Q.Let $a_1, a_2, \ldots, a_n$ be fixed real numbers and define a function $f(x) = (x - a_1)(x - a_2)\cdots(x - a_n)$. What is $\lim_{x\to a_1} f(x)$? For some $a \neq a_1, a_2, \ldots, a_n$, compute $\lim_{x\to a} f(x)$.

Showing the 12 most recent of 32 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.$\int_{\pi/4}^{3\pi/4} \frac{x \sin x}{1 + \cos 2x} dx =$ (A) $\frac{\pi}{\sqrt{2}}$ (B) $-\frac{\pi}{\sqrt{2}}$ (C) $\sqrt{2\pi}$ (D) $-\sqrt{2\pi}$

   ›Reveal solutionSolution

   Use the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$ to exploit symmetry, then simplify the integrand using $\cos 2x = 2\cos^2 x - 1$. The integral evaluates to $\frac{\pi}{\sqrt{2}}$.

   The key insight here is that the integrand looks messy, but the limits $\pi/4$ to $3\pi/4$ are symmetric about $\pi/2$. Whenever you see an integral with $x$ multiplied by a function of $\sin x$ and $\cos x$ over symmetric limits, the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$ is your best friend. It often lets you replace $x$ with something simpler, like $\pi - x$ in this case, and then add the two forms to cancel the $x$ factor.

   Let’s also simplify the denominator first. $\cos 2x = 2\cos^2 x - 1$, so $1 + \cos 2x = 2\cos^2 x$. That’s clean — the integrand becomes $\frac{x \sin x}{2\cos^2 x} = \frac{x}{2} \cdot \frac{\sin x}{\cos^2 x} = \frac{x}{2} \sec x \tan x$. That derivative form is a hint, but the $x$ outside still makes direct integration tricky. Symmetry to the rescue.

   1. Apply the symmetry property. Let $I = \int_{\pi/4}^{3\pi/4} \frac{x \sin x}{1 + \cos 2x} dx$. Using $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$, here $a+b = \pi/4 + 3\pi/4 = \pi$. So replace $x$ by $\pi - x$:

   $I = \int_{\pi/4}^{3\pi/4} \frac{(\pi - x) \sin(\pi - x)}{1 + \cos 2(\pi - x)} dx.$

   Since $\sin(\pi - x) = \sin x$ and $\cos(2\pi - 2x) = \cos 2x$, the denominator stays the same. Thus:

   $I = \int_{\pi/4}^{3\pi/4} \frac{(\pi - x) \sin x}{1 + \cos 2x} dx.$

   2. Add the two expressions for $I$. We have:

   $I = \int_{\pi/4}^{3\pi/4} \frac{x \sin x}{1 + \cos 2x} dx \quad \text{and} \quad I = \int_{\pi/4}^{3\pi/4} \frac{(\pi - x) \sin x}{1 + \cos 2x} dx.$

   Adding them:

   $2I = \int_{\pi/4}^{3\pi/4} \frac{[x + (\pi - x)] \sin x}{1 + \cos 2x} dx = \int_{\pi/4}^{3\pi/4} \frac{\pi \sin x}{1 + \cos 2x} dx.$

   The $x$ cancels beautifully. So:

   $I = \frac{\pi}{2} \int_{\pi/4}^{3\pi/4} \frac{\sin x}{1 + \cos 2x} dx.$

   3. Simplify the remaining integrand. As noted, $1 + \cos 2x = 2\cos^2 x$, so:

   $\frac{\sin x}{1 + \cos 2x} = \frac{\sin x}{2\cos^2 x} = \frac{1}{2} \sec x \tan x.$

   Therefore:

   $I = \frac{\pi}{2} \int_{\pi/4}^{3\pi/4} \frac{1}{2} \sec x \tan x \, dx = \frac{\pi}{4} \int_{\pi/4}^{3\pi/4} \sec x \tan x \, dx.$

   4. Integrate. …

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2. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.

   [!FORMULA] $\int \frac{\tan 2x}{\cos^4 x} dx =$

   (A) $\tan^2 x - \log \left( 1 - \tan^2 x \right)^2 + c$ (B) $-\tan^2 x - \log \left( 1 - \tan^2 x \right)^2 + c$ (C) $-\tan^2 x + \log \left( 1 - \tan^2 x \right)^2 + c$ (D) $\tan^2 x + \log \left( 1 - \tan^2 x \right)^2 + c$

   ›Reveal solutionSolution

   The key is to rewrite $\tan 2x$ in terms of $\tan x$ and $\cos^4 x$ in terms of $\sec^4 x$, then substitute $t = \tan x$. The integral simplifies to $-\tan^2 x - \log(1 - \tan^2 x)^2 + c$, which matches option (B).

   The problem asks for an indefinite integral, and the options are all expressed in terms of $\tan x$ and a logarithm. That’s a strong hint: the integrand should be rewritten so that a substitution $t = \tan x$ becomes natural. Let’s see why that works.

   We have $\tan 2x = \frac{2\tan x}{1 - \tan^2 x}$, and $\cos^4 x = (\cos^2 x)^2 = \frac{1}{\sec^4 x} = \frac{1}{(1 + \tan^2 x)^2}$. So the integrand becomes a rational function of $\tan x$ times $dx$. And since $d(\tan x) = \sec^2 x \, dx$, we need to account for that factor — but here we have $\frac{1}{\cos^4 x} = \sec^4 x$, which is $\sec^2 x \cdot \sec^2 x$. One $\sec^2 x$ will combine with $dx$ to give $dt$, and the other remains as $1 + t^2$. Perfect.

   1. Rewrite the integrand in terms of $\tan x$.

   $\tan 2x = \frac{2\tan x}{1 - \tan^2 x}, \quad \frac{1}{\cos^4 x} = \sec^4 x = (1 + \tan^2 x)^2.$

   So

   $\frac{\tan 2x}{\cos^4 x} = \frac{2\tan x}{1 - \tan^2 x} \cdot (1 + \tan^2 x)^2.$

   2. Substitute $t = \tan x$. Then $dt = \sec^2 x \, dx = (1 + t^2) dx$, so $dx = \frac{dt}{1 + t^2}$. The integral becomes

   $\int \frac{2t}{1 - t^2} \cdot (1 + t^2)^2 \cdot \frac{dt}{1 + t^2} = \int \frac{2t}{1 - t^2} \cdot (1 + t^2) \, dt.$

   3. Simplify the integrand.

   $\frac{2t(1 + t^2)}{1 - t^2} = \frac{2t + 2t^3}{1 - t^2}.$

   Perform polynomial division: divide $2t + 2t^3$ by $1 - t^2$.

   $\frac{2t + 2t^3}{1 - t^2} = -2t + \frac{4t}{1 - t^2}.$

   (Check: $-2t(1 - t^2) = -2t + 2t^3$, subtract from $2t + 2t^3$ gives $4t$, so remainder $4t$ over $1 - t^2$.)

   4. Integrate term by term. $\int \left( -2t + \frac{4t}{1 - t^2} \right) dt = -2 \int t \, dt + 4 \int \frac{t}{1 - t^2} \, dt.$ …

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3. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.

   [!FORMULA] $\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \frac{x + \frac{\pi}{2}}{2 - \sin^2 x} dx =$

   (A) $\frac{\pi^2}{6\sqrt{3}}$ (B) $\frac{\pi}{\sqrt{2}} \tan^{-1} \left( \frac{\sqrt{3}}{\sqrt{2}} \right)$ (C) $\frac{\pi^2}{3\sqrt{2}}$ (D) $\frac{\pi}{2} \tan^{-1} \left( \frac{\sqrt{3}}{2} \right)$

   ›Reveal solutionSolution

   Use the property $\int_{-a}^{a} f(x)\,dx = \int_{0}^{a} [f(x)+f(-x)]\,dx$ to simplify the integrand; the odd part vanishes and the even part reduces to a standard integral, yielding $\frac{\pi^2}{6\sqrt{3}}$.

   The key insight is symmetry. The limits are symmetric about zero, so we can split the integrand into even and odd parts. The denominator $2 - \sin^2 x$ is even because $\sin^2 x$ is even. The numerator $x + \frac{\pi}{2}$ is the sum of an odd part ($x$) and an even part ($\frac{\pi}{2}$). Over a symmetric interval, the odd part integrates to zero, leaving only the even part to handle.

   1. Write the integral as

   $I = \int_{-\pi/3}^{\pi/3} \frac{x + \frac{\pi}{2}}{2 - \sin^2 x}\,dx.$

   Use the property for symmetric limits:

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

   Here $f(x) = \frac{x + \pi/2}{2 - \sin^2 x}$, so

   $f(x) + f(-x) = \frac{x + \pi/2}{2 - \sin^2 x} + \frac{-x + \pi/2}{2 - \sin^2 x} = \frac{\pi}{2 - \sin^2 x}.$

   The $x$ terms cancel because $\sin^2(-x) = \sin^2 x$. Therefore

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

   2. Now we need to evaluate $J = \int_{0}^{\pi/3} \frac{dx}{2 - \sin^2 x}$. A standard technique for integrals of rational functions of $\sin^2 x$ is to divide numerator and denominator by $\cos^2 x$, turning the integrand into a function of $\tan x$. Since $\cos^2 x > 0$ on $[0, \pi/3]$, this is safe.

   $\frac{1}{2 - \sin^2 x} = \frac{1/\cos^2 x}{2/\cos^2 x - \tan^2 x} = \frac{\sec^2 x}{2(1+\tan^2 x) - \tan^2 x} = \frac{\sec^2 x}{2 + \tan^2 x}.$

   So

   $J = \int_{0}^{\pi/3} \frac{\sec^2 x}{2 + \tan^2 x}\,dx.$

   3. Substitute $t = \tan x$, so $dt = \sec^2 x\,dx$. When $x=0$, $t=0$; when $x=\pi/3$, $t = \tan(\pi/3) = \sqrt{3}$. Then

   $J = \int_{0}^{\sqrt{3}} \frac{dt}{2 + t^2}.$

   This is a standard form: $\int \frac{dt}{a^2 + t^2} = \frac{1}{a} \tan^{-1}\left(\frac{t}{a}\right) + C$. Here $a = \sqrt{2}$, so

   $$$$

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4. TG EAPCET 2026Set eng-2026-05-11-AN1 markMCQ
   Q.If in $(0, 2\pi)$, $2\cos x + k = 3\sec x$ and $\cot x = -1$, then the sum of squares of all possible values of $k$ is (A) $0$ (B) $4\sqrt{2}$ (C) $8$ (D) $16$

   ›Reveal solutionSolution

   $\cot x=-1$ gives $x=\tfrac{3\pi}{4},\tfrac{7\pi}{4}$; solving $2\cos x+k=3\sec x$ yields $k=\mp2\sqrt2$, so $\sum k^2=8+8=16$.

   In $(0,2\pi)$, $\cot x=-1\Rightarrow x=\dfrac{3\pi}{4}$ or $x=\dfrac{7\pi}{4}$.

   At $x=\tfrac{3\pi}{4}$: $\cos x=-\tfrac{1}{\sqrt2},\ \sec x=-\sqrt2$.

   $2\left(-\tfrac{1}{\sqrt2}\right)+k=3(-\sqrt2)\Rightarrow -\sqrt2+k=-3\sqrt2\Rightarrow k=-2\sqrt2,\quad k^2=8.$ …

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5. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.$\int_{0}^{\pi/4} x^2 \sin 2x dx =$ (A) $\dfrac{\pi^2 - 2}{8}$ (B) $\dfrac{\pi(\pi - 2)}{8}$ (C) $\dfrac{\pi - 2}{8}$ (D) $\dfrac{\pi + 2}{8}$

   ›Reveal solutionSolution

   This definite integral is solved by integration by parts twice, yielding the value $\frac{\pi^2 - 2}{8}$, which corresponds to option (A).

   We are evaluating

   $\int_{0}^{\pi/4} x^2 \sin 2x \, dx.$

   The integrand is a product of a polynomial ($x^2$) and a trigonometric function ($\sin 2x$). Integration by parts is the natural choice: it lets us reduce the power of $x$ step by step until only a simple trigonometric integral remains.

   ---

   1. First integration by parts

   Let

   $u = x^2, \quad dv = \sin 2x \, dx.$

   Then

   $du = 2x \, dx, \quad v = -\frac{1}{2} \cos 2x.$

   Integration by parts gives

   $\int x^2 \sin 2x \, dx = -\frac{1}{2} x^2 \cos 2x - \int \left(-\frac{1}{2} \cos 2x\right) (2x \, dx) = -\frac{1}{2} x^2 \cos 2x + \int x \cos 2x \, dx.$

   ---

   2. Second integration by parts

   Now we need $\int x \cos 2x \, dx$. Set

   $u = x, \quad dv = \cos 2x \, dx.$

   Then

   $du = dx, \quad v = \frac{1}{2} \sin 2x.$

   So

   $\int x \cos 2x \, dx = \frac{1}{2} x \sin 2x - \int \frac{1}{2} \sin 2x \, dx = \frac{1}{2} x \sin 2x + \frac{1}{4} \cos 2x.$

   ---

   3. Combine the results

   Substituting back,

   $\int x^2 \sin 2x \, dx = -\frac{1}{2} x^2 \cos 2x + \frac{1}{2} x \sin 2x + \frac{1}{4} \cos 2x + C.$

   ---

   4. Evaluate the definite integral from $0$ to $\pi/4$

   $\left[ -\frac{1}{2} x^2 \cos 2x + \frac{1}{2} x \sin 2x + \frac{1}{4} \cos 2x \right]_{0}^{\pi/4}.$

   At $x = \pi/4$:

   $\cos(2 \cdot \pi/4) = \cos(\pi/2) = 0$,

   $\sin(2 \cdot \pi/4) = \sin(\pi/2) = 1$.

   So the value is

   $-\frac{1}{2} \left(\frac{\pi}{4}\right)^2 \cdot 0 + \frac{1}{2} \cdot \frac{\pi}{4} \cdot 1 + \frac{1}{4} \cdot 0 = \frac{\pi}{8}.$

   At $x = 0$:

   $\cos 0 = 1$, $\sin 0 = 0$.

   So the value is

   $-\frac{1}{2} \cdot 0^2 \cdot 1 + \frac{1}{2} \cdot 0 \cdot 0 + \frac{1}{4} \cdot 1 = \frac{1}{4}.$

   Thus the definite integral is

   $\frac{\pi}{8} - \frac{1}{4} = \frac{\pi}{8} - \frac{2}{8} = \frac{\pi - 2}{8}.$ …

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6. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.If $\dfrac{5\pi}{4} < x < \dfrac{7\pi}{4}$, then $\int \dfrac{\sqrt{1-\sin 2x}}{\sqrt{1+\sin 2x}} dx =$ (A) $-\sec^2\left(\dfrac{\pi}{4} - x\right) + c$ (B) $-\log\left|\sec\left(\dfrac{\pi}{4} - x\right)\right| + c$ (C) $\sec^2\left(\dfrac{\pi}{4} - x\right) + c$ (D) $\log\left|\sec\left(\dfrac{\pi}{4} - x\right)\right| + c$

   ›Reveal solutionSolution

   On $\left(\tfrac{5\pi}{4},\tfrac{7\pi}{4}\right)$ the integrand reduces to $\tan\!\left(x-\tfrac{\pi}{4}\right)$, whose integral is $\log\left|\sec\!\left(\tfrac{\pi}{4}-x\right)\right|+c$. Answer: (D).

   Factor the surds. Since $1\mp\sin 2x=(\cos x\mp\sin x)^2$,

   $\sqrt{\frac{1-\sin 2x}{1+\sin 2x}}=\frac{|\cos x-\sin x|}{|\cos x+\sin x|}.$

   Signs on the interval. For $x\in\left(\tfrac{5\pi}{4},\tfrac{7\pi}{4}\right)$: $\cos x+\sin x=\sqrt2\sin\!\left(x+\tfrac{\pi}{4}\right)<0$, while $\cos x-\sin x=\sqrt2\cos\!\left(x+\tfrac{\pi}{4}\right)>0$. Hence

   $\frac{|\cos x-\sin x|}{|\cos x+\sin x|}=\frac{\cos x-\sin x}{-(\cos x+\sin x)}=-\frac{1-\tan x}{1+\tan x}=-\tan\!\left(\frac{\pi}{4}-x\right)=\tan\!\left(x-\frac{\pi}{4}\right).$

   Integrate.

   $\int\tan\!\left(x-\frac{\pi}{4}\right)dx=-\log\left|\cos\!\left(x-\frac{\pi}{4}\right)\right|+c=\log\left|\sec\!\left(\frac{\pi}{4}-x\right)\right|+c,$ …

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7. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.If $\dfrac{5\pi}{4} < x < \dfrac{7\pi}{4}$, then $\int \dfrac{\sqrt{1-\sin 2x}}{1+\sin 2x} dx =$ (A) $-\log \left| \sec \left( \dfrac{\pi}{4} - x \right) \right| + c$ (B) $-\sec^2 \left( \dfrac{\pi}{4} - x \right) + c$ (C) $\log \left| \sec \left( \dfrac{\pi}{4} - x \right) \right| + c$ (D) $\sec^2 \left( \dfrac{\pi}{4} - x \right) + c$

   ›Reveal solutionSolution

   Both $1-\sin 2x$ and $1+\sin 2x$ are perfect squares, so the integrand collapses to $\left|\tan\left(\frac{\pi}{4}-x\right)\right|$. On $\left(\frac{5\pi}{4},\frac{7\pi}{4}\right)$ that modulus opens as $-\tan\left(\frac{\pi}{4}-x\right)$, whose integral is $\log\left|\sec\left(\frac{\pi}{4}-x\right)\right|+c$ — option (C).

   The concept first

   Whenever you see $1\pm\sin 2x$ under a root, do not reach for a substitution — reach for the identity. Since $\sin 2x=2\sin x\cos x$ and $\sin^2x+\cos^2x=1$,

   $1-\sin 2x=\cos^2x-2\sin x\cos x+\sin^2x=(\cos x-\sin x)^2,$

   $1+\sin 2x=(\cos x+\sin x)^2.$

   A square root of a square is a modulus, not the bracket itself. That is exactly why the question bothers to tell you the interval $\frac{5\pi}{4}<x<\frac{7\pi}{4}$: the interval decides the sign.

   Step-by-step

   Step 1 — rewrite the integrand.

   $\frac{\sqrt{1-\sin 2x}}{\sqrt{1+\sin 2x}}=\frac{|\cos x-\sin x|}{|\cos x+\sin x|}=\left|\frac{\cos x-\sin x}{\cos x+\sin x}\right|.$

   Step 2 — recognise the tangent. Divide numerator and denominator by $\cos x$:

   $\frac{\cos x-\sin x}{\cos x+\sin x}=\frac{1-\tan x}{1+\tan x}=\tan\!\left(\frac{\pi}{4}-x\right).$

   Step 3 — fix the sign on the given interval. Write both brackets as single sinusoids:

   $\cos x-\sin x=\sqrt2\cos\!\left(x+\frac{\pi}{4}\right),\qquad \cos x+\sin x=\sqrt2\sin\!\left(x+\frac{\pi}{4}\right).$

   If $\frac{5\pi}{4}<x<\frac{7\pi}{4}$ then $x+\frac{\pi}{4}\in\left(\frac{3\pi}{2},2\pi\right)$ — the fourth quadrant. There $\cos>0$ and $\sin<0$, so

   $\cos x-\sin x>0,\qquad \cos x+\sin x<0 .$

   The quotient is therefore negative, and its modulus is

   $\left|\tan\!\left(\frac{\pi}{4}-x\right)\right|=-\tan\!\left(\frac{\pi}{4}-x\right).$

   Step 4 — integrate. Put $t=\frac{\pi}{4}-x$, so $dt=-dx$, i.e. $dx=-dt$: …

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8. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.$\int e^{-2x} \left( \tan 2x - 2 \sec^2 2x \tan 2x \right) dx =$ (A) $e^{-2x} \tan 2x + c$ (B) $-\dfrac{e^{-2x}}{2} \left[ \sec^2 2x + \tan 2x \right] + c$ (C) $-\dfrac{e^{-2x}}{2} \left[ \tan 2x - \sec^2 2x \right] + c$ (D) $e^{-2x} \sec^2 2x + c$

   ›Reveal solutionSolution

   The integrand is the exact derivative of $-\tfrac{1}{2}e^{-2x}\big(\sec^2 2x + \tan 2x\big)$ — option (B).

   Test option (B) by differentiating it. Let

   $F(x) = -\frac{1}{2}e^{-2x}\left(\sec^{2}2x + \tan 2x\right).$

   Using the product rule with $\dfrac{d}{dx}\sec^{2}2x = 4\sec^{2}2x\,\tan 2x$ and $\dfrac{d}{dx}\tan 2x = 2\sec^{2}2x$:

   $F'(x) = -\frac{1}{2}\Big[-2e^{-2x}\big(\sec^{2}2x + \tan 2x\big) + e^{-2x}\big(4\sec^{2}2x\,\tan 2x + 2\sec^{2}2x\big)\Big].$

   Simplify inside the bracket:

   $-2\sec^{2}2x - 2\tan 2x + 4\sec^{2}2x\,\tan 2x + 2\sec^{2}2x = -2\tan 2x + 4\sec^{2}2x\,\tan 2x.$

   So …

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9. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.$\int (\sqrt{1-\sin x} + \sqrt{1+\sin x}) dx = f(x) + c$, where $c$ is the constant of integration. If $\frac{5\pi}{2} < x < \frac{7\pi}{2}$ and $f\left(\frac{8\pi}{3}\right) = -2$, then $f'\left(\frac{8\pi}{3}\right) =$ (A) $1$ (B) $\sqrt{3}$ (C) $0$ (D) $-1$

   ›Reveal solutionSolution

   The key is to simplify the integrand using trigonometric identities, then integrate piecewise with the correct sign based on the given interval. The derivative of the antiderivative is just the original integrand evaluated at $x = 8\pi/3$, which simplifies to $1$.

   Concept & Intuition

   We are given that $\int (\sqrt{1-\sin x} + \sqrt{1+\sin x}) dx = f(x) + c$, so $f'(x)$ is exactly the integrand:

   $f'(x) = \sqrt{1-\sin x} + \sqrt{1+\sin x}.$

   The problem asks for $f'(8\pi/3)$, which is simply the value of this expression at $x = 8\pi/3$. The tricky part is simplifying the square roots correctly for the given interval $\frac{5\pi}{2} < x < \frac{7\pi}{2}$, because $\sqrt{1 \pm \sin x}$ can be rewritten using half-angle formulas, but the sign of the resulting cosine or sine terms depends on the quadrant.

   Step-by-step solution

   1. Rewrite the integrand using half-angle identities. Recall:

   $1 - \sin x = (\sin(x/2) - \cos(x/2))^2, \quad 1 + \sin x = (\sin(x/2) + \cos(x/2))^2.$

   Therefore,

   $\sqrt{1 - \sin x} = |\sin(x/2) - \cos(x/2)|, \quad \sqrt{1 + \sin x} = |\sin(x/2) + \cos(x/2)|.$

   2. Determine the sign of these expressions on the given interval. The interval is $\frac{5\pi}{2} < x < \frac{7\pi}{2}$, so dividing by 2:

   $\frac{5\pi}{4} < \frac{x}{2} < \frac{7\pi}{4}.$

   This places $x/2$ in the third and fourth quadrants (angles between $225^\circ$ and $315^\circ$).

   • On this range, $\sin(x/2)$ is negative (from $-\sqrt{2}/2$ to $-1$ and back to $-\sqrt{2}/2$).
   • $\cos(x/2)$ is also negative in the third quadrant but becomes positive in the fourth quadrant. Let’s check the sum and difference:

   For $x/2$ just above $5\pi/4$ (e.g., $230^\circ$): both $\sin$ and $\cos$ are negative, so $\sin + \cos$ is negative, $\sin - \cos$ is negative (since $\sin$ is more negative than $\cos$).

   For $x/2$ just below $7\pi/4$ (e.g., $310^\circ$): $\sin$ negative, $\cos$ positive, so $\sin + \cos$ could be negative or positive? At $7\pi/4$, $\sin = -\sqrt{2}/2$, $\cos = \sqrt{2}/2$, so sum = 0, difference = $-\sqrt{2}$.

   Actually, we need the sign of $\sin(x/2) \pm \cos(x/2)$ over the whole interval. A clean way:

   • $\sin(x/2) + \cos(x/2) = \sqrt{2} \sin(x/2 + \pi/4)$. For $x/2 \in (5\pi/4, 7\pi/4)$, the argument $x/2 + \pi/4 \in (3\pi/2, 2\pi)$. Sine is negative there (except at $2\pi$ where it's 0), so $\sin(x/2) + \cos(x/2) \le 0$.
   • $\sin(x/2) - \cos(x/2) = \sqrt{2} \sin(x/2 - \pi/4)$. For $x/2 \in (5\pi/4, 7\pi/4)$, the argument $x/2 - \pi/4 \in (\pi, 3\pi/2)$. Sine is negative there (except at $\pi$ and $3\pi/2$ where it's 0), so $\sin(x/2) - \cos(x/2) \le 0$. Hence both expressions are non-positive on the entire interval. So: $|\sin(x/2) - \cos(x/2)| = -(\sin(x/2) - \cos(x/2)) = \cos(x/2) - \sin(x/2),$ …

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10. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
    Q.In triangle ABC, if $a = 4$, $b = 3$, $c = 2$ then $2(a - b \cos C)(a - c \sec B) =$ (A) $0$ (B) $1$ (C) $2$ (D) $3$

    ›Reveal solutionSolution

    With $a=4,\,b=3,\,c=2$, the Law of Cosines gives $a-b\cos C=\tfrac{11}{8}$ and $a-c\sec B=\tfrac{12}{11}$, so $2(a-b\cos C)(a-c\sec B)=3$ — option (D).

    First factor. $\cos C=\dfrac{a^2+b^2-c^2}{2ab}=\dfrac{16+9-4}{24}=\dfrac78$, so

    $a-b\cos C=4-3\cdot\frac78=4-\frac{21}{8}=\frac{11}{8}.$

    Second factor. $\cos B=\dfrac{a^2+c^2-b^2}{2ac}=\dfrac{16+4-9}{16}=\dfrac{11}{16}$, so $\sec B=\dfrac{16}{11}$ and …

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11. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
    Q.$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\sec x}{1 + 2 \sin^2 x} \, dx =$ (A) $\frac{1}{3} \log(\sqrt{2} + 1) + \frac{\pi \sqrt{2}}{12}$ (B) $\frac{2}{3} \log(\sqrt{2} + 1) + \frac{\pi \sqrt{2}}{6}$ (C) $\frac{1}{6} \log(\sqrt{2} - 1) + \frac{\pi}{12}$ (D) $\frac{1}{4} \log(\sqrt{2} - 1) - \frac{\pi \sqrt{3}}{6}$

    ›Reveal solutionSolution

    Substitute $u=\sin x$, split by partial fractions, and the integral evaluates to $\tfrac{1}{3}\log(\sqrt2+1)+\tfrac{\pi\sqrt2}{12}$.

    Reduce to an algebraic integral. Multiply numerator and denominator by $\cos x$:

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

    Let $u=\sin x$, $du=\cos x\,dx$. Over the interval the limits run $u:0\to\tfrac{1}{\sqrt2}$, so

    $I=\int_{0}^{1/\sqrt2}\frac{du}{(1-u^2)(1+2u^2)}.$

    Partial fractions (in $t=u^2$): $\dfrac{1}{(1-t)(1+2t)}=\dfrac{1/3}{1-t}+\dfrac{2/3}{1+2t}$, so

    $\frac{1}{(1-u^2)(1+2u^2)}=\frac{1}{3}\cdot\frac{1}{1-u^2}+\frac{2}{3}\cdot\frac{1}{1+2u^2}.$

    Integrate each piece.

    $\frac{1}{3}\int_{0}^{1/\sqrt2}\frac{du}{1-u^2}=\frac{1}{3}\,\big[\operatorname{artanh}u\big]_{0}^{1/\sqrt2}=\frac{1}{3}\cdot\frac{1}{2}\log\frac{1+\tfrac1{\sqrt2}}{1-\tfrac1{\sqrt2}}=\frac{1}{3}\log(\sqrt2+1),$

    using $\dfrac{\sqrt2+1}{\sqrt2-1}=(\sqrt2+1)^2$. …

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12. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
    Q.$\int_{0}^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} \, dx =$ (A) $\sqrt{2} \log(\sqrt{2} + 1)$ (B) $\frac{1}{\sqrt{2}} \log(\sqrt{2} + 1)$ (C) $\log(\sqrt{2} + 1)$ (D) $\frac{1}{\sqrt{2}} \log(\sqrt{2} - 1)$

    ›Reveal solutionSolution

    The key idea is to exploit symmetry: replace $x$ by $\frac{\pi}{2} - x$ and add the two forms to simplify the denominator. The integral evaluates to $\frac{1}{\sqrt{2}}\log(\sqrt{2}+1)$, which corresponds to option (B).

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    Concept and intuition

    When you see a trigonometric integral over $[0, \pi/2]$ with both $\sin x$ and $\cos x$ in the denominator, a classic trick is to use the substitution $x \to \frac{\pi}{2} - x$. This swaps $\sin x$ and $\cos x$, often creating a simpler sum. Here, the denominator $\sin x + \cos x$ becomes symmetric, and the numerator $\sin^2 x$ becomes $\cos^2 x$. Adding the original integral and its transformed version gives a new integral with a constant denominator, which is easy to handle.

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    Step-by-step solution

    1. Define the integral and apply the symmetry substitution Let

    $I = \int_{0}^{\pi/2} \frac{\sin^2 x}{\sin x + \cos x} \, dx.$

    Substitute $x = \frac{\pi}{2} - t$. Then $dx = -dt$, and when $x=0$, $t=\pi/2$; when $x=\pi/2$, $t=0$. So

    $I = \int_{\pi/2}^{0} \frac{\sin^2(\pi/2 - t)}{\sin(\pi/2 - t) + \cos(\pi/2 - t)} \, (-dt) = \int_{0}^{\pi/2} \frac{\cos^2 t}{\cos t + \sin t} \, dt.$

    Renaming $t$ back to $x$, we have

    $I = \int_{0}^{\pi/2} \frac{\cos^2 x}{\sin x + \cos x} \, dx.$

    2. Add the two expressions for $I$ Adding the original and the transformed version:

    $2I = \int_{0}^{\pi/2} \frac{\sin^2 x + \cos^2 x}{\sin x + \cos x} \, dx = \int_{0}^{\pi/2} \frac{1}{\sin x + \cos x} \, dx.$

    So

    $I = \frac12 \int_{0}^{\pi/2} \frac{1}{\sin x + \cos x} \, dx.$

    3. Simplify the denominator using a trigonometric identity Recall that $\sin x + \cos x = \sqrt{2} \sin\!\left(x + \frac{\pi}{4}\right)$. Thus

    $I = \frac12 \int_{0}^{\pi/2} \frac{1}{\sqrt{2} \sin\!\left(x + \frac{\pi}{4}\right)} \, dx = \frac{1}{2\sqrt{2}} \int_{0}^{\pi/2} \csc\!\left(x + \frac{\pi}{4}\right) dx.$

    4. Perform a simple substitution Let $u = x + \frac{\pi}{4}$. Then $du = dx$, and when $x=0$, $u = \frac{\pi}{4}$; when $x=\frac{\pi}{2}$, $u = \frac{3\pi}{4}$. So

    $I = \frac{1}{2\sqrt{2}} \int_{\pi/4}^{3\pi/4} \csc u \, du.$

    5. Evaluate the integral of $\csc u$ The standard antiderivative is

    $\int \csc u \, du = \log|\csc u - \cot u| + C.$

    Hence

    $I = \frac{1}{2\sqrt{2}} \left[ \log|\csc u - \cot u| \right]_{\pi/4}^{3\pi/4}.$

    6. Compute the values at the bounds
       • At $u = \frac{3\pi}{4}$: $\csc(3\pi/4) = \sqrt{2}$, $\cot(3\pi/4) = -1$. So $\csc - \cot = \sqrt{2} - (-1) = \sqrt{2} + 1$. …

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