Q.Choose the correct answer: $\int \frac{\cos 2x}{(\sin x+\cos x)^2}\,dx$ is equal to (A) $\frac{-1}{\sin x+\cos x}+C$ (B) $\log|\sin x+\cos x|+C$ (C) $\log|\sin x-\cos x|+C$ (D) $\frac{1}{(\sin x+\cos x)^2}$

Showing the 12 most recent of 51 on this concept.

1. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{1}{\left(\sin x + \cos x + \sqrt{2}\sqrt{\sin 2x}\right)^2}\,dx =$ (A) $\dfrac{-(1+3\sqrt{\tan x})}{(3+\tan^2 x)^3} + C$ (B) $\dfrac{-(1+3\sqrt{\tan x})}{3(1+\sqrt{\tan x})^3} + C$ (C) $\dfrac{-(1+\sqrt{\tan x})}{3(1+3\sqrt{\tan x})^2} + C$ (D) $\dfrac{1}{(1+3\sqrt{\tan x})^3} + C$

   ›Reveal solutionSolution

   Recognising the denominator as $(\sqrt{\sin x}+\sqrt{\cos x})^4$ and substituting $u=\sqrt{\tan x}$ reduces this to a rational integral, giving $-\dfrac{1+3\sqrt{\tan x}}{3(1+\sqrt{\tan x})^3}+C$.

   Concept and Intuition

   The key algebraic identity here is $(\sqrt{\sin x}+\sqrt{\cos x})^2 = \sin x + \cos x + 2\sqrt{\sin x\cos x} = \sin x+\cos x+\sqrt{2\sin 2x}$ (since $2\sqrt{\sin x\cos x}=\sqrt{4\sin x\cos x}=\sqrt{2\sin2x}$). That matches the given denominator's base exactly, turning a scary-looking radical expression into a clean fourth power.

   Step-by-Step Solution

   1. Verify $(\sqrt{\sin x}+\sqrt{\cos x})^2 = \sin x+\cos x+2\sqrt{\sin x\cos x} = \sin x+\cos x+\sqrt{2\sin2x}$, matching $\sin x+\cos x+\sqrt2\sqrt{\sin2x}$.
   2. So the denominator is $(\sqrt{\sin x}+\sqrt{\cos x})^4$.
   3. Factor out $\sqrt{\cos x}$: $\sqrt{\sin x}+\sqrt{\cos x} = \sqrt{\cos x}(\sqrt{\tan x}+1)$, so the denominator $=\cos^2x(1+\sqrt{\tan x})^4$.
   4. Integral becomes $\displaystyle\int \dfrac{\sec^2x\,dx}{(1+\sqrt{\tan x})^4}$. Let $t=\tan x$, $dt=\sec^2x\,dx$: $\displaystyle\int \dfrac{dt}{(1+\sqrt t)^4}$.
   5. Let $u=\sqrt t$, $t=u^2$, $dt=2u\,du$: $\displaystyle\int \dfrac{2u\,du}{(1+u)^4}$.
   6. Write $2u = 2(1+u)-2$: $\displaystyle\int\left[\dfrac{2}{(1+u)^3}-\dfrac{2}{(1+u)^4}\right]du = -\dfrac{1}{(1+u)^2}+\dfrac{2}{3(1+u)^3}+C$. …

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2. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{\sec^2 x}{(\sec x+\tan x)^{5/2}}\,dx=$ (A) $-\dfrac{(\sec x+\tan x)^{5/2}}{5}-\dfrac{(\sec x+\tan x)^{7/2}}{7}+c$ (B) $-\dfrac{(\sec x-\tan x)^{5/2}}{5}-\dfrac{(\sec x-\tan x)^{7/2}}{7}+c$ (C) $-\dfrac{(\sec x+\tan x)^{3/2}}{3}-\dfrac{(\sec x+\tan x)^{7/2}}{7}+c$ (D) $-\dfrac{(\sec x-\tan x)^{3/2}}{3}-\dfrac{(\sec x-\tan x)^{7/2}}{7}+c$

   ›Reveal solutionSolution

   A substitution $t=\sec x+\tan x$ (which pairs neatly with $\sec x-\tan x=1/t$) reduces this odd-looking integral to a simple power-rule integral, whose answer re-expresses in terms of $\sec x-\tan x$.

   Concept and Intuition

   Whenever $\sec x+\tan x$ appears, remember its reciprocal identity $(\sec x+\tan x)(\sec x-\tan x)=1$, and that $\dfrac{d}{dx}(\sec x+\tan x)=\sec x(\sec x+\tan x)$ — this makes $t=\sec x+\tan x$ a natural substitution whenever $\sec x\,dx$ multiplies a function of $t$.

   Step-by-Step Solution

   1. Let $t=\sec x+\tan x$. Then $dt=\sec x(\sec x+\tan x)\,dx=\sec x\cdot t\,dx$, so $\sec x\,dx=\dfrac{dt}{t}$.
   2. Also $\sec x-\tan x=\dfrac1t$, so $\sec x=\dfrac{t+1/t}{2}=\dfrac{t^2+1}{2t}$.
   3. The integral $\displaystyle\int\frac{\sec^2x}{t^{5/2}}dx=\int\frac{\sec x}{t^{5/2}}\cdot(\sec x\,dx)=\int\frac{\sec x}{t^{5/2}}\cdot\frac{dt}{t}=\int\frac{\sec x}{t^{7/2}}dt$.
   4. Substitute $\sec x=\dfrac{t^2+1}{2t}$: integral $=\displaystyle\int\frac{t^2+1}{2t^{9/2}}dt=\frac12\int\left(t^{-5/2}+t^{-9/2}\right)dt$.
   5. $=\dfrac12\left[-\dfrac{2}{3}t^{-3/2}-\dfrac27 t^{-7/2}\right]+c=-\dfrac{t^{-3/2}}{3}-\dfrac{t^{-7/2}}{7}+c$. …

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3. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{\cos x - \sin x}{10 + \sin 2x}\, dx =$ (A) $\dfrac{1}{2}\log(10 + \sin 2x) + c$ (B) $\dfrac{1}{3}\log(10 + \sin 2x) + c$ (C) $\dfrac{1}{3}\, \mathrm{Tan}^{-1}\left(\dfrac{\sin x + \cos x}{3}\right) + c$ (D) $\dfrac{1}{3}\, \mathrm{Tan}^{-1}(10 + \sin 2x) + c$

   ›Reveal solutionSolution

   The numerator $\cos x-\sin x$ is exactly $d(\sin x+\cos x)$, and the denominator rewrites in terms of $u=\sin x+\cos x$ via $\sin2x=u^2-1$. That collapses the integral to a standard $\int du/(a^2+u^2)$ arctangent form. Answer: $\dfrac13\mathrm{Tan}^{-1}\!\left(\dfrac{\sin x+\cos x}{3}\right)+c$.

   Concept and Intuition

   Whenever an integrand contains both $\sin x-\cos x$ (or $\cos x-\sin x$) and $\sin2x$, it is worth trying $u=\sin x+\cos x$ (or $\sin x-\cos x$) as the substitution, because $u^2=1\pm\sin2x$ links the two.

   Step-by-Step Solution

   1. Let $u=\sin x+\cos x$. Then $du=(\cos x-\sin x)\,dx$ — this is exactly the numerator times $dx$.
   2. Also $u^2=\sin^2x+\cos^2x+2\sin x\cos x=1+\sin2x$, so $\sin2x=u^2-1$.
   3. Denominator: $10+\sin2x=10+u^2-1=9+u^2$.
   4. The integral becomes $\displaystyle\int\frac{du}{9+u^2}=\frac13\mathrm{Tan}^{-1}\!\left(\frac u3\right)+c$. …

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4. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{4+5\cos x} =$ (A) $-\dfrac{1}{3}\log\left|\dfrac{3+\tan\frac{x}{2}}{3-\tan\frac{x}{2}}\right|+C$ (B) $\dfrac{1}{3}\log\left|\dfrac{3+\tan\frac{x}{2}}{3-\tan\frac{x}{2}}\right|+C$ (C) $-\dfrac{1}{9}\log\left|\dfrac{3-\tan\frac{x}{2}}{3+\tan\frac{x}{2}}\right|+C$ (D) $\dfrac{1}{9}\log\left|\dfrac{3-\tan\frac{x}{2}}{3+\tan\frac{x}{2}}\right|+C$

   ›Reveal solutionSolution

   The Weierstrass substitution $t=\tan(x/2)$ turns this into a standard $\int dt/(a^2-t^2)$ integral, giving option (B).

   Concept and Intuition

   For $\int\dfrac{dx}{a+b\cos x}$ the substitution $t=\tan(x/2)$ (so $\cos x=\tfrac{1-t^2}{1+t^2}$, $dx=\tfrac{2\,dt}{1+t^2}$) always converts the integral into a rational function of $t$ alone.

   Step-by-Step Solution

   1. $\cos x=\dfrac{1-t^2}{1+t^2}$, $dx=\dfrac{2\,dt}{1+t^2}$, with $t=\tan(x/2)$.
   2. $4+5\cos x=4+5\cdot\dfrac{1-t^2}{1+t^2}=\dfrac{4(1+t^2)+5(1-t^2)}{1+t^2}=\dfrac{9-t^2}{1+t^2}$.
   3. Integral becomes $\int\dfrac{2\,dt/(1+t^2)}{(9-t^2)/(1+t^2)}=\int\dfrac{2\,dt}{9-t^2}$.
   4. Using $\int\dfrac{dt}{a^2-t^2}=\dfrac{1}{2a}\log\left|\dfrac{a+t}{a-t}\right|+C$ with $a=3$: $\int\dfrac{dt}{9-t^2}=\dfrac16\log\left|\dfrac{3+t}{3-t}\right|+C$. …

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5. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
   Q.$\int \cos\sqrt{x}\, dx =$ (A) $2\sqrt{x}\sin\sqrt{x} + 2\cos\sqrt{x} + c$ (B) $2\sqrt{x}\sin\sqrt{x} + 2\sin\sqrt{x} + c$ (C) $2\sqrt{x}\sin\sqrt{x} - 2\cos\sqrt{x} + c$ (D) $\sqrt{x}\cos\sqrt{x} - 2\sin\sqrt{x} + c$

   ›Reveal solutionSolution

   Substitute $t=\sqrt x$ to turn the integral into a standard integration-by-parts problem. Answer: $2\sqrt x\sin\sqrt x+2\cos\sqrt x+c$.

   Concept and Intuition

   Whenever you see $\sqrt x$ trapped inside a trig or exponential function, substituting $t=\sqrt x$ converts it into a polynomial-times-trig integral solvable by parts.

   Step-by-Step Solution

   1. Let $t=\sqrt x \Rightarrow x=t^2,\ dx=2t\,dt$.
   2. $\int\cos\sqrt x\,dx = \int\cos t\cdot 2t\,dt = 2\int t\cos t\,dt$.
   3. Integrate by parts: $\int t\cos t\,dt = t\sin t - \int\sin t\,dt = t\sin t+\cos t$. …

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6. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{1}{\cos 4x \cos 2x}\,dx = \dfrac{1}{4\sqrt{2}}\log\left(\dfrac{1+f(x)}{1-f(x)}\right) - \dfrac{1}{2}\log g(x) + C$, then $g\left(\dfrac{\pi}{6}\right) - \sqrt{2}\,f\left(\dfrac{\pi}{6}\right) =$ (A) $\dfrac{\pi}{2\sqrt{2}}$ (B) $\pi + 3$ (C) $2$ (D) $1$

   ›Reveal solutionSolution

   Solving the integral via $t=\sin2x$ identifies $f(x)=\sqrt2\sin2x$ and $g(x)=\sqrt{\frac{1+\sin2x}{1-\sin2x}}$; evaluating at $x=\pi/6$ gives $g(\pi/6)-\sqrt2f(\pi/6)=2$.

   Concept and Intuition

   The integral $\int\dfrac{dx}{\cos4x\cos2x}$ is tackled by substituting $t=\sin2x$, since $\cos4x=1-2\sin^2 2x=1-2t^2$ turns the whole integrand into a rational function of $t$, solvable by partial fractions into two logarithmic terms — one built from $1-t^2$ and one from $1-2t^2$, matching exactly the two-log structure given in the problem.

   Step-by-Step Solution

   1. Let $t=\sin2x$, so $dt=2\cos2x\,dx$ and $\cos4x=1-2t^2$.
   2. Rewriting the integral in terms of $t$: $\displaystyle\int\dfrac{dx}{\cos4x\cos2x} = \int\dfrac{dt}{2(1-t^2)(1-2t^2)}$.
   3. Partial fractions: $\dfrac{1}{(1-t^2)(1-2t^2)} = \dfrac{-1}{1-t^2}+\dfrac{2}{1-2t^2}$.
   4. Integrating each piece gives standard log forms: one in $\dfrac{1+t}{1-t}$ (from the $1-t^2$ term) and one in $\dfrac{1+\sqrt2 t}{1-\sqrt2 t}$ (from the $1-2t^2$ term), exactly matching the pattern $\dfrac{1}{4\sqrt2}\log\dfrac{1+f}{1-f} - \dfrac12\log g$ with $f(x)=\sqrt2\sin2x$ and $g(x)=\sqrt{\dfrac{1+\sin2x}{1-\sin2x}}$.
   5. Evaluate at $x=\pi/6$: $2x=\pi/3$, $\sin(\pi/3)=\dfrac{\sqrt3}{2}$. …

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7. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{\sqrt{\sin^3 x \cos(x-\alpha)}} =$ (A) $\dfrac{1}{\cos\alpha}\sqrt{\cot x+\tan\alpha}+c$ (B) $\dfrac{1}{\sqrt{\cos\alpha}}\sqrt{\cot x-\tan\alpha}+c$ (C) $\dfrac{-1}{\sqrt{\sin\alpha}}\sqrt{\cot x+\tan\alpha}+c$ (D) $\dfrac{-2}{\sqrt{\cos\alpha}}\sqrt{\cot x+\tan\alpha}+c$

   ›Reveal solutionSolution

   Expanding $\cos(x-\alpha)$ and substituting $u=\cot x$ reduces this to a simple square-root integral; final answer is (D).

   Concept and Intuition

   The key move is expanding $\cos(x-\alpha)$ and factoring out $\sin^4x$ from inside the square root so that a substitution $u=\cot x$ (whose differential is $-\csc^2x\,dx$, conveniently matching what's left outside) linearizes the whole thing.

   Step-by-Step Solution

   1. $\cos(x-\alpha)=\cos x\cos\alpha+\sin x\sin\alpha$.
   2. $\sin^3x\cos(x-\alpha)=\sin^3x\cos x\cos\alpha+\sin^4x\sin\alpha = \sin^4x\left(\dfrac{\cos x}{\sin x}\cos\alpha+\sin\alpha\right)=\sin^4x(\cot x\cos\alpha+\sin\alpha)$.
   3. So $\sqrt{\sin^3x\cos(x-\alpha)}=\sin^2x\sqrt{\cot x\cos\alpha+\sin\alpha}$ (taking $\sin^2x>0$ outside the root).
   4. The integral is $\displaystyle\int\dfrac{dx}{\sin^2x\sqrt{\cos\alpha\cot x+\sin\alpha}}=\int\dfrac{\csc^2x\,dx}{\sqrt{\cos\alpha\cot x+\sin\alpha}}$.
   5. Let $u=\cot x$, $du=-\csc^2x\,dx$. Integral becomes $-\displaystyle\int\dfrac{du}{\sqrt{\cos\alpha\,u+\sin\alpha}}=-\dfrac{2\sqrt{\cos\alpha\,u+\sin\alpha}}{\cos\alpha}+c$. …

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8. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{\sin^2 x \cos^2 x}{\cos^6 x + \sin^6 x}\,dx =$ (A) $\dfrac{1}{2}\mathrm{Tan}^{-1}(\tan^2 x) + c$ (B) $\dfrac{1}{3}\mathrm{Tan}^{-1}(\tan^2 x) + c$ (C) $\dfrac{1}{3}\mathrm{Tan}^{-1}(\tan^3 x) + c$ (D) $\mathrm{Tan}^{-1}(\tan^3 x) + c$

   ›Reveal solutionSolution

   Dividing through by $\cos^6x$ to introduce $\tan x$, then a double substitution ($t=\tan x$, then $u=t^3$), reduces the integral to a standard arctangent form, giving $\dfrac13\mathrm{Tan}^{-1}(\tan^3x)+c$.

   Concept and Intuition

   When an integrand has $\sin$ and $\cos$ appearing only in even powers that can be grouped, dividing everything by the highest power of $\cos x$ converts the whole expression into a rational function purely of $\tan x$ — a very common trick that opens the door to the substitution $t=\tan x$. Here, since the denominator naturally forms $1+\tan^6x$, a further substitution $u=t^3$ turns it into the standard $\int\frac{du}{1+u^2}$ arctangent integral.

   Step-by-Step Solution

   1. Divide numerator and denominator of $\dfrac{\sin^2x\cos^2x}{\cos^6x+\sin^6x}$ by $\cos^6x$: numerator becomes $\dfrac{\sin^2x\cos^2x}{\cos^6x} = \dfrac{\sin^2x}{\cos^4x} = \tan^2x\sec^2x$; denominator becomes $1 + \tan^6x$.
   2. The integral is now $\displaystyle\int \dfrac{\tan^2x\sec^2x}{1+\tan^6x}\,dx$.
   3. Substitute $t=\tan x$, so $dt = \sec^2x\,dx$. The integral becomes $\displaystyle\int\dfrac{t^2}{1+t^6}\,dt$.
   4. Substitute $u = t^3$, so $du = 3t^2\,dt$, i.e. $t^2\,dt = \dfrac{du}{3}$. The integral becomes $\dfrac13\displaystyle\int\dfrac{du}{1+u^2}$. …

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9. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.$\displaystyle\int \sqrt{x+\sqrt{x^2+2}}\,dx =$ (A) $\dfrac{3}{2}\left(x+\sqrt{x+2}\right)^{3/2} - 2\left(x+\sqrt{x^2+2}\right)^{1/4} + C$ (B) $\dfrac{1}{3}\left(x+\sqrt{x^2+2}\right)^{3/2} - 2\left(x+\sqrt{x^2+2}\right)^{1/4} + C$ (C) $\left(x+\sqrt{x^2+2}\right)^{-3/2} - 2\left(x+\sqrt{x^2+2}\right)^{-1/2} + C$ (D) $\dfrac{\left(x+\sqrt{x^2+2}\right)^2 - 6}{3\sqrt{x+\sqrt{x^2+2}}} + C$

   ›Reveal solutionSolution

   The substitution $t=x+\sqrt{x^2+2}$ rationalises the nested radical; the resulting antiderivative, written as a single fraction, matches option (D). Answer: option (D).

   Concept and Intuition

   Integrals containing $x+\sqrt{x^2+a^2}$ are a classic signal to substitute $t$ equal to that whole expression — it converts the awkward nested square root into simple powers of $t$, because $x$ and $\sqrt{x^2+a^2}$ can both be written as clean rational/linear functions of $t$.

   Step-by-Step Solution

   1. Let $t=x+\sqrt{x^2+2}$. Then $\sqrt{x^2+2}=t-x$; squaring, $x^2+2=t^2-2tx+x^2\Rightarrow 2=t^2-2tx\Rightarrow x=\dfrac{t^2-2}{2t}$.
   2. Then $\sqrt{x^2+2}=t-x=t-\dfrac{t^2-2}{2t}=\dfrac{2t^2-t^2+2}{2t}=\dfrac{t^2+2}{2t}$.
   3. Differentiate $t$ w.r.t. $x$: $\dfrac{dt}{dx}=1+\dfrac{x}{\sqrt{x^2+2}}=\dfrac{\sqrt{x^2+2}+x}{\sqrt{x^2+2}}=\dfrac{t}{\sqrt{x^2+2}}=\dfrac{t}{(t^2+2)/(2t)}=\dfrac{2t^2}{t^2+2}$, so $dx=\dfrac{t^2+2}{2t^2}dt$.
   4. Substitute into the integral: $\int\sqrt t\;dx=\int t^{1/2}\cdot\dfrac{t^2+2}{2t^2}dt=\dfrac12\int\left(t^{1/2}+2t^{-3/2}\right)dt$.
   5. Integrate: $\dfrac12\left[\dfrac23t^{3/2}-4t^{-1/2}\right]+C=\dfrac13t^{3/2}-2t^{-1/2}+C$. …

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10. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
    Q.If $\displaystyle\int \dfrac{1}{\sin^3 x+\cos^3 x}\,dx = A\log\left|\dfrac{\sqrt{2}+t}{\sqrt{2}-t}\right|+B\,\mathrm{Tan}^{-1}(t)+c$, then $\left(\dfrac{B}{A},t\right)=$ (A) $(2\sqrt{2},\sin x+\cos x)$ (B) $\left(\dfrac{\sqrt{2}}{9},\sin x+\cos x\right)$ (C) $\left(\dfrac{\sqrt{2}}{9},\sin x-\cos x\right)$ (D) $(2\sqrt{2},\sin x-\cos x)$

    ›Reveal solutionSolution

    Factoring the sum of cubes and substituting $t=\sin x-\cos x$ turns the trigonometric integral into a clean rational-function integral in $t$, giving $B/A=2\sqrt2$ with $t=\sin x-\cos x$.

    Concept and Intuition

    $\sin^3x+\cos^3x$ factors as a sum of cubes: $(\sin x+\cos x)(1-\sin x\cos x)$. Both $\sin x+\cos x$ and $1-\sin x\cos x$ can be written purely in terms of $u=\sin x-\cos x$, because $(\sin x+\cos x)^2+(\sin x-\cos x)^2=2$ and $1-\sin x\cos x=\tfrac{1+(\sin x-\cos x)^2}{2}$. Crucially, $\dfrac{du}{dx}=\cos x+\sin x$, which is exactly the factor left over after using the second identity — so the whole integral collapses into a rational function of $u$ alone.

    Step-by-Step Solution

    1. Factor: $\sin^3x+\cos^3x=(\sin x+\cos x)(1-\sin x\cos x)$.
    2. Let $u=\sin x-\cos x$. Then $u^2=1-2\sin x\cos x$, so $1-\sin x\cos x=\dfrac{1+u^2}{2}$.
    3. Also $(\sin x+\cos x)^2=1+2\sin x\cos x=2-u^2$, so $\sin x+\cos x=\sqrt{2-u^2}$ (taking the appropriate branch), and $\dfrac{du}{dx}=\cos x+\sin x=\sqrt{2-u^2}$.
    4. So $\sin^3x+\cos^3x=\sqrt{2-u^2}\cdot\dfrac{1+u^2}{2}$, and

    $I=\int\frac{dx}{\sin^3x+\cos^3x}=\int\frac{2\,dx}{\sqrt{2-u^2}(1+u^2)}=\int\frac{2}{\sqrt{2-u^2}(1+u^2)}\cdot\frac{du}{\sqrt{2-u^2}}=\int\frac{2\,du}{(2-u^2)(1+u^2)}.$

    5. Partial fractions (by symmetry, only even terms survive): $\dfrac{2}{(2-u^2)(1+u^2)}=\dfrac{2/3}{2-u^2}+\dfrac{2/3}{1+u^2}$. …

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11. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
    Q.If $f(x) = \dfrac{x}{(1 + nx^n)^{1/n}}$ for $n \geq 2$, then $\displaystyle\int x^{n-2} f(x)\, dx =$ (A) $\dfrac{1}{n(n-1)} (1 + nx^n)^{1 - \frac{1}{n}} + C$ (B) $\dfrac{1}{n-1} (1 + nx^n)^{1 - \frac{1}{n}} + C$ (C) $\dfrac{1}{n(n-1)} (1 + nx^n)^{1 + \frac{1}{n}} + C$ (D) $\dfrac{1}{n+1} (1 + nx^n)^{1 + \frac{1}{n}} + C$

    ›Reveal solutionSolution

    A direct substitution $u=1+nx^n$ turns the integral into a simple power rule.

    Concept and Intuition

    The integrand's power of $x$ (namely $x^{n-1}$) is exactly proportional to the derivative of $u=1+nx^n$, which is the classic signal to substitute.

    Step-by-Step Solution

    1. $x^{n-2}f(x)=x^{n-2}\cdot\dfrac{x}{(1+nx^n)^{1/n}}=\dfrac{x^{n-1}}{(1+nx^n)^{1/n}}$.
    2. Let $u=1+nx^n$. Then $du=n\cdot n x^{n-1}dx=n^2x^{n-1}dx$, so $x^{n-1}dx=\dfrac{du}{n^2}$.
    3. Integral becomes $\displaystyle\int u^{-1/n}\cdot\dfrac{du}{n^2}=\dfrac{1}{n^2}\cdot\dfrac{u^{1-1/n}}{1-1/n}+C$.
    4. Simplify: $\dfrac{1}{n^2}\cdot\dfrac{1}{\frac{n-1}{n}}=\dfrac{1}{n^2}\cdot\dfrac{n}{n-1}=\dfrac{1}{n(n-1)}$. …

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12. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
    Q.If $\displaystyle\int \dfrac{dx}{\sin^3 x+\cos^3 x} = A\log\left|\dfrac{\sqrt{2}+t}{\sqrt{2}-t}\right|+B\mathrm{Tan}^{-1}(t)+c$, then $\left(\dfrac{B}{A},t\right) =$ (A) $(3\sqrt{2},\sin x-\cos x)$ (B) $(2\sqrt{2},\sin x-\cos x)$ (C) $\left(\dfrac{\sqrt{2}}{3},\sin x-\cos x\right)$ (D) $\left(\dfrac{3}{\sqrt{2}},\sin x+\cos x\right)$

    ›Reveal solutionSolution

    The standard $\sin^3x+\cos^3x$ integral, solved via the substitution $t=\sin x-\cos x$; matching to the given $A\log|\cdots|+B\tan^{-1}t$ form gives $B/A=2\sqrt2$ with $t=\sin x-\cos x$.

    Concept and Intuition

    $\sin^3x+\cos^3x$ factors as a sum of cubes, $(\sin x+\cos x)(\sin^2x-\sin x\cos x+\cos^2x)=(\sin x+\cos x)(1-\sin x\cos x)$. Both remaining factors can be expressed in terms of $t=\sin x-\cos x$ (since $t^2=1-2\sin x\cos x$ links $\sin x\cos x$ to $t$, and $dt=(\sin x+\cos x)\,dx$ conveniently cancels the other factor).

    Step-by-Step Solution

    1. Factor: $\sin^3x+\cos^3x = (\sin x+\cos x)(1-\sin x\cos x)$.
    2. Let $t=\sin x-\cos x \Rightarrow dt = (\cos x+\sin x)\,dx$, and $t^2 = 1-2\sin x\cos x \Rightarrow \sin x\cos x = \dfrac{1-t^2}{2}$.
    3. So $1-\sin x\cos x = 1-\dfrac{1-t^2}{2} = \dfrac{1+t^2}{2}$.
    4. Also $(\sin x+\cos x)^2 = 1+2\sin x\cos x = 1+(1-t^2) = 2-t^2$, so $\sin x+\cos x=\sqrt{2-t^2}$.
    5. Rewrite the integral:

    $\int\frac{dx}{(\sin x+\cos x)(1-\sin x\cos x)} = \int\frac{dx}{\sqrt{2-t^2}\cdot\frac{1+t^2}{2}}.$

    Since $dx = \dfrac{dt}{\sin x+\cos x} = \dfrac{dt}{\sqrt{2-t^2}}$:

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

    6. Split using $1=\dfrac{(2-t^2)+(1+t^2)}{3}$:

    $\frac{2}{(2-t^2)(1+t^2)} = \frac23\cdot\frac{(2-t^2)+(1+t^2)}{(2-t^2)(1+t^2)} = \frac23\left[\frac{1}{1+t^2}+\frac{1}{2-t^2}\right].$

    7. Integrate each piece: $\displaystyle\int\frac{dt}{1+t^2}=\tan^{-1}t$; $\displaystyle\int\frac{dt}{2-t^2}=\frac{1}{2\sqrt2}\log\left|\frac{\sqrt2+t}{\sqrt2-t}\right|$ (standard form $\int\frac{dx}{a^2-x^2}=\frac{1}{2a}\log\left|\frac{a+x}{a-x}\right|$ with $a=\sqrt2$).
    8. So the integral is …

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