Q.Evaluate the definite integral: $\int_{-1}^1 (x+1) \ dx$

Showing the 12 most recent of 48 on this concept.

1. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.$\displaystyle\int_{-1}^{1} \left(\sqrt{1+x+x^2} - \sqrt{1-x+x^2}\right) dx =$ (A) $2$ (B) $4$ (C) $0$ (D) $8$

   ›Reveal solutionSolution

   The integrand is an odd function of $x$, so its integral over the symmetric interval $[-1,1]$ vanishes.

   Concept and Intuition

   $\int_{-a}^a f(x)dx=0$ whenever $f$ is odd, i.e. $f(-x)=-f(x)$. Recognising this symmetry avoids a painful direct integration of the square roots.

   Step-by-Step Solution

   1. Let $g(x)=\sqrt{1+x+x^2}$. Then $g(-x)=\sqrt{1-x+x^2}$.
   2. The integrand is $f(x)=g(x)-g(-x)$.
   3. Check parity: $f(-x) = g(-x)-g(x) = -\big(g(x)-g(-x)\big) = -f(x)$ — so $f$ is odd.
   4. Hence $\int_{-1}^1 f(x)\,dx = 0$.

   Common Mistakes

   • Attempting brute-force integration of each square root term (very messy) instead of noticing the odd symmetry.
   • Mis-identifying which term maps to which under $x\to-x$.

   ✓Final answer

   The correct option is (C) — $0$.

   ANSWER: C

2. AP EAPCET 2021Set eng-2021-08-24-AN1 markMCQ
   Q.$\displaystyle\int_{-\pi/4}^{\pi/4} x\tan(1+x^2)\,dx=$ (A) 0 (B) $\dfrac{\pi}{4}$ (C) $\dfrac{-\pi}{4}$ (D) 1

   ›Reveal solutionSolution

   This is a symmetric-limits definite integral where checking odd/even parity instantly gives the answer without doing any actual antiderivative work: it is $0$.

   Concept and Intuition

   For $\displaystyle\int_{-a}^{a} f(x)\,dx$: if $f$ is odd ($f(-x)=-f(x)$) the integral is $0$; if $f$ is even it equals $2\int_0^a f(x)dx$. Any function built as (odd function of $x$) $\times$ (function of $x^2$) is automatically odd, because a function of $x^2$ never changes sign under $x\to -x$.

   Step-by-Step Solution

   1. Let $f(x) = x\tan(1+x^2)$.
   2. Replace $x$ by $-x$: $f(-x) = (-x)\tan(1+(-x)^2) = -x\tan(1+x^2) = -f(x)$.
   3. So $f$ is odd on the symmetric interval $\left[-\dfrac{\pi}{4},\dfrac{\pi}{4}\right]$.
   4. Therefore $\displaystyle\int_{-\pi/4}^{\pi/4} x\tan(1+x^2)\,dx = 0$.

   Common Mistakes

   • Trying to find the actual antiderivative instead of noticing the much faster parity shortcut.
   • Misjudging $1+x^2$ as odd — it is even, since squaring removes the sign, and adding a constant keeps it even.

   ✓Final answer

   The correct option is (A) — $0$.

   ANSWER: A

3. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.$\displaystyle\int_{-\pi}^{\pi} \dfrac{x \sin x}{1 + \cos^2 x}\,dx =$ (A) $\dfrac{3\pi^2}{4}$ (B) $\dfrac{\pi}{2} + 1$ (C) $\dfrac{\pi^2}{4}$ (D) $\dfrac{\pi^2}{2}$

   ›Reveal solutionSolution

   Combine the even-function property over $[-\pi,\pi]$ with the classic $x\to\pi-x$ symmetry trick for integrals of $x\,g(\sin x,\cos x)$. Answer: $\pi^2/2$.

   Concept and Intuition

   First check parity: since $\sin(-x)=-\sin x$ and $\cos^2(-x)=\cos^2x$, the integrand $f(x)=\dfrac{x\sin x}{1+\cos^2x}$ satisfies $f(-x)=f(x)$ — it's even, so the integral over $[-\pi,\pi]$ is twice the integral over $[0,\pi]$. Then, for integrals of $x$ times a function of $\sin x,\cos x$ over $[0,\pi]$, the substitution $x\to \pi-x$ is the standard tool to eliminate the explicit $x$.

   Step-by-Step Solution

   1. Since $f(-x)=f(x)$: $\displaystyle\int_{-\pi}^{\pi} f\,dx = 2\int_0^\pi f\,dx = 2J$, where $J=\displaystyle\int_0^\pi \frac{x\sin x}{1+\cos^2x}dx$.
   2. Substitute $x\to\pi-x$ in $J$: $\sin(\pi-x)=\sin x$, $\cos(\pi-x)=-\cos x$ (so $\cos^2$ unchanged): $J=\int_0^\pi \frac{(\pi-x)\sin x}{1+\cos^2x}dx = \pi\int_0^\pi\frac{\sin x}{1+\cos^2x}dx - J.$
   3. So $2J = \pi K$ where $K=\displaystyle\int_0^\pi \frac{\sin x}{1+\cos^2x}dx$.
   4. Compute $K$ via $c=\cos x$, $dc=-\sin x\,dx$: $K=\displaystyle\int_{-1}^{1}\frac{dc}{1+c^2} = [\tan^{-1}c]_{-1}^1 = \frac\pi4-\left(-\frac\pi4\right)=\frac\pi2$.
   5. So $2J=\pi\cdot\frac\pi2 = \frac{\pi^2}{2} \Rightarrow J=\frac{\pi^2}{4}$.
   6. Final answer: $\displaystyle\int_{-\pi}^\pi f\,dx = 2J = \frac{\pi^2}{2}$.

   Common Mistakes

   • Skipping the even-function step and only computing $J$ over $[0,\pi]$, then forgetting to double it for the full $[-\pi,\pi]$ range.
   • Sign slip when substituting $\cos(\pi-x)=-\cos x$ into $1+\cos^2x$ (squaring removes the sign, which is easy to mishandle).

   ✓Final answer

   The correct option is (D) — $\dfrac{\pi^2}{2}$.

   ANSWER: D

4. AP EAPCET 2025Set eng-2025-05-23-AN1 markMCQ
   Q.$\displaystyle\int_{-\pi/2}^{\pi/2} \sin^2 x \cos^2 x (\sin x + \cos x)\,dx =$ (A) 0 (B) $\dfrac{2}{15}$ (C) $\dfrac{4}{15}$ (D) $\dfrac{2}{5}$

   ›Reveal solutionSolution

   This tests using odd/even symmetry over a symmetric interval to kill half the integral, then a simple $u=\sin x$ substitution for the rest. The answer is $\dfrac{4}{15}$, option (C).

   Concept and Intuition

   Over a symmetric interval $[-a,a]$, an odd integrand integrates to zero and an even integrand integrates to twice the integral over $[0,a]$. Expanding $(\sin x+\cos x)$ splits the problem cleanly into one odd term and one even term, so only the even term survives — turning a seemingly complicated integral into a single elementary substitution.

   Step-by-Step Solution

   1. Expand: $\sin^2x\cos^2x(\sin x+\cos x)=\sin^3x\cos^2x+\sin^2x\cos^3x$.
   2. $g(x)=\sin^3x\cos^2x$: since $\sin^3(-x)=-\sin^3x$ and $\cos^2(-x)=\cos^2x$, $g(-x)=-g(x)$ — odd. So $\int_{-\pi/2}^{\pi/2}g(x)\,dx=0$.
   3. $h(x)=\sin^2x\cos^3x$: both factors are even, so $h$ is even, and $\int_{-\pi/2}^{\pi/2}h\,dx=2\int_0^{\pi/2}h\,dx$.
   4. Write $h(x)=\sin^2x\cos^2x\cos x=\sin^2x(1-\sin^2x)\cos x$. Let $u=\sin x$, $du=\cos x\,dx$; limits $0\to1$.
   5. $\int_0^{\pi/2}h\,dx=\int_0^1u^2(1-u^2)\,du=\left[\dfrac{u^3}{3}-\dfrac{u^5}{5}\right]_0^1=\dfrac13-\dfrac15=\dfrac{2}{15}$.
   6. So the full integral $=2\times\dfrac{2}{15}=\dfrac{4}{15}$.

   Common Mistakes

   • Missing the odd/even split and trying to integrate the whole product directly, which is much more error-prone.
   • Forgetting the factor of 2 from doubling the $[0,\pi/2]$ integral for the even part.

   ✓Final answer

   The correct option is (C) — $\dfrac{4}{15}$.

   ANSWER: C

5. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
   Q.If $A=\int_0^{\infty}\dfrac{1+x^2}{1+x^4}dx$, $B=\int_0^1\dfrac{1+x^2}{1+x^4}dx$, then (A) $2A=B$ (B) $A=B$ (C) $2B=A$ (D) $2B+A=0$

   ›Reveal solutionSolution

   The reciprocal substitution $x=1/t$ maps $[1,\infty)$ onto $[0,1]$ and leaves this particular integrand invariant, so the tail $\int_1^\infty$ equals $B$ itself, giving $A=2B$.

   Concept and Intuition

   For integrands of the form $\dfrac{1+x^2}{1+x^4}$, the substitution $x\to 1/x$ is special: both the numerator/denominator structure and the $dx=-dt/t^2$ factor combine to reproduce the SAME function of the new variable. Recognizing this self-similarity avoids ever computing the (messy) closed form of the integral.

   Step-by-Step Solution

   1. Write $A=\displaystyle\int_0^1\frac{1+x^2}{1+x^4}dx+\int_1^\infty\frac{1+x^2}{1+x^4}dx = B+\int_1^\infty\frac{1+x^2}{1+x^4}dx$.
   2. In the second integral, substitute $x=\dfrac1t$, so $dx=-\dfrac{dt}{t^2}$; as $x:1\to\infty$, $t:1\to0$.
   3. $1+x^2=1+\dfrac1{t^2}=\dfrac{t^2+1}{t^2}$, and $1+x^4=1+\dfrac1{t^4}=\dfrac{t^4+1}{t^4}$, so $\dfrac{1+x^2}{1+x^4}=\dfrac{(t^2+1)t^2}{t^4+1}$.
   4. Multiplying by $dx=-dt/t^2$: the integrand becomes $\dfrac{(t^2+1)t^2}{t^4+1}\cdot\left(-\dfrac{1}{t^2}\right)dt=-\dfrac{t^2+1}{t^4+1}dt$.
   5. So $\displaystyle\int_1^\infty\frac{1+x^2}{1+x^4}dx=\int_{t=1}^{0}-\frac{t^2+1}{t^4+1}dt=\int_0^1\frac{1+t^2}{1+t^4}dt=B$.
   6. Therefore $A=B+B=2B$.

   Common Mistakes

   • Losing track of the sign flips: the substitution flips the limits AND introduces a minus sign, and the two cancel — many students keep only one and get $A=0$ or similar nonsense.
   • Trying to evaluate the improper integral $A$ directly (it can be done, but it's slower and error-prone compared to this symmetry argument).

   ✓Final answer

   The correct option is (C) — $2B=A$.

   ANSWER: C

6. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.$\displaystyle\int_0^{\pi/2} \dfrac{\sin^3 x}{\sin x + \cos x}\, dx =$ (A) $\dfrac{\pi + 1}{4}$ (B) $\dfrac{\pi - 2}{2}$ (C) $\dfrac{\pi - 2}{4}$ (D) $\dfrac{\pi - 1}{4}$

   ›Reveal solutionSolution

   By symmetry $I=J$ (the $\cos^3$ analogue) and $I+J=\pi/2-1/2$ via a sum-of-cubes factorisation, giving $I=(\pi-1)/4$.

   Concept and Intuition

   Define $J$ as the same integral with $\sin^3x$ replaced by $\cos^3x$. The substitution $x\to\pi/2-x$ shows $I=J$ exactly (not just numerically), because it swaps sine and cosine while leaving $\sin x+\cos x$ and the interval unchanged. Then the sum-of-cubes identity $\sin^3x+\cos^3x = (\sin x+\cos x)(1-\sin x\cos x)$ lets $I+J$ collapse to an elementary integral, and since $I=J$, each equals half of that sum.

   Step-by-Step Solution

   1. Let $I=\displaystyle\int_0^{\pi/2}\dfrac{\sin^3x}{\sin x+\cos x}dx$ and $J=\displaystyle\int_0^{\pi/2}\dfrac{\cos^3x}{\sin x+\cos x}dx$.
   2. Substituting $u=\pi/2-x$ in $I$: $\sin x\to\cos u$, $\cos x\to\sin u$, limits unchanged, giving $I=\displaystyle\int_0^{\pi/2}\dfrac{\cos^3u}{\cos u+\sin u}du = J$. So $I=J$.
   3. $I+J = \displaystyle\int_0^{\pi/2}\dfrac{\sin^3x+\cos^3x}{\sin x+\cos x}dx$. Using $\sin^3x+\cos^3x=(\sin x+\cos x)(\sin^2x-\sin x\cos x+\cos^2x)=(\sin x+\cos x)(1-\sin x\cos x)$, this simplifies to $\displaystyle\int_0^{\pi/2}(1-\sin x\cos x)\,dx$.
   4. $\displaystyle\int_0^{\pi/2}1\,dx=\dfrac{\pi}{2}$; $\displaystyle\int_0^{\pi/2}\sin x\cos x\,dx=\dfrac12\displaystyle\int_0^{\pi/2}\sin2x\,dx=\dfrac12\left[-\dfrac{\cos2x}{2}\right]_0^{\pi/2}=\dfrac12\cdot\dfrac{1-(-1)}{2}=\dfrac12$.
   5. So $I+J = \dfrac{\pi}{2}-\dfrac12$.
   6. Since $I=J$: $2I=\dfrac{\pi}{2}-\dfrac12 \Rightarrow I=\dfrac{\pi}{4}-\dfrac14=\dfrac{\pi-1}{4}$.

   Common Mistakes

   • Assuming $I=J$ only "by symmetry of appearance" without justifying it via the actual $x\to\pi/2-x$ substitution.
   • Sign/factor slip evaluating $\int_0^{\pi/2}\sin x\cos x\,dx$ (easy to lose the $\tfrac12$ from the double-angle rewrite).

   ✓Final answer

   The correct option is (D) — $\dfrac{\pi-1}{4}$.

   ANSWER: D

7. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.$\displaystyle\int_{\frac{\pi}{2}}^{\frac{4051\pi}{2}} \dfrac{\cos^2 2x}{1+\sin 2x}\,dx =$ (A) $2026\pi$ (B) $2047\pi$ (C) $2027\pi$ (D) $2025\pi$

   ›Reveal solutionSolution

   The Pythagorean identity collapses $\cos^2 2x/(1+\sin2x)$ to $1-\sin2x$; integrating this over the given huge range gives $2025\pi$ after the cosine boundary terms cancel.

   Concept and Intuition

   A fraction like $\dfrac{\cos^2\theta}{1+\sin\theta}$ almost always simplifies using $\cos^2\theta = 1-\sin^2\theta = (1-\sin\theta)(1+\sin\theta)$, cancelling the $(1+\sin\theta)$ factor. This turns a seemingly hard rational-trig integral into a trivial polynomial-in-trig integral. The huge integration range is a distractor meant to make direct integration look unpleasant — the simplification removes that entirely.

   Step-by-Step Solution

   1. Simplify the integrand:

   $\frac{\cos^2 2x}{1+\sin 2x} = \frac{1-\sin^2 2x}{1+\sin2x} = \frac{(1-\sin2x)(1+\sin2x)}{1+\sin2x} = 1-\sin2x$

   (this holds wherever $1+\sin2x\neq 0$, which is almost everywhere, so it doesn't affect the definite integral).

   2. Integrate: $\displaystyle\int (1-\sin2x)\,dx = x + \frac{1}{2}\cos2x + C$.

   3. Evaluate at the upper limit $x = \dfrac{4051\pi}{2}$: here $2x = 4051\pi$. Since $4051$ is odd, $\cos(4051\pi) = -1$.

   So the antiderivative's value is $\dfrac{4051\pi}{2} - \dfrac12$.

   4. Evaluate at the lower limit $x=\dfrac{\pi}{2}$: here $2x=\pi$, so $\cos\pi=-1$.

   The antiderivative's value is $\dfrac{\pi}{2}-\dfrac12$.

   5. Subtract: $\left(\dfrac{4051\pi}{2}-\dfrac12\right) - \left(\dfrac{\pi}{2}-\dfrac12\right) = \dfrac{4051\pi-\pi}{2} = \dfrac{4050\pi}{2} = 2025\pi$.

   Common Mistakes

   • Attempting to integrate the original rational-trig form directly instead of simplifying with the Pythagorean identity first.
   • Miscounting whether $4051$ is odd (it is), which determines whether $\cos(4051\pi)=-1$ or $+1$ and whether the cosine terms cancel or double.

   ✓Final answer

   The correct option is (D) — $2025\pi$.

   ANSWER: D

8. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.$\displaystyle\int_0^5 x(5 - x)^{20}\, dx =$ (A) $\dfrac{5^{20}}{441}$ (B) $\dfrac{5^{21}}{441}$ (C) $\dfrac{5^{21}}{462}$ (D) $\dfrac{5^{22}}{462}$

   ›Reveal solutionSolution

   The substitution $u=5-x$ turns the integral into two elementary power integrals, giving $\dfrac{5^{22}}{462}$.

   Concept and Intuition

   Whenever an integral has a linear factor times a high power of $(a-x)$, the substitution $u=a-x$ swaps the roles so the high power becomes the simple variable, and the linear factor becomes $(a-u)$ — turning the whole thing into two standard power-rule integrals.

   Step-by-Step Solution

   1. Let $u = 5-x$, so $x=5-u$ and $dx=-du$. Limits: $x=0\Rightarrow u=5$; $x=5\Rightarrow u=0$.
   2. $\displaystyle\int_0^5 x(5-x)^{20}dx = \int_5^0 (5-u)u^{20}(-du) = \int_0^5(5-u)u^{20}\,du$.
   3. Split: $= 5\displaystyle\int_0^5 u^{20}\,du - \int_0^5 u^{21}\,du = 5\left[\dfrac{u^{21}}{21}\right]_0^5 - \left[\dfrac{u^{22}}{22}\right]_0^5 = \dfrac{5\cdot5^{21}}{21} - \dfrac{5^{22}}{22} = \dfrac{5^{22}}{21}-\dfrac{5^{22}}{22}$.
   4. Common denominator: $\dfrac{5^{22}}{21}-\dfrac{5^{22}}{22} = 5^{22}\cdot\dfrac{22-21}{21\times22} = \dfrac{5^{22}}{462}$.

   Common Mistakes

   • Sign error on flipping the limits after substitution (forgetting the extra $-1$ from $dx=-du$, which cancels with the limit flip).
   • Arithmetic slip combining $\tfrac{1}{21}-\tfrac{1}{22}$ or losing a power of 5.

   ✓Final answer

   The correct option is (D) — $\dfrac{5^{22}}{462}$.

   ANSWER: D

9. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.$\displaystyle\int_{-a}^{a} f(x)\,dx - \int_{0}^{a} f(-x)\,dx =$ (A) $\displaystyle\int_{-a}^{a} f(a-x)\,dx$ (B) $\displaystyle\int_{-a}^{a} f(x) + f(a-x)\,dx$ (C) $\displaystyle\int_{0}^{a} f(x) + f(a-x)\,dx$ (D) $\displaystyle\int_{0}^{a} f(a-x)\,dx$

   ›Reveal solutionSolution

   The expression simplifies to $\int_0^a f(x)dx$, which by the standard reflection property equals $\int_0^a f(a-x)dx$.

   Concept and Intuition

   Split the symmetric integral at the origin and use the substitution $x\to -x$ on the negative half to relate it to $\int_0^a f(-x)dx$ — this is exactly the term being subtracted, so it cancels, leaving a plain $\int_0^a f(x)dx$. Then apply the classic property $\int_0^a f(x)dx=\int_0^a f(a-x)dx$ (true for any integrable $f$, by substituting $x\to a-x$) to match it to the given answer choices.

   Step-by-Step Solution

   1. Split: $\displaystyle\int_{-a}^a f(x)dx = \int_{-a}^0 f(x)dx + \int_0^a f(x)dx$.
   2. In $\int_{-a}^0 f(x)dx$, substitute $x=-t$, $dx=-dt$; limits $x=-a\to t=a$, $x=0\to t=0$: $\displaystyle\int_{-a}^0 f(x)dx = \int_a^0 f(-t)(-dt) = \int_0^a f(-t)dt = \int_0^a f(-x)dx$.
   3. So $\displaystyle\int_{-a}^a f(x)dx = \int_0^a f(-x)dx + \int_0^a f(x)dx$.
   4. Subtracting $\int_0^a f(-x)dx$ from both sides (as required by the question) leaves exactly $\displaystyle\int_0^a f(x)dx$.
   5. Now use the property $\displaystyle\int_0^a f(x)dx = \int_0^a f(a-x)dx$ (substitute $x\to a-x$ in either integral — the limits swap and flip sign twice, leaving the same value), which is option (D).

   Common Mistakes

   • Assuming option (C), $\int_0^a[f(x)+f(a-x)]dx$, is the answer — that expression actually equals $2\int_0^af(x)dx$, twice too large.
   • Forgetting the sign flip when substituting $x=-t$ in the lower half of the symmetric integral.

   ✓Final answer

   The correct option is (D) — $\displaystyle\int_{0}^{a} f(a-x)\,dx$.

   ANSWER: D

10. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
    Q.$\displaystyle\int_0^{\pi/4} \log(1 + \tan x)\,dx =$ (A) $\pi \log 2 + 1$ (B) $\dfrac{\pi}{2}\log 2 + 1$ (C) $\dfrac{\pi}{4}\log 2$ (D) $\dfrac{\pi}{8}\log 2$

    ›Reveal solutionSolution

    A textbook symmetry trick ($x \to \pi/4-x$) doubles the integral into something trivial to evaluate. Answer: $\dfrac{\pi}{8}\log2$.

    Concept and Intuition

    Whenever the limits are $0$ to $a$ and the integrand involves $\tan x$, trying the substitution $x \to a-x$ is a classic move — here it converts $\log(1+\tan x)$ into $\log\!\left(\dfrac{2}{1+\tan x}\right)$, and adding the original and transformed integrals collapses most of the complexity.

    Step-by-Step Solution

    1. Let $I=\displaystyle\int_0^{\pi/4}\log(1+\tan x)\,dx$.
    2. Substitute $x\to \dfrac{\pi}{4}-x$: $\tan\!\left(\dfrac{\pi}{4}-x\right)=\dfrac{1-\tan x}{1+\tan x}$, so $1+\tan\!\left(\dfrac\pi4-x\right) = \dfrac{2}{1+\tan x}$.
    3. So $I = \displaystyle\int_0^{\pi/4}\log\!\left(\frac{2}{1+\tan x}\right)dx = \int_0^{\pi/4}\log2\,dx - \int_0^{\pi/4}\log(1+\tan x)\,dx = \frac{\pi}{4}\log2 - I$.
    4. So $2I = \dfrac{\pi}{4}\log2 \Rightarrow I=\dfrac{\pi}{8}\log2$.

    Common Mistakes

    • Forgetting to flip the limits' effect correctly and losing the minus sign on the second integral, which would double $I$ instead of cancelling it.
    • Miscomputing $\tan(\pi/4-x)$ from the tangent subtraction formula.

    ✓Final answer

    The correct option is (D) — $\dfrac{\pi}{8}\log 2$.

    ANSWER: D

11. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
    Q.$\displaystyle\int_{-\pi}^{\pi} \dfrac{2x(1 + \sin x)}{1 + \cos^2 x}\,dx =$ (A) $2\pi$ (B) $\pi^2$ (C) $\pi + 2$ (D) $\pi/2$

    ›Reveal solutionSolution

    Splitting the integrand by parity kills the odd piece immediately; the remaining even piece is handled with the classic "$\int_0^\pi x f(\sin x, \cos x)dx=\frac{\pi}{2}\int_0^\pi f(\sin x,\cos x)dx$"-style symmetry trick, collapsing to a simple arctangent integral.

    Concept and Intuition

    Whenever a definite integral is over a symmetric interval like $[-\pi,\pi]$, always check parity first: odd integrands vanish, and even integrands can be doubled and computed over $[0,\pi]$ only. For integrals of the form $\int_0^\pi x\,g(\sin x,\cos x)\,dx$, the substitution $x\to\pi-x$ often converts the "$x$" factor into "$\pi - x$", letting you solve for the integral in terms of a simpler one without the $x$ multiplier.

    Step-by-Step Solution

    1. Write $\dfrac{2x(1+\sin x)}{1+\cos^2x}=\dfrac{2x}{1+\cos^2x}+\dfrac{2x\sin x}{1+\cos^2x}$.
    2. The first term, $\dfrac{2x}{1+\cos^2x}$, is odd (odd/even = odd), so $\int_{-\pi}^\pi$ of it is $0$.
    3. The second term, $\dfrac{2x\sin x}{1+\cos^2x}$, is even (odd$\times$odd = even, divided by even), so

    $\int_{-\pi}^{\pi}\frac{2x\sin x}{1+\cos^2x}dx = 2\int_0^\pi\frac{2x\sin x}{1+\cos^2x}dx=4\int_0^\pi\frac{x\sin x}{1+\cos^2x}dx.$

    4. Let $J=\displaystyle\int_0^\pi\frac{x\sin x}{1+\cos^2x}dx$. Substituting $x\to\pi-x$ (using $\sin(\pi-x)=\sin x,\ \cos(\pi-x)=-\cos x$, and $\cos^2$ unchanged):

    $J=\int_0^\pi\frac{(\pi-x)\sin x}{1+\cos^2x}dx=\pi\int_0^\pi\frac{\sin x}{1+\cos^2x}dx - J.$

    5. So $2J=\pi\displaystyle\int_0^\pi\frac{\sin x}{1+\cos^2x}dx$. Substitute $u=\cos x,\ du=-\sin x\,dx$: this integral becomes $\displaystyle\int_{-1}^1\frac{du}{1+u^2}=[\arctan u]_{-1}^1=\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)=\frac{\pi}{2}$.
    6. So $2J=\pi\cdot\dfrac{\pi}{2}=\dfrac{\pi^2}{2}\Rightarrow J=\dfrac{\pi^2}{4}$.
    7. The even-part integral is $4J=4\cdot\dfrac{\pi^2}{4}=\pi^2$. Adding the (zero) odd part, the total is $\pi^2$.

    Common Mistakes

    • Forgetting to check parity first and instead attempting direct (much harder) integration.
    • Sign slips in the $x\to\pi-x$ substitution, especially with $\cos(\pi-x)=-\cos x$ but $\cos^2(\pi-x)=\cos^2x$ (unchanged).

    ✓Final answer

    The correct option is (B) — $\pi^2$.

    ANSWER: B

12. AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
    Q.$\int_{\pi/11}^{9\pi/22} \dfrac{dx}{1 + \sqrt{\tan x}} =$ (A) $\pi/4$ (B) $\pi/22$ (C) $\pi/11$ (D) $7\pi/44$

    ›Reveal solutionSolution

    This tests the classic King's-rule substitution $x\to a+b-x$ for definite integrals whose limits sum to $\pi/2$, applied to a $1/(1+\sqrt{\tan x})$ integrand.

    Concept and Intuition

    Whenever the limits of integration sum to a value that makes $\tan x \to \cot x$ under the substitution $x\to a+b-x$ (here $a+b=\pi/2$, so $\tan(\pi/2-x)=\cot x$), adding the original and transformed integrals often produces a simple constant integrand.

    Step-by-Step Solution

    1. Check $a+b$: $\dfrac{\pi}{11}+\dfrac{9\pi}{22} = \dfrac{2\pi}{22}+\dfrac{9\pi}{22}=\dfrac{11\pi}{22}=\dfrac{\pi}{2}$.
    2. Substitute $x\to \dfrac\pi2-x$: $\tan\left(\dfrac\pi2-x\right)=\cot x = \dfrac1{\tan x}$, so the transformed integrand is $\dfrac{1}{1+1/\sqrt{\tan x}}=\dfrac{\sqrt{\tan x}}{1+\sqrt{\tan x}}$.
    3. So $I=\displaystyle\int_a^b\dfrac{\sqrt{\tan x}}{1+\sqrt{\tan x}}dx$ as well (same value as the original, by the substitution property).
    4. Add the original expression for $I$ to this: $2I = \displaystyle\int_a^b\left[\dfrac1{1+\sqrt{\tan x}}+\dfrac{\sqrt{\tan x}}{1+\sqrt{\tan x}}\right]dx = \int_a^b 1\,dx = b-a$.
    5. $b-a = \dfrac{9\pi}{22}-\dfrac{\pi}{11} = \dfrac{9\pi}{22}-\dfrac{2\pi}{22}=\dfrac{7\pi}{22}$.
    6. $I = \dfrac{7\pi}{44}$.

    Common Mistakes

    • Forgetting to check that $a+b=\pi/2$ first (the trick only works because of this specific limit sum).

    ✓Final answer

    The correct option is (D) — $\dfrac{7\pi}{44}$.

    ANSWER: D