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 22 on this concept.

1. 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 $\int_0^{\pi/2} \cos^{2n} x \, dx = \frac{(2n-1)!!}{(2n)!!} \cdot \frac{\pi}{2}$.

      For $n=2$: $\int_0^{\pi/2} \cos^4 x \, dx = \frac{3!!}{4!!} \cdot \frac{\pi}{2} = \frac{3 \cdot 1}{4 \cdot 2} \cdot \frac{\pi}{2} = \frac{3\pi}{16}$.

      So $\int_0^\pi \cos^4 x \, dx = 2 \cdot \frac{3\pi}{16} = \frac{3\pi}{8}$.

      Alternatively, use $\cos^4 x = \frac{3}{8} + \frac{1}{2}\cos 2x + \frac{1}{8}\cos 4x$, integrate from $0$ to $\pi$:

      $\int_0^\pi \frac{3}{8} dx = \frac{3\pi}{8}$, the cosine terms integrate to zero. Same result.

   5. Sum the parts

   $I = \frac{5\pi}{16} + \frac{8}{15} + \frac{3\pi}{8}.$

   Convert $\frac{3\pi}{8}$ to sixteenths: $\frac{3\pi}{8} = \frac{6\pi}{16}$.

   So

   $I = \frac{5\pi}{16} + \frac{6\pi}{16} + \frac{8}{15} = \frac{11\pi}{16} + \frac{8}{15}.$

   Watch out

   A common mistake is forgetting the factor of 2 from the middle term or misapplying the symmetry for $\sin^3 x \cos^2 x$ — always check parity or use substitution to be safe.

   Tip

   For integrals of $\sin^m x \cos^n x$ over $[0,\pi]$, if either exponent is odd, a $u$-substitution often works cleanly. For even powers, reduction formulas or double-angle identities are your friends.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

2. 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. The function $f(\sin x)$ is symmetric about $x = \pi/2$ because $\sin(\pi - x) = \sin x$. Therefore the integral from $0$ to $\pi$ is twice the integral from $0$ to $\pi/2$:

   $\int_{0}^{\pi} f(\sin x) \, dx = 2 \int_{0}^{\pi/2} f(\sin x) \, dx.$

   Substituting back:

   $I = \frac{\pi}{2} \cdot 2 \int_{0}^{\pi/2} f(\sin x) \, dx = \pi \int_{0}^{\pi/2} f(\sin x) \, dx.$

   Watch out

   A common mistake is to forget the factor of $\pi/2$ from the addition step and jump directly to $\pi \int_0^\pi f(\sin x)\,dx$ without halving. Always add the two forms carefully.

   Tip

   The property $\int_0^a x f(x)\,dx = \frac{a}{2} \int_0^a f(x)\,dx$ when $f(a-x)=f(x)$ is a powerful shortcut. Here $a=\pi$ and $f(\sin x)$ satisfies the symmetry, so you can write the answer immediately.

   ✓Final answer

   The correct option is (D): $\pi \int_{0}^{\pi/2} f(\sin x) \, dx$.

3. 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 $\pi/2$:

   $\int_{-\pi/2}^{\pi/2} \sin^2 x \cos^3 x \, dx = 2 \int_{0}^{\pi/2} \sin^2 x \cos^3 x \, dx.$

   Use substitution $u = \sin x$, $du = \cos x \, dx$. Then $\cos^2 x = 1 - u^2$, so $\cos^3 x = (1-u^2)\cos x$. The integral becomes:

   $2 \int_{0}^{1} u^2 (1-u^2) \, du = 2 \int_{0}^{1} (u^2 - u^4) \, du = 2\left[ \frac{u^3}{3} - \frac{u^5}{5} \right]_0^1 = 2\left( \frac{1}{3} - \frac{1}{5} \right) = 2 \cdot \frac{2}{15} = \frac{4}{15}.$

   5. Combine results: Total integral = $0 + \frac{4}{15} = \frac{4}{15}$.

   Watch out

   A common mistake is to think the whole integrand is odd because $\sin x + \cos x$ looks “odd-ish,” but $\cos x$ is even. Always check each term separately.

   Tip

   Even functions can be integrated from $0$ to $a$ and doubled; odd functions vanish. This saves half the work.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

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

   $B\!\left(\frac{5}{2}, \frac{7}{2}\right) = \frac{\frac{3\sqrt{\pi}}{4} \cdot \frac{15\sqrt{\pi}}{8}}{120} = \frac{45\pi}{32 \cdot 120} = \frac{45\pi}{3840} = \frac{3\pi}{256}.$

   4. Combine with the factor from step 1.

   $\int_{0}^{\pi/2} \sin^4 x \cos^6 x \, dx = \frac{1}{2} \cdot \frac{3\pi}{256} = \frac{3\pi}{512}.$

   Then the original integral is

   $\int_{-2\pi}^{2\pi} \sin^4 x \cos^6 x \, dx = 8 \cdot \frac{3\pi}{512} = \frac{24\pi}{512} = \frac{3\pi}{64}.$

   Tip

   A common shortcut: for $\int_0^{2\pi} \sin^{2m}x \cos^{2n}x\,dx$, the result is $2\pi \cdot \frac{(2m-1)!!(2n-1)!!}{(2m+2n)!!}$, but careful with double factorials and the range. Here the range is $4\pi$, so it's double that.

   Watch out

   A classic mistake is forgetting that the period is $\pi$, not $2\pi$, for even powers. If you mistakenly treat the period as $2\pi$, you'd get half the correct value.

   ✓Final answer

   The correct option is (D).

   ANSWER: D

5. 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.

   $I=\int_{0}^{\pi/15}\cos 5x\,dx=\left[\frac{\sin 5x}{5}\right]_0^{\pi/15}=\frac{1}{5}\sin\!\left(\frac{\pi}{3}\right)=\frac{1}{5}\cdot\frac{\sqrt3}{2}=\frac{\sqrt3}{10}.$

   ✓Final answer

   $\displaystyle\int_{-\pi/15}^{\pi/15}\frac{\cos 5x}{1+e^{5x}}\,dx=\frac{\sqrt3}{10}$ — option (B).

6. 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 $0\to\pi/2$:

   $I=\frac14\int_0^{\pi/2}\sin^4u\,du=\frac14\cdot\frac{3\pi}{16}=\frac{3\pi}{64},$

   using the Wallis value $\displaystyle\int_0^{\pi/2}\sin^4u\,du=\frac{3\cdot1}{4\cdot2}\cdot\frac{\pi}{2}=\frac{3\pi}{16}$.

   ✓Final answer

   $\displaystyle\int_{-\pi/8}^{\pi/8}\frac{\sin^4(4x)}{1+e^{4x}}\,dx=\frac{3\pi}{64}$ — option (C).

7. 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:

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

   The denominators are identical, so the sum of numerators is $\sqrt{x} + \sqrt{9-x}$, which cancels the denominator:

   $2I = \int_{3}^{6} 1 \, dx.$

   4. Evaluate the simple integral

   $\int_{3}^{6} 1 \, dx = 6 - 3 = 3.$

   Hence $2I = 3$, so $I = \frac{3}{2}$.

   Tip

   This trick works whenever the integrand is of the form $\frac{f(x)}{f(x)+f(a+b-x)}$ over $[a,b]$. The result is always $\frac{b-a}{2}$, independent of $f$ (as long as $f$ is positive and integrable).

   Watch out

   A common mistake is to try direct substitution or partial fractions — that leads to messy algebra. The symmetry method is far cleaner and avoids any heavy computation.

   ✓Final answer

   The correct option is (B).

   ANSWER: B

8. 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

   $2I = \int_0^1 1\,dx = \bigl[x\bigr]_{0}^{1} = 1.$

   4. Solve for $I$.

   $I = \frac12.$

   Watch out

   A common mistake is to assume $f$ is constant. The condition only forces the sum at symmetric points to be $1$, not that $f$ itself is constant. For example, $f(x)=x$ works because $x+(1-x)=1$, and its integral is indeed $\frac12$. The method above works for any function satisfying the condition.

   Tip

   This is a classic “symmetric sum” trick: whenever you see $f(x)+f(a-x)=c$, the integral over $[0,a]$ is $\frac{c a}{2}$. Here $a=1$, $c=1$, so the answer is $\frac12$ immediately.

   ✓Final answer

   The correct option is (C).

   ANSWER: C

9. 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:

   $\int_0^3\cos\!\left(\tfrac{\pi}{3}x\right)dx=\left[\dfrac{3}{\pi}\sin\!\left(\tfrac{\pi}{3}x\right)\right]_0^3=\dfrac3\pi\big(\sin\pi-\sin0\big)=0.$

   Therefore

   $\int_0^3\left[\sin\!\left(\tfrac{\pi}{3}x\right)-\cos\!\left(\tfrac{\pi}{3}x\right)\right]dx=\dfrac6\pi-0=\dfrac6\pi.$

   ✓Final answer

   $\dfrac{6}{\pi}$ — Option (D).

10. 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 $\tfrac{\pi}{2}$:

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

    With $u=\cos x$, each piece gives $\int_0^1 \sqrt{u}\,du = \left[\tfrac{2}{3}u^{3/2}\right]_0^1 = \tfrac{2}{3}$. Total $= \tfrac{2}{3}+\tfrac{2}{3} = \tfrac{4}{3}$.

    Full integral and result.

    $\int_0^{25\pi}\sqrt{|\cos x|}\,|\sin x|\,dx = 25\cdot\frac{4}{3} = \frac{100}{3}.$

    $\frac{3}{25}\int_0^{25\pi}\sqrt{|\cos x - \cos^3 x|}\,dx = \frac{3}{25}\cdot\frac{100}{3} = 4.$

    ✓Final answer

    The value is $4$ — option (B).

11. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
    Q.$\int_{0}^{2} x^3 (2-x)^4 \, dx =$ (A) $\dfrac{128}{105}$ (B) $\dfrac{16}{35}$ (C) $\dfrac{256}{105}$ (D) $\dfrac{32}{35}$

    ›Reveal solutionSolution

    Substitute $u=2-x$ and expand; the integral evaluates to $\dfrac{32}{35}$.

    Setup. Let $u = 2-x$, so $x = 2-u$ and $dx = -du$. The limits map $x:0\to2$ into $u:2\to0$:

    $\int_{0}^{2} x^3(2-x)^4\,dx = \int_{0}^{2} (2-u)^3\,u^4\,du.$

    Expand $(2-u)^3 = 8 - 12u + 6u^2 - u^3$, so the integrand becomes

    $8u^4 - 12u^5 + 6u^6 - u^7.$

    Integrate term-by-term from $0$ to $2$:

    $\left[\frac{8}{5}u^5 - 2u^6 + \frac{6}{7}u^7 - \frac{1}{8}u^8\right]_0^2.$

    At $u=2$: $\dfrac{8}{5}(32) - 2(64) + \dfrac{6}{7}(128) - \dfrac{1}{8}(256) = \dfrac{256}{5} - 128 + \dfrac{768}{7} - 32.$

    Combining over a common denominator $35$:

    $\frac{1792}{35} - \frac{5600}{35} + \frac{3840}{35} - \frac{1120}{35} = \frac{-1088}{35}+\frac{4832}{35}=\frac{32}{35}.$

    ✓Final answer

    $\displaystyle\int_{0}^{2} x^3(2-x)^4\,dx = \dfrac{32}{35}$ — option (D).

12. 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 $p=\tfrac52,\ q=\tfrac92$:

    $\int_{0}^{\pi/2}\sin^4\theta\cos^8\theta\,d\theta=\tfrac12\,\frac{\Gamma\!\left(\tfrac52\right)\Gamma\!\left(\tfrac92\right)}{\Gamma(7)}.$

    With $\Gamma\!\left(\tfrac52\right)=\tfrac{3\sqrt\pi}{4}$, $\Gamma\!\left(\tfrac92\right)=\tfrac{105\sqrt\pi}{16}$, $\Gamma(7)=720$:

    $\frac{\Gamma\!\left(\tfrac52\right)\Gamma\!\left(\tfrac92\right)}{\Gamma(7)}=\frac{\tfrac{3\sqrt\pi}{4}\cdot\tfrac{105\sqrt\pi}{16}}{720}=\frac{\tfrac{315\pi}{64}}{720}=\frac{7\pi}{1024}.$

    So the integral is $\tfrac12\cdot\tfrac{7\pi}{1024}=\dfrac{7\pi}{2048}$.

    4. Combine.

    $I=8192\cdot\frac{7\pi}{2048}=4\cdot7\pi=28\pi.$

    ✓Final answer

    $\displaystyle\int_{-2}^{2}x^{4}(4-x^2)^{7/2}\,dx=28\pi$. Correct option: (C).

    ANSWER: C