Q.Evaluate the definite integral: $\int_0^1 \frac{dx}{1+x^2}$

Showing the 12 most recent of 16 on this concept.

1. 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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2. TG EAPCET 2026Set eng-2026-05-10-FN1 markMCQ
   Q.$\int \frac{dx}{10\cos^2 x + \cos 2x + 5} =$ (A) $\frac{1}{4} \tan^{-1} \left( \frac{\tan x}{2} \right) + c$ (B) $\frac{1}{8} \tan^{-1} \left( \frac{\tan x}{2} \right) + c$ (C) $\frac{1}{4} \tan^{-1} \left( \frac{\tan x}{5} \right) + c$ (D) $\frac{1}{8} \tan^{-1} \left( \frac{\tan x}{5} \right) + c$

   ›Reveal solutionSolution

   Use the identity $\cos 2x = 2\cos^2 x - 1$ to rewrite the denominator entirely in terms of $\cos^2 x$, then divide numerator and denominator by $\cos^2 x$ to obtain a standard $\int \frac{du}{a^2+u^2}$ form. The answer is $\frac{1}{4}\tan^{-1}\left(\frac{\tan x}{2}\right)+c$.

   The key is to notice that the denominator is a quadratic in $\cos^2 x$ once we replace $\cos 2x$ using the double-angle identity. That lets us factor out $\cos^2 x$ and substitute $u = \tan x$, turning the integral into a familiar arctan form.

   1. Rewrite $\cos 2x$ in terms of $\cos^2 x$ Use $\cos 2x = 2\cos^2 x - 1$. The denominator becomes:

   $10\cos^2 x + (2\cos^2 x - 1) + 5 = 12\cos^2 x + 4.$

   So the integral is

   $\int \frac{dx}{12\cos^2 x + 4}.$

   2. Factor out the constant

   $= \int \frac{dx}{4(3\cos^2 x + 1)} = \frac14 \int \frac{dx}{3\cos^2 x + 1}.$

   3. Divide numerator and denominator by $\cos^2 x$ This is the classic trick for integrals of the form $\int \frac{dx}{a\cos^2 x + b}$:

   $\frac14 \int \frac{\sec^2 x \, dx}{3 + \sec^2 x}.$

   But $\sec^2 x = 1 + \tan^2 x$, so the denominator becomes $3 + (1 + \tan^2 x) = 4 + \tan^2 x$.

   Hence

   $\frac14 \int \frac{\sec^2 x \, dx}{4 + \tan^2 x}.$

   4. Substitute $u = \tan x$ Then $du = \sec^2 x \, dx$, and the integral becomes

   $\frac14 \int \frac{du}{4 + u^2}.$

   5. Recognise the standard form …

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3. 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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4. 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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5. 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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6. 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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7. TG EAPCET 2025Set eng-2025-05-04-AN1 markMCQ
   Q.If $\int_{0}^{\pi/2} \tan^{14}\left(\frac{x}{2}\right) dx = 2 \left[ \sum_{n=1}^{7} f(n) - \frac{\pi}{4} \right]$, then $f(n) =$ (A) $\dfrac{(-1)^n}{n-1}$ (B) $\dfrac{(-1)^n}{2n+1}$ (C) $\dfrac{(-1)^{n+1}}{2n-1}$ (D) $\dfrac{(-1)^{n+1}}{n+1}$

   ›Reveal solutionSolution

   The integral is transformed via the substitution $t = \tan(x/2)$, turning it into a rational integral that expands into an alternating series; matching the given form shows $f(n) = \frac{(-1)^{n+1}}{2n-1}$, which is option (C).

   We start with the integral

   $I = \int_{0}^{\pi/2} \tan^{14}\left(\frac{x}{2}\right) dx.$

   The presence of $\tan(x/2)$ strongly suggests the standard tangent half-angle substitution: let

   $t = \tan\left(\frac{x}{2}\right), \quad \text{so that} \quad x = 2\arctan t, \quad dx = \frac{2}{1+t^2}\, dt.$

   When $x = 0$, $t = 0$; when $x = \pi/2$, $t = \tan(\pi/4) = 1$.

   Thus

   $I = \int_{0}^{1} t^{14} \cdot \frac{2}{1+t^2}\, dt = 2 \int_{0}^{1} \frac{t^{14}}{1+t^2}\, dt.$

   Now the integrand is a rational function. Since the denominator is $1+t^2$, we can perform polynomial division or use the geometric series expansion for $|t|<1$ (valid on $[0,1]$ except at the endpoint, but the integral converges). Write

   $\frac{t^{14}}{1+t^2} = t^{14} \cdot \frac{1}{1+t^2} = t^{14} \sum_{k=0}^{\infty} (-1)^k t^{2k} = \sum_{k=0}^{\infty} (-1)^k t^{14+2k}.$

   Integrating termwise from 0 to 1 gives

   $\int_{0}^{1} \frac{t^{14}}{1+t^2}\, dt = \sum_{k=0}^{\infty} (-1)^k \frac{1}{14+2k+1} = \sum_{k=0}^{\infty} \frac{(-1)^k}{2k+15}.$

   So

   $I = 2 \sum_{k=0}^{\infty} \frac{(-1)^k}{2k+15}.$

   The problem states that

   $I = 2\left[ \sum_{n=1}^{7} f(n) - \frac{\pi}{4} \right].$

   We need to relate the infinite series to a finite sum minus $\pi/4$. Recall the Leibniz series for $\pi/4$:

   $\frac{\pi}{4} = \sum_{m=0}^{\infty} \frac{(-1)^m}{2m+1}.$

   Our series starts at $k=0$ with denominator $2k+15 = 2(k+7)+1$. Let $m = k+7$, then $k = m-7$ and when $k=0$, $m=7$. So

   $\sum_{k=0}^{\infty} \frac{(-1)^k}{2k+15} = \sum_{m=7}^{\infty} \frac{(-1)^{m-7}}{2m+1} = (-1)^{-7} \sum_{m=7}^{\infty} \frac{(-1)^m}{2m+1}.$

   Since $(-1)^{-7} = -1$ (because $(-1)^7 = -1$), we have

   $\sum_{k=0}^{\infty} \frac{(-1)^k}{2k+15} = - \sum_{m=7}^{\infty} \frac{(-1)^m}{2m+1}.$

   Now the full Leibniz series is

   $\frac{\pi}{4} = \sum_{m=0}^{\infty} \frac{(-1)^m}{2m+1} = \sum_{m=0}^{6} \frac{(-1)^m}{2m+1} + \sum_{m=7}^{\infty} \frac{(-1)^m}{2m+1}.$

   Thus

   $\sum_{m=7}^{\infty} \frac{(-1)^m}{2m+1} = \frac{\pi}{4} - \sum_{m=0}^{6} \frac{(-1)^m}{2m+1}.$

   Substituting back,

   $\sum_{k=0}^{\infty} \frac{(-1)^k}{2k+15} = - \left( \frac{\pi}{4} - \sum_{m=0}^{6} \frac{(-1)^m}{2m+1} \right) = \sum_{m=0}^{6} \frac{(-1)^m}{2m+1} - \frac{\pi}{4}.$

   Therefore

   $$$$

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8. TG EAPCET 2025Set eng-2025-05-04-FN1 markMCQ
   Q.If $\int_{0}^{\frac{\pi}{2}} \tan^{14}\left(\frac{x}{2}\right) dx = 2 \left[ \sum_{n=1}^{7} f(n) - \frac{\pi}{4} \right]$, then $f(n) =$ (A) $\dfrac{(-1)^{n+1}}{n+1}$ (B) $\dfrac{(-1)^{n+1}}{2n-1}$ (C) $\dfrac{(-1)^n}{n-1}$ (D) $\dfrac{(-1)^n}{2n+1}$

   ›Reveal solutionSolution

   The integral reduces to a sum of alternating reciprocals of odd numbers; comparing forms gives $f(n) = \frac{(-1)^{n+1}}{2n-1}$, which is option (B).

   The problem asks you to match a definite integral to a given summation form. The integral has a high power of $\tan(x/2)$ — that’s a strong hint to use the substitution $t = \tan(x/2)$, which turns trigonometric integrals into rational ones. Once you do that, the integral becomes a standard rational function that expands into an alternating series of odd reciprocals.

   Let’s walk through it.

   1. Substitution Put $t = \tan\frac{x}{2}$. Then $x = 2\tan^{-1}t$, so $dx = \frac{2}{1+t^2}\,dt$. When $x=0$, $t=0$; when $x=\frac{\pi}{2}$, $t = \tan\frac{\pi}{4} = 1$. The integral becomes

   $I = \int_{0}^{\pi/2} \tan^{14}\!\left(\frac{x}{2}\right) dx = \int_{0}^{1} t^{14} \cdot \frac{2}{1+t^2}\,dt.$

   2. Rational function expansion We now have

   $I = 2 \int_{0}^{1} \frac{t^{14}}{1+t^2}\,dt.$

   Since the degree of the numerator (14) is higher than the denominator’s (2), we divide. But there’s a cleaner trick: write

   $\frac{t^{14}}{1+t^2} = t^{12} - t^{10} + t^{8} - t^{6} + t^{4} - t^{2} + 1 - \frac{1}{1+t^2}.$

   Check: multiply $(1+t^2)$ by the polynomial $t^{12} - t^{10} + \cdots + 1$; you get $t^{14} + 1$, so subtracting $\frac{1}{1+t^2}$ recovers the original fraction. This alternating pattern comes from the geometric series formula:

   $\frac{1}{1+t^2} = 1 - t^2 + t^4 - t^6 + \cdots \quad\text{(for }|t|<1\text{)},$

   rearranged.

   3. Integrate term by term

   $I = 2 \int_{0}^{1} \left( t^{12} - t^{10} + t^{8} - t^{6} + t^{4} - t^{2} + 1 - \frac{1}{1+t^2} \right) dt.$

   Each power integrates to $\frac{1}{k+1}$ from 0 to 1, and $\int_0^1 \frac{dt}{1+t^2} = \frac{\pi}{4}$. So

   $I = 2 \left[ \frac{1}{13} - \frac{1}{11} + \frac{1}{9} - \frac{1}{7} + \frac{1}{5} - \frac{1}{3} + 1 - \frac{\pi}{4} \right].$

   4. Match to the given form The problem states $$$$

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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 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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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).

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    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.

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    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 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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