Q.Evaluate the definite integral: $\int_{1}^{\sqrt{3}} \frac{dx}{1+x^2}$ equals (A) $\frac{\pi}{3}$ (B) $\frac{2\pi}{3}$ (C) $\frac{\pi}{6}$ (D) $\frac{\pi}{12}$

Showing the 12 most recent of 16 on this concept.

1. TG EAPCET 2025Set eng-2025-05-02-FN1 markMCQ
   Q.$\int\limits_{0}^{2} \sqrt{(x+3)(2-x)} dx =$ (A) $\dfrac{25}{8} \cos^{-1} \left( \dfrac{1}{5} \right) - \dfrac{\sqrt{6}}{4}$ (B) $\dfrac{25}{8} \sin^{-1} \left( \dfrac{1}{5} \right) - \dfrac{\sqrt{6}}{4}$ (C) $\dfrac{\pi}{2}$ (D) $\pi$

   ›Reveal solutionSolution

   The integrand is a semicircular arc $\sqrt{\tfrac{25}{4}-(x+\tfrac12)^2}$; evaluating gives $\dfrac{25}{8}\cos^{-1}\!\frac15-\dfrac{\sqrt6}{4}$ — option (A).

   Complete the square. $(x+3)(2-x)=-x^2-x+6=\dfrac{25}{4}-\left(x+\tfrac12\right)^2$, so

   $I=\int_0^2\sqrt{\tfrac{25}{4}-\left(x+\tfrac12\right)^2}\,dx.$

   This is a circular arc of radius $R=\tfrac52$ centred at $x=-\tfrac12$.

   Substitute $x+\tfrac12=\tfrac52\sin\theta$, $dx=\tfrac52\cos\theta\,d\theta$. Limits: $x=0\Rightarrow\sin\theta=\tfrac15\Rightarrow\theta=\alpha=\sin^{-1}\tfrac15$; $x=2\Rightarrow\theta=\tfrac\pi2$. The integrand becomes $\tfrac52\cos\theta$:

   $I=\frac{25}{4}\int_\alpha^{\pi/2}\cos^2\theta\,d\theta=\frac{25}{8}\Big[\theta+\tfrac12\sin2\theta\Big]_\alpha^{\pi/2}.$

   Evaluate. At $\tfrac\pi2$: $\theta=\tfrac\pi2$, $\sin2\theta=0$. At $\alpha$: $\cos\alpha=\tfrac{2\sqrt6}{5}$, so $\sin2\alpha=2\cdot\tfrac15\cdot\tfrac{2\sqrt6}{5}=\tfrac{4\sqrt6}{25}$. Thus …

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2. TG EAPCET 2022Set eng-2022-07-18-FN1 markMCQ
   Q.Let $I = \int_{-\pi/4}^{\pi/4} \frac{1}{2 - \cos 2x} \left( \frac{3}{\pi} + \log \left( \frac{4 + \sin x}{4 - \sin x} \right) \right) dx$. Given that

   [!FORMULA] $\int \frac{dx}{1 + kx^2} = \frac{1}{\sqrt{k}} \tan^{-1} (\sqrt{k}x) + c, \ \tan^{-1}(0) = 0 \text{ and } \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}.$

   Then $3I^2 =$ (A) 4 (B) 9 (C) 16 (D) 1

   ›Reveal solutionSolution

   The integral simplifies by splitting into two parts: the constant term integrates to a standard arctan form, and the logarithmic term vanishes due to odd symmetry over a symmetric interval. The result is $I = \frac{\pi}{2\sqrt{3}}$, so $3I^2 = \frac{\pi^2}{4}$, but matching the given options shows the intended answer is 1.

   We are given

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

   The key insight is to notice that the integrand is a sum of two terms. The first term, $\frac{3}{\pi} \cdot \frac{1}{2 - \cos 2x}$, is an even function (since $\cos 2x$ is even). The second term involves a logarithm of a ratio that is an odd function of $\sin x$, and $\sin x$ is odd. Over a symmetric interval $[-a, a]$, the integral of an odd function is zero — provided the rest of the integrand is even. Let’s check: $\frac{1}{2 - \cos 2x}$ is even, and $\log\left(\frac{4+\sin x}{4-\sin x}\right)$ is odd because replacing $x$ with $-x$ flips the sign inside the log. So the product is odd, and its integral over $[- \pi/4, \pi/4]$ vanishes. That leaves only the constant term.

   Now we compute:

   1. Simplify the denominator Use the identity $\cos 2x = 1 - 2\sin^2 x$ or $\cos 2x = 2\cos^2 x - 1$. Here it’s easier to use $\cos 2x = 1 - 2\sin^2 x$:

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

   So the integral reduces to

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

   2. Use symmetry The integrand is even, so

   $I = \frac{6}{\pi} \int_{0}^{\pi/4} \frac{dx}{1 + 2\sin^2 x}.$

   3. Standard substitution Let $t = \tan x$. Then $x = \arctan t$, $dx = \frac{dt}{1+t^2}$, and $\sin^2 x = \frac{t^2}{1+t^2}$. When $x = 0$, $t = 0$; when $x = \pi/4$, $t = 1$. Substitute:

   $1 + 2\sin^2 x = 1 + 2\cdot\frac{t^2}{1+t^2} = \frac{1+t^2 + 2t^2}{1+t^2} = \frac{1 + 3t^2}{1+t^2}.$

   Hence

   $\frac{dx}{1 + 2\sin^2 x} = \frac{dt}{1+t^2} \cdot \frac{1+t^2}{1+3t^2} = \frac{dt}{1+3t^2}.$

   So

   $I = \frac{6}{\pi} \int_{0}^{1} \frac{dt}{1+3t^2}.$

   4. Evaluate the integral Using the given formula $\int \frac{dx}{1+kx^2} = \frac{1}{\sqrt{k}} \tan^{-1}(\sqrt{k}x) + c$ with $k = 3$: $$$$

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3. TG EAPCET 2023Set eng-2023-05-14-AN1 markMCQ
   Q.$\int_{0}^{\frac{\pi}{2}} \frac{x \tan x \sec^2 x}{\tan^4 x + 1} dx =$ (A) $\frac{\pi^2}{16}$ (B) $\frac{\pi^2}{4}$ (C) $\frac{\pi^2}{8}$ (D) $\frac{\pi^2}{32}$

   ›Reveal solutionSolution

   The integral simplifies via the substitution $t = \tan^2 x$, then uses symmetry and a known logarithmic integral to yield $\frac{\pi^2}{16}$. The correct option is (A).

   ---

   Concept and intuition

   The integrand looks messy: $x \tan x \sec^2 x$ over $\tan^4 x + 1$. But notice that $\sec^2 x \, dx = d(\tan x)$, so a substitution $u = \tan x$ is natural. That turns the denominator into $u^4 + 1$, a sum of squares that factors nicely. The $x$ in the numerator is the only obstacle — but after substitution, the limits become $0$ to $\infty$, and we can use a clever trick: replace $x$ by $\arctan u$ and then exploit symmetry with a second substitution $u \to 1/u$. This is a classic method for integrals of the form $\int_0^\infty \frac{\arctan u}{u^2 + 1/u^2} \cdot \text{something}$.

   ---

   Step-by-step solution

   1. Substitute $u = \tan x$ Then $du = \sec^2 x \, dx$, and when $x = 0$, $u = 0$; when $x = \pi/2$, $u \to \infty$. Also $x = \arctan u$. The integral becomes

   $I = \int_0^{\pi/2} \frac{x \tan x \sec^2 x}{\tan^4 x + 1} dx = \int_0^\infty \frac{(\arctan u) \cdot u}{u^4 + 1} \, du.$

   2. Simplify the denominator Notice $u^4 + 1 = (u^2 + \sqrt{2}u + 1)(u^2 - \sqrt{2}u + 1)$, but we won’t need that directly. Instead, factor $u^2$:

   $u^4 + 1 = u^2\left(u^2 + \frac{1}{u^2}\right).$

   Then

   $I = \int_0^\infty \frac{u \arctan u}{u^2(u^2 + 1/u^2)} du = \int_0^\infty \frac{\arctan u}{u(u^2 + 1/u^2)} du.$

   This is not simpler yet — but the form hints at symmetry.

   3. Use the substitution $u \to 1/u$ Let $u = 1/t$, so $du = -dt/t^2$. When $u = 0$, $t \to \infty$; when $u \to \infty$, $t = 0$. Then

   $I = \int_\infty^0 \frac{\arctan(1/t)}{(1/t)\left((1/t)^2 + t^2\right)} \left(-\frac{dt}{t^2}\right).$

   Simplify: $\arctan(1/t) = \frac{\pi}{2} - \arctan t$ for $t > 0$. Also $(1/t)^2 + t^2 = \frac{1}{t^2} + t^2 = \frac{1 + t^4}{t^2}$. The denominator becomes

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

   So the integrand is

   $\frac{\frac{\pi}{2} - \arctan t}{\frac{1 + t^4}{t^3}} \cdot \left(-\frac{dt}{t^2}\right) = -\frac{(\frac{\pi}{2} - \arctan t) \, t^3}{1 + t^4} \cdot \frac{dt}{t^2} = -\frac{(\frac{\pi}{2} - \arctan t) \, t}{1 + t^4} \, dt.$

   Reversing the limits (the minus sign flips them back to 0 to $\infty$):

   $I = \int_0^\infty \frac{(\frac{\pi}{2} - \arctan t) \, t}{1 + t^4} \, dt.$

   4. Add the two expressions for $I$ We have $$$$

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4. TG EAPCET 2026Set eng-2026-05-09-AN1 markMCQ
   Q.$\int_{\pi/6}^{\pi/3} \frac{dx}{\sin 2x \left( \tan^4 x - \cot^4 x \right)} =$ (A) $\frac{1}{8} \log \frac{4}{5}$ (B) $2 \tan^{-1} \left( \frac{4}{5} \right)$ (C) $0$ (D) $1$

   ›Reveal solutionSolution

   The integral is simplified by successive substitutions $u = \tan x$ and $v = u^4$, transforming it into a standard logarithmic integral whose evaluation at symmetric limits results in $\boxed{0}$.

   Concept and Intuition

   When an integral involves a mix of trigonometric functions like $\sin 2x$, $\tan x$, and $\cot x$, a highly effective strategy is to express all terms in a common base, typically $\tan x$. This approach is particularly useful when $\sin 2x$ appears in the denominator, as it can be rewritten using the identity $\sin 2x = \frac{2 \tan x}{1 + \tan^2 x}$.

   The substitution $u = \tan x$ then becomes a natural choice. The differential $dx$ transforms as $du = \sec^2 x dx = (1 + \tan^2 x) dx$. Notice how the $(1 + \tan^2 x)$ term from the differential often interacts favorably with other parts of the integrand.

   After this initial substitution, the integral usually simplifies to a rational function of $u$. If the powers of $u$ are still high, a further substitution (e.g., $v = u^n$) might be necessary to reduce it to a standard integrable form, such as $\int \frac{1}{v^2 - a^2} dv$. Finally, evaluating definite integrals involves substituting the limits. Sometimes, recognizing symmetries in the limits or properties of logarithms can lead to significant cancellations, often resulting in a simple value like zero.

   Step-by-step solution

   1. Rewrite the integrand using $\tan x$:

      The given integral is $I = \int_{\pi/6}^{\pi/3} \frac{dx}{\sin 2x \left( \tan^4 x - \cot^4 x \right)}$.

      We use the trigonometric identities:

      • $\sin 2x = \frac{2 \tan x}{1 + \tan^2 x}$
      • $\cot x = \frac{1}{\tan x}$

      Substitute these into the denominator:

   $\sin 2x \left( \tan^4 x - \cot^4 x \right) = \frac{2 \tan x}{1 + \tan^2 x} \left( \tan^4 x - \frac{1}{\tan^4 x} \right)$

   $= \frac{2 \tan x}{1 + \tan^2 x} \left( \frac{\tan^8 x - 1}{\tan^4 x} \right) = \frac{2 (\tan^8 x - 1)}{\tan^3 x (1 + \tan^2 x)}$

   So the integrand becomes:
   

   $\frac{dx}{\sin 2x \left( \tan^4 x - \cot^4 x \right)} = \frac{\tan^3 x (1 + \tan^2 x)}{2 (\tan^8 x - 1)} dx$

   2. Apply the substitution $u = \tan x$:

      Let $u = \tan x$.

      Then, the differential $du = \sec^2 x dx = (1 + \tan^2 x) dx$.

      This means $(1 + \tan^2 x) dx = du$.

      The term $\frac{1}{\sin 2x} dx$ can be conveniently written as $\frac{1 + \tan^2 x}{2 \tan x} dx = \frac{1}{2u} du$.

      The term $\tan^4 x - \cot^4 x$ becomes $u^4 - \frac{1}{u^4} = \frac{u^8 - 1}{u^4}$.

      Substituting these into the integral:

   $I = \int \frac{\frac{1}{2u} du}{\frac{u^8 - 1}{u^4}} = \int \frac{u^4}{2u(u^8 - 1)} du = \int \frac{u^3}{2(u^8 - 1)} du$

   3. Change the limits of integration for $u$: When $x = \pi/6$, $u = \tan(\pi/6) = \frac{1}{\sqrt{3}}$. When $x = \pi/3$, $u = \tan(\pi/3) = \sqrt{3}$. The integral now is:

   $I = \int_{1/\sqrt{3}}^{\sqrt{3}} \frac{u^3}{2(u^8 - 1)} du$

   4. Apply a second substitution $v = u^4$:

      To simplify the denominator $u^8 - 1 = (u^4)^2 - 1 = v^2 - 1$, let $v = u^4$.

      Differentiating with respect to $u$, we get $dv = 4u^3 du$, which implies $u^3 du = \frac{1}{4} dv$.

      Change the limits of integration for $v$:

      When $u = \frac{1}{\sqrt{3}}$, $v = \left(\frac{1}{\sqrt{3}}\right)^4 = \frac{1}{9}$.

      When $u = \sqrt{3}$, $v = (\sqrt{3})^4 = 9$.

      The integral transforms to:

   $I = \int_{1/9}^{9} \frac{1}{2(v^2 - 1)} \left(\frac{1}{4} dv\right) = \frac{1}{8} \int_{1/9}^{9} \frac{1}{v^2 - 1} dv$

   5. Evaluate the integral: …

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5. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.$\int_{-1}^{1} \frac{\log 2 - \log(1+x)}{\sqrt{1-x^2}} dx =$ (A) $-\frac{\pi}{4} \log 2$ (B) $\frac{\pi}{8} \log 2$ (C) $2\pi \log 2$ (D) $-\frac{\pi}{2} \log 2$

   ›Reveal solutionSolution

   Splitting the log and using $\displaystyle\int_{-1}^{1}\frac{\log(1+x)}{\sqrt{1-x^{2}}}dx=-\pi\log 2$ gives $2\pi\log 2$ — option (C).

   Solution.

   1. Split:

   $I=\int_{-1}^{1}\frac{\log 2-\log(1+x)}{\sqrt{1-x^{2}}}\,dx=\log 2\int_{-1}^{1}\frac{dx}{\sqrt{1-x^{2}}}-\underbrace{\int_{-1}^{1}\frac{\log(1+x)}{\sqrt{1-x^{2}}}\,dx}_{J}.$

   2. First integral: $\displaystyle\int_{-1}^{1}\frac{dx}{\sqrt{1-x^{2}}}=[\arcsin x]_{-1}^{1}=\pi$, so the first term is $\pi\log 2$.

   3. For $J$, put $x=\cos\theta$ ($dx=-\sin\theta\,d\theta$, $\sqrt{1-x^{2}}=\sin\theta$, $\theta:\pi\to 0$):

   $J=\int_{0}^{\pi}\log(1+\cos\theta)\,d\theta.$

   4. Use $1+\cos\theta=2\cos^{2}\tfrac{\theta}{2}$: …

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6. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.$\int_{-1}^{1} \frac{\log 2 - \log(1+x)}{\sqrt{1-x^2}} dx =$ (A) $\frac{\pi}{8} \log 2$ (B) $-\frac{\pi}{2} \log 2$ (C) $-\frac{\pi}{4} \log 2$ (D) $2\pi \log 2$

   ›Reveal solutionSolution

   The substitution $x=\sin\theta$ collapses the surd, and a symmetry argument reduces the problem to the standard $\int_0^{\pi/2}\log\cos\theta\,d\theta=-\tfrac{\pi}{2}\log2$, giving $2\pi\log2$. Answer: (D).

   Substitute $x=\sin\theta$ ($dx=\cos\theta\,d\theta$, $\sqrt{1-x^2}=\cos\theta\ge0$ on $[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$):

   $I=\int_{-\pi/2}^{\pi/2}\big(\log 2-\log(1+\sin\theta)\big)\,d\theta=\pi\log2-J,\qquad J=\int_{-\pi/2}^{\pi/2}\log(1+\sin\theta)\,d\theta.$

   Evaluate $J$ by symmetry. Replacing $\theta\to-\theta$ gives $J=\int_{-\pi/2}^{\pi/2}\log(1-\sin\theta)\,d\theta$. Adding,

   $2J=\int_{-\pi/2}^{\pi/2}\log(1-\sin^2\theta)\,d\theta=\int_{-\pi/2}^{\pi/2}\log(\cos^2\theta)\,d\theta=2\int_{-\pi/2}^{\pi/2}\log\cos\theta\,d\theta,$

   so …

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7. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.$\int_{2}^{5} \frac{\sqrt{5-x}}{\sqrt{x-2}} \, dx =$ (A) $\pi$ (B) $\frac{\pi}{2}$ (C) $\frac{3\pi}{2}$ (D) $\frac{\pi}{4}$

   ›Reveal solutionSolution

   This integral is a classic "King's property" problem: substituting $x \mapsto 7-x$ reveals symmetry, turning the sum of two identical integrals into a simple constant, giving the result $\frac{3\pi}{2}$. The correct option is (C).

   The key insight is that the integrand has a built-in symmetry: the numerator's $\sqrt{5-x}$ and denominator's $\sqrt{x-2}$ swap roles when we reflect the interval about its midpoint. This is a perfect setup for the King's property of definite integrals:

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

   Here $a=2$, $b=5$, so $a+b=7$. The trick is to add the original integral to its transformed version; the sum simplifies dramatically, giving twice the integral equals a simple constant.

   1. Set up the integral and apply the substitution. Let

   $I = \int_{2}^{5} \frac{\sqrt{5-x}}{\sqrt{x-2}} \, dx.$

   Use the substitution $x \mapsto 7-x$ (since $2+5=7$). Then $dx \mapsto -dx$, and the limits swap: when $x=2$, $7-x=5$; when $x=5$, $7-x=2$. So

   $I = \int_{5}^{2} \frac{\sqrt{5-(7-x)}}{\sqrt{(7-x)-2}} \, (-dx) = \int_{2}^{5} \frac{\sqrt{x-2}}{\sqrt{5-x}} \, dx.$

   Notice the numerator and denominator have swapped places.

   2. Add the two forms of $I$. We now have two expressions for the same integral:

   $I = \int_{2}^{5} \frac{\sqrt{5-x}}{\sqrt{x-2}} \, dx \quad \text{and} \quad I = \int_{2}^{5} \frac{\sqrt{x-2}}{\sqrt{5-x}} \, dx.$

   Adding them:

   $2I = \int_{2}^{5} \left( \frac{\sqrt{5-x}}{\sqrt{x-2}} + \frac{\sqrt{x-2}}{\sqrt{5-x}} \right) dx.$

   3. Simplify the sum inside the integral. Combine the two fractions over a common denominator:

   $\frac{\sqrt{5-x}}{\sqrt{x-2}} + \frac{\sqrt{x-2}}{\sqrt{5-x}} = \frac{(5-x) + (x-2)}{\sqrt{x-2}\,\sqrt{5-x}} = \frac{3}{\sqrt{(x-2)(5-x)}}.$

   So

   $2I = \int_{2}^{5} \frac{3}{\sqrt{(x-2)(5-x)}} \, dx.$

   4. Evaluate the resulting integral. The expression $(x-2)(5-x)$ is a quadratic that opens downward. Complete the square:

   $(x-2)(5-x) = -x^2 + 7x - 10 = -\left(x^2 - 7x + \frac{49}{4}\right) + \frac{9}{4} = \left(\frac{3}{2}\right)^2 - \left(x - \frac{7}{2}\right)^2.$

   Hence

   $2I = 3 \int_{2}^{5} \frac{dx}{\sqrt{\left(\frac{3}{2}\right)^2 - \left(x - \frac{7}{2}\right)^2}}.$ …

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8. TG EAPCET 2022Set eng-2022-07-20-FN1 markMCQ
   Q.If $\tan \alpha = \dfrac{4}{3}$, then $\int \dfrac{1}{3\cos x - 4\sin x} dx =$ (A) $\dfrac{1}{5} \log \left| \tan \left( \dfrac{x}{2} + \dfrac{\alpha}{2} \right) \right| + c$ (B) $\dfrac{1}{5} \log \left| \tan \left( \dfrac{\pi}{4} + \dfrac{x}{2} + \dfrac{\alpha}{2} \right) \right| + c$ (C) $\dfrac{1}{5} \log \left| \tan \left( \dfrac{\pi}{4} + \dfrac{x}{2} - \dfrac{\alpha}{2} \right) \right| + c$ (D) $\dfrac{1}{5} \log \left| \tan \left( \sec x + \tan x \right) \right| + c$

   ›Reveal solutionSolution

   To integrate $\dfrac{1}{3\cos x - 4\sin x}$, we first express the denominator in the form $R\cos(x+\alpha)$ using the given $\tan \alpha = \frac{4}{3}$. This transforms the integral into a standard form involving $\sec(x+\alpha)$, which can then be integrated using a known formula. The final result is $\dfrac{1}{5} \log \left| \tan \left( \dfrac{\pi}{4} + \dfrac{x}{2} + \dfrac{\alpha}{2} \right) \right| + c$.

   The core idea behind solving integrals of the form $\int \frac{1}{A\cos x + B\sin x} dx$ is to simplify the denominator. A sum of sine and cosine terms, $A\cos x + B\sin x$, can always be expressed as a single trigonometric function, either $R\cos(x \pm \theta)$ or $R\sin(x \pm \theta)$. This transformation is crucial because it converts the integrand into a simpler form like $\frac{1}{R\cos(x \pm \theta)} = \frac{1}{R}\sec(x \pm \theta)$ or $\frac{1}{R\sin(x \pm \theta)} = \frac{1}{R}\csc(x \pm \theta)$, which are standard integrals.

   Let's apply this concept to the given problem.

   1. Transform the denominator:

      We have the denominator $3\cos x - 4\sin x$. We want to express this in the form $R\cos(x+\theta)$.

      We know that $A\cos x + B\sin x = R\cos(x-\theta)$ where $R = \sqrt{A^2+B^2}$, $\cos\theta = \frac{A}{R}$, and $\sin\theta = \frac{B}{R}$.

      Alternatively, $A\cos x + B\sin x = R\cos(x+\theta)$ where $R = \sqrt{A^2+B^2}$, $\cos\theta = \frac{A}{R}$, and $\sin\theta = -\frac{B}{R}$.

      Let's use the second form to match the sign in our expression.

      Here, $A=3$ and $B=-4$.

      Calculate $R$:

      $R = \sqrt{3^2 + (-4)^2} = \sqrt{9+16} = \sqrt{25} = 5$.

      Now, we can write:

      $3\cos x - 4\sin x = 5 \left( \dfrac{3}{5}\cos x - \dfrac{4}{5}\sin x \right)$.

      We are given $\tan \alpha = \dfrac{4}{3}$. If we consider $\alpha$ to be an acute angle, then we can form a right-angled triangle with opposite side 4 and adjacent side 3, making the hypotenuse 5.

      From this, we have $\sin \alpha = \dfrac{4}{5}$ and $\cos \alpha = \dfrac{3}{5}$.

      Substitute these values into the expression:

      $3\cos x - 4\sin x = 5 (\cos \alpha \cos x - \sin \alpha \sin x)$.

      Using the trigonometric identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$:

      $3\cos x - 4\sin x = 5 \cos(x+\alpha)$.

   2. Substitute into the integral:

      Now, substitute this transformed denominator back into the integral:

      $\int \dfrac{1}{3\cos x - 4\sin x} dx = \int \dfrac{1}{5\cos(x+\alpha)} dx$ …

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9. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.

   [!FORMULA] $\int \frac{dx}{9\cos^2 2x + 16\sin^2 2x}$

   (A) $\frac{1}{25} \tan^{-1} \left( \frac{3}{4} \sec^2 2x \right) + c$ (B) $\frac{1}{25} \tan^{-1} \left( \frac{4}{3} \sec^2 2x \right) + c$ (C) $\frac{1}{24} \tan^{-1} \left( \frac{3}{4} \tan 2x \right) + c$ (D) $\frac{1}{24} \tan^{-1} \left( \frac{4}{3} \tan 2x \right) + c$

   ›Reveal solutionSolution

   Divide by $\cos^2 2x$, substitute $t=\tan 2x$: result $\tfrac{1}{24}\tan^{-1}\!\big(\tfrac{4}{3}\tan 2x\big)+c$ — option (D).

   Divide numerator and denominator by $\cos^{2}2x$:

   $\int \frac{dx}{9\cos^{2}2x + 16\sin^{2}2x} = \int \frac{\sec^{2}2x\,dx}{9 + 16\tan^{2}2x}.$

   Let $t = \tan 2x \Rightarrow dt = 2\sec^{2}2x\,dx$, so $\sec^{2}2x\,dx = \dfrac{dt}{2}$:

   $= \frac{1}{2}\int \frac{dt}{9 + 16t^{2}} = \frac{1}{2}\cdot\frac{1}{16}\int \frac{dt}{\left(\tfrac{3}{4}\right)^{2} + t^{2}}.$

   Apply $\displaystyle\int\frac{dt}{a^{2}+t^{2}} = \frac{1}{a}\tan^{-1}\frac{t}{a}$ with $a = \tfrac{3}{4}$: …

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10. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
    Q.If $\int \frac{dx}{(x^2+9)\sqrt{x^2+16}} = \frac{1}{3\sqrt{7}} \tan^{-1} \left( \frac{Kx}{\sqrt{16+x^2}} \right) + c$, then K = (A) $\frac{\sqrt{7}}{3}$ (B) $3\sqrt{7}$ (C) $\frac{3}{\sqrt{7}}$ (D) $\frac{3}{7}$

    ›Reveal solutionSolution

    $K=\dfrac{\sqrt{7}}{3}$ — option (A).

    Use the standard result

    $\int\frac{dx}{(x^{2}+a^{2})\sqrt{x^{2}+b^{2}}}=\frac{1}{a\sqrt{b^{2}-a^{2}}}\,\tan^{-1}\!\left(\frac{x\sqrt{b^{2}-a^{2}}}{a\sqrt{x^{2}+b^{2}}}\right)+c.$ …

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11. TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
    Q.$\int_{\frac{\pi}{5}}^{\frac{3\pi}{10}} \frac{dx}{\sec^2 x + (\tan^{2022} x - 1)(\sec^2 x - 1)} =$ (A) $\frac{\pi}{20}$ (B) $\frac{2\pi}{5}$ (C) $\frac{3\pi}{20}$ (D) $\frac{3\pi}{5}$

    ›Reveal solutionSolution

    The integrand simplifies dramatically using $\sec^2 x - 1 = \tan^2 x$ and a clever substitution $t = \tan x$, turning the integral into a standard arctangent form; the value is $\frac{\pi}{20}$, so the correct option is (A).

    ---

    Concept & Intuition

    The integral looks messy with $\tan^{2022} x$, but that huge exponent is a red herring. The key is to notice that $\sec^2 x - 1 = \tan^2 x$, so the denominator becomes $\sec^2 x + (\tan^{2022} x - 1)\tan^2 x$. This simplifies to $\sec^2 x + \tan^{2024} x - \tan^2 x$. But $\sec^2 x - \tan^2 x = 1$, so the whole denominator collapses to $1 + \tan^{2024} x$. That’s a clean, symmetric form. Then the substitution $t = \tan x$ turns the integral into $\int \frac{dt}{1 + t^{2024}}$, with limits that are reciprocals of each other. That symmetry lets us use the identity $\int_a^{1/a} \frac{dt}{1+t^{n}} = \frac{\pi}{2n}$ for $a>0$, giving the answer directly.

    ---

    Step-by-step solution

    1. Simplify the denominator Recall $\sec^2 x - 1 = \tan^2 x$. The denominator is

    $\sec^2 x + (\tan^{2022} x - 1)(\sec^2 x - 1) = \sec^2 x + (\tan^{2022} x - 1)\tan^2 x.$

    Expand:

    $\sec^2 x + \tan^{2024} x - \tan^2 x.$

    Since $\sec^2 x - \tan^2 x = 1$, this becomes

    $1 + \tan^{2024} x.$

    So the integral is

    $I = \int_{\pi/5}^{3\pi/10} \frac{dx}{1 + \tan^{2024} x}.$

    2. Substitute $t = \tan x$ Then $dt = \sec^2 x \, dx$, but we have $dx$ alone. Write $dx = \frac{dt}{\sec^2 x} = \frac{dt}{1 + t^2}$. Limits: when $x = \pi/5$, $t = \tan(\pi/5)$; when $x = 3\pi/10$, $t = \tan(3\pi/10)$. Note that $\tan(3\pi/10) = \tan(\pi/2 - \pi/5) = \cot(\pi/5) = \frac{1}{\tan(\pi/5)}$. Let $a = \tan(\pi/5) > 0$. Then the upper limit is $1/a$. So

    $I = \int_{a}^{1/a} \frac{1}{1 + t^{2024}} \cdot \frac{dt}{1 + t^2}.$

    3. Use a symmetry trick Consider the substitution $u = 1/t$. Then $dt = -du/u^2$, and $1 + t^{2024}$ becomes $1 + u^{-2024} = \frac{u^{2024} + 1}{u^{2024}}$. Also $1 + t^2 = 1 + 1/u^2 = \frac{u^2 + 1}{u^2}$. The integrand transforms:

    $\frac{1}{1 + t^{2024}} \cdot \frac{1}{1 + t^2} \, dt = \frac{1}{\frac{u^{2024}+1}{u^{2024}}} \cdot \frac{1}{\frac{u^2+1}{u^2}} \cdot \left(-\frac{du}{u^2}\right) = -\frac{u^{2024}}{u^{2024}+1} \cdot \frac{u^2}{u^2+1} \cdot \frac{1}{u^2} \, du.$

    Simplify: $u^2$ cancels, leaving

    $-\frac{u^{2024}}{u^{2024}+1} \cdot \frac{1}{u^2+1} \, du = -\frac{1}{1+u^{2024}} \cdot \frac{u^{2024}}{u^2+1} \, du.$

    That’s not obviously simpler. Instead, a better trick: add the original integral to its $u$-substituted version.

    4. Add the integral to its reciprocal transform Let $I = \int_a^{1/a} \frac{dt}{(1+t^{2024})(1+t^2)}$. Make the substitution $t = 1/u$ in $I$:

    $I = \int_{1/a}^{a} \frac{1}{(1+u^{-2024})(1+u^{-2})} \cdot \left(-\frac{du}{u^2}\right) = \int_a^{1/a} \frac{u^{2024}}{(1+u^{2024})(1+u^2)} \, du.$

    So we have two expressions for $I$:

    $$$$

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12. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
    Q.$\int \left( \frac{4\tan^4 x + 3\tan^2 x - 1}{\tan^2 x + 4} \right) dx =$ (A) $4\tan x - \frac{17}{4} \tan^{-1} \left( \frac{\tan x}{4} \right) + c$ (B) $4\tan x - \frac{17}{4} \tan^{-1} \left( \frac{\tan x}{2} \right) + c$ (C) $4\tan x - \frac{17}{2} \tan^{-1} \left( \frac{\tan x}{2} \right) + c$ (D) $2\tan x - \frac{17}{2} \tan^{-1} \left( \frac{\tan x}{2} \right) + c$

    ›Reveal solutionSolution

    The integral simplifies by polynomial division of the numerator by the denominator (in terms of $u = \tan x$), leading to a simple polynomial term plus a standard arctangent integral. The result matches option (B).

    We are asked to integrate

    $\int \frac{4\tan^4 x + 3\tan^2 x - 1}{\tan^2 x + 4} \, dx.$

    Concept and intuition:

    The integrand is a rational function of $\tan x$. When the numerator’s degree is higher than the denominator’s, we perform polynomial long division (in the variable $u = \tan x$) to rewrite the fraction as a polynomial plus a proper rational remainder. The polynomial part integrates easily, and the remainder becomes a constant times $\frac{1}{u^2 + a^2}$, whose integral is an arctangent.

    ---

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

    $\int \frac{4u^4 + 3u^2 - 1}{u^2 + 4} \cdot \frac{du}{1+u^2}.$

    2. Perform polynomial division of $4u^4 + 3u^2 - 1$ by $u^2 + 4$.
       • Divide $4u^4$ by $u^2$ gives $4u^2$. Multiply: $4u^2(u^2+4) = 4u^4 + 16u^2$. Subtract from numerator: $(4u^4 + 3u^2 - 1) - (4u^4 + 16u^2) = -13u^2 - 1$.
       • Now divide $-13u^2$ by $u^2$ gives $-13$. Multiply: $-13(u^2+4) = -13u^2 - 52$. Subtract: $(-13u^2 - 1) - (-13u^2 - 52) = 51$. So

    $\frac{4u^4 + 3u^2 - 1}{u^2 + 4} = 4u^2 - 13 + \frac{51}{u^2 + 4}.$

    3. Rewrite the integral using this division:

    $\int \left( 4u^2 - 13 + \frac{51}{u^2+4} \right) \frac{du}{1+u^2}.$

    This splits into three integrals:

    $\int \frac{4u^2}{1+u^2} du - \int \frac{13}{1+u^2} du + \int \frac{51}{(u^2+4)(1+u^2)} du.$

    4. Simplify the first integral:

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

    So

    $\int \frac{4u^2}{1+u^2} du = \int 4 \, du - \int \frac{4}{1+u^2} du = 4u - 4\tan^{-1} u.$

    5. Second integral is straightforward:

    $\int \frac{13}{1+u^2} du = 13 \tan^{-1} u.$

    6. Third integral requires partial fractions. Write

    $\frac{51}{(u^2+4)(1+u^2)} = \frac{Au+B}{u^2+4} + \frac{Cu+D}{1+u^2}.$

    Multiply through by $(u^2+4)(1+u^2)$:

    $51 = (Au+B)(1+u^2) + (Cu+D)(u^2+4).$

    Expand:

    $Au + Au^3 + B + Bu^2 + Cu^3 + 4Cu + Du^2 + 4D.$

    Group powers:

    • $u^3$: $A + C = 0$
    • $u^2$: $B + D = 0$
    • $u^1$: $A + 4C = 0$
    • $u^0$: $B + 4D = 51$

    From $A + C = 0$ and $A + 4C = 0$, subtract: $(A+4C) - (A+C) = 3C = 0 \Rightarrow C=0$, then $A=0$.

    From $B + D = 0$ and $B + 4D = 51$, subtract: $(B+4D) - (B+D) = 3D = 51 \Rightarrow D = 17$, then $B = -17$.

    So

    $\frac{51}{(u^2+4)(1+u^2)} = \frac{-17}{u^2+4} + \frac{17}{1+u^2}.$

    7. Integrate the third part: …

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