left integral In the first integral $\int_{-a}^{0} f(x)\, dx$, substitute $x = -u$. Then $dx = -du$, and when $x = -a$, $u = a$; when $x = 0$, $u = 0$:

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

   Renaming the dummy variable $u \to x$:

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

   2. Combine the two halves Our original integral becomes:

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

   3. When does this equal zero? The integral vanishes if and only if the integrand is identically zero:

   $f(-x) + f(x) = 0 \quad \Longleftrightarrow \quad f(-x) = -f(x)$

   This is precisely the definition of an odd function.

   Tip

   Geometrically, an odd function has rotational symmetry about the origin: if you rotate the graph 180° about $(0,0)$, it looks identical. The area under the curve from $-a$ to $0$ is the negative of the area from $0$ to $a$, so they cancel.

   4. Check the other options
      • (A) $f(-x) = f(x)$ (even function): The integral becomes $\int_{0}^{a} 2f(x)\, dx$, which is generally not zero.
      • (C) and (D) involve $f(a-x)$, which relates values within the interval but doesn't create the necessary cancellation across the origin. These conditions don't guarantee symmetry about $x = 0$.

   Watch out

   Don't confuse even/odd symmetry (about the origin) with symmetry about other points like $x = a/2$. Only odd symmetry about the center of the integration interval guarantees cancellation.

   ✓Final answer

   The correct option is (B): $\int_{-a}^{a} f(x)\, dx = 0$ when $f(-x) = -f(x)$.

6. 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,

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

   Hence $\displaystyle\int_{-\pi/2}^{\pi/2}\sin^7 x\,dx=0.$

   ✓Final answer

   (b) $0$

7. 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 $0$.

   Since $\sin(-x)=-\sin x$, we have $\sin^5(-x)=(-\sin x)^5=-\sin^5 x$, so $\sin^5 x$ is odd. For any odd function $f$, $\int_{-a}^{a}f(x)\,dx=0$. Therefore

   $\int_{-\pi}^{\pi}\sin^5 x\,dx=0$.

   ✓Final answer

   (D) $0$.

8. 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.

   A key property of definite integrals states that for any odd function $g$,

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

   (intuitively, the contributions from $x$ and $-x$ cancel exactly).

   ✓Final answer

   The correct option is (a) $0$.

9. 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 $\cos^4(-x)=\cos^4x$.)

   So $f(x)$ is an odd function.

   Apply the property $\displaystyle\int_{-a}^{a} f(x)\,dx = 0$ whenever $f$ is odd:

   $\int_{-1}^{1} \sin^5x\cos^4x\,dx = 0$

   ✓Final answer

   $0$

10. 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),

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

    so $f$ is odd. By the property $\int_{-a}^{a}f(x)\,dx=0$ for an odd $f$,

    $\int_{-1}^{1}\sin^7x\cos^{13}x\,dx=0.$

    ✓Final answer

    (A) $0$.

11. 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

    $\int_{\alpha}^{\beta}\phi(x)\,dx+\int_{\beta}^{\alpha}\phi(x)\,dx=\int_{\alpha}^{\beta}\phi(x)\,dx-\int_{\alpha}^{\beta}\phi(x)\,dx=0.$

    ✓Final answer

    (C) $0$.

12. 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 $\cos(-x) = \cos x$:

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

    so $f$ is odd. The integral of an odd function over a symmetric interval $[-a,a]$ is $0$:

    $\int_{-1}^{1} \sin^{13}x\cos^{12}x\,dx = 0.$

    ✓Final answer

    (A) $0$.