Q.Evaluate the definite integral $\int_{0}^{\pi/4}\frac{\sin x\cos x}{\cos^4 x+\sin^4 x}\,dx$

Showing the 12 most recent of 51 on this concept.

1. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{\cos^3 x}{(1 + \sin x)^4}\, dx =$ (A) $-\dfrac{\cos^4 x}{5(1 + \sin x)^5} + c$ (B) $\dfrac{\cos^4 x}{5(1 + \sin x)^5} + c$ (C) $\dfrac{\cos^4 x}{4(1 + \sin x)^4} + c$ (D) $-\dfrac{\cos^4 x}{4(1 + \sin x)^4} + c$

   ›Reveal solutionSolution

   Factor $\cos^3x$ using $\cos^2x=(1-\sin x)(1+\sin x)$ and substitute $t=\sin x$; the resulting antiderivative can equivalently be written in the $\cos^4x/(1+\sin x)^4$ form given in the options (they differ only by an added constant). Answer: $-\dfrac{\cos^4x}{4(1+\sin x)^4}+c$.

   Concept and Intuition

   Integrals of $\cos^{\text{odd}}x$ over powers of $(1+\sin x)$ are handled by peeling off one factor of $\cos x$ to pair with $dx$ (making $d(\sin x)$) and expressing the remaining even power of $\cos x$ in terms of $\sin x$. Since the MCQ options are phrased in terms of $\cos^4x$ rather than $\sin x$ directly, it is often faster (and safer against sign traps) to guess-and-check an antiderivative of that shape by differentiating a general form $A\cos^4x(1+\sin x)^{-n}$ and matching powers/coefficients — this is exactly how the printed option is confirmed.

   Step-by-Step Solution

   1. $\cos^3x=\cos x\cdot\cos^2x=\cos x(1-\sin^2x)=\cos x(1-\sin x)(1+\sin x)$.
   2. Integrand $=\dfrac{\cos x(1-\sin x)(1+\sin x)}{(1+\sin x)^4}=\dfrac{\cos x(1-\sin x)}{(1+\sin x)^3}$.
   3. Let $t=\sin x$, $dt=\cos x\,dx$: $I=\displaystyle\int\frac{1-t}{(1+t)^3}dt$. Writing $1-t=2-(1+t)$: $I=\displaystyle\int\left(\frac{2}{(1+t)^3}-\frac{1}{(1+t)^2}\right)dt=-\frac{1}{(1+t)^2}+\frac{1}{1+t}+c=\frac{t}{(1+t)^2}+c$.
   4. So $I=\dfrac{\sin x}{(1+\sin x)^2}+c$ is one valid closed form. …

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2. 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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3. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.If $\displaystyle\int \dfrac{\cos^3 x}{\sin^2 x + \sin^4 x}dx = c - \mathrm{cosec}\,x - f(x)$, then $f\left(\dfrac{\pi}{2}\right)=$ (A) 1 (B) 0 (C) $\dfrac{\pi}{2}$ (D) $\pi$

   ›Reveal solutionSolution

   Substituting $s=\sin x$ and partial-fractioning gives $-\mathrm{cosec}\,x-2\mathrm{Tan}^{-1}(\sin x)$, so $f(x)=2\mathrm{Tan}^{-1}(\sin x)$ and $f(\pi/2)=\pi/2$.

   Concept and Intuition

   Writing $\cos^3x\,dx = \cos^2x\cdot\cos x\,dx=(1-\sin^2x)\,d(\sin x)$ converts a trig integral into an algebraic one in $s=\sin x$, which is then handled by ordinary partial fractions.

   Step-by-Step Solution

   1. Let $s=\sin x$, so $ds=\cos x\,dx$, and $\cos^3x\,dx=(1-s^2)\,ds$.
   2. The integral becomes $\displaystyle\int\frac{1-s^2}{s^2+s^4}\,ds=\int\frac{1-s^2}{s^2(1+s^2)}\,ds$.
   3. Partial fractions (in $u=s^2$): $\dfrac{1-u}{u(1+u)}=\dfrac1u-\dfrac2{1+u}$, so the integrand is $\dfrac1{s^2}-\dfrac2{1+s^2}$.
   4. Integrating: $\displaystyle\int\left(\frac1{s^2}-\frac2{1+s^2}\right)ds=-\frac1s-2\,\mathrm{Tan}^{-1}s+C=-\mathrm{cosec}\,x-2\,\mathrm{Tan}^{-1}(\sin x)+C$. …

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4. AP EAPCET 2021Set eng-2021-10-05-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{\sin x}{\sin 3x}\, dx =$ (A) $\dfrac{1}{2\sqrt{3}} \log\left|\dfrac{\sqrt{3} + \tan x}{\sqrt{3} - \tan x}\right| + c$ (B) $\dfrac{1}{2\sqrt{3}} \log\left|\dfrac{\sqrt{3} - \tan x}{\sqrt{3} + \tan x}\right| + c$ (C) $\dfrac{1}{4\sqrt{3}} \log\left|\dfrac{\sqrt{3} + \tan x}{\sqrt{3} - \tan x}\right| + c$ (D) $\dfrac{1}{4\sqrt{3}} \log\left|\dfrac{\sqrt{3} - \tan x}{\sqrt{3} + \tan x}\right| + c$

   ›Reveal solutionSolution

   Use the triple-angle identity to cancel $\sin x$, rewrite in terms of $\cos2x$, then apply the Weierstrass-type substitution $t=\tan x$ to reduce to a standard rational integral. The answer is (A).

   Concept and Intuition

   $\sin3x$ factors as $\sin x(3-4\sin^2x)$, so $\sin x$ cancels immediately with the numerator, turning a trigonometric-looking integral into a much simpler one in $\sin^2x$ (hence in $\cos2x$), which is a textbook target for the $t=\tan x$ substitution.

   Step-by-Step Solution

   1. $\sin3x = 3\sin x - 4\sin^3x = \sin x(3-4\sin^2x)$.
   2. $\dfrac{\sin x}{\sin3x} = \dfrac{1}{3-4\sin^2x}$.
   3. Using $\sin^2x=\dfrac{1-\cos2x}2$: $4\sin^2x = 2-2\cos2x$, so $3-4\sin^2x = 1+2\cos2x$.
   4. Integral becomes $\displaystyle\int\dfrac{dx}{1+2\cos2x}$.
   5. Substitute $t=\tan x$, $dx=\dfrac{dt}{1+t^2}$, $\cos2x=\dfrac{1-t^2}{1+t^2}$: $1+2\cos2x = \dfrac{(1+t^2)+2(1-t^2)}{1+t^2} = \dfrac{3-t^2}{1+t^2}$.
   6. Integral $= \displaystyle\int \dfrac{1+t^2}{3-t^2}\cdot\dfrac{dt}{1+t^2} = \int\dfrac{dt}{3-t^2}$. …

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5. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{x^2(x^4+1)^{3/4}}=$ (A) $\left(1+\dfrac{1}{x^4}\right)^{3/4}+c$ (B) $\left(1+\dfrac{1}{x^6}\right)^{1/2}+c$ (C) $-\left(1+\dfrac{1}{x^4}\right)^{-1/4}+c$ (D) $-\left(1+\dfrac{1}{x^4}\right)^{1/4}+c$

   ›Reveal solutionSolution

   Pulling $x^4$ out from under the radical and substituting $t=1+x^{-4}$ reduces the integral to a simple power rule, giving $-(1+1/x^4)^{1/4}+c$.

   Concept and Intuition

   When the integrand has $(x^n+1)^p$, it often helps to factor out the highest power of $x$ from inside the bracket so that a substitution like $t = 1+x^{-n}$ produces a clean differential matching the rest of the integrand.

   Step-by-Step Solution

   1. Write $(x^4+1)^{3/4} = \left(x^4\left(1+\dfrac{1}{x^4}\right)\right)^{3/4} = x^3\left(1+\dfrac{1}{x^4}\right)^{3/4}$.
   2. So $\dfrac{1}{x^2(x^4+1)^{3/4}} = \dfrac{1}{x^2\cdot x^3\left(1+\frac{1}{x^4}\right)^{3/4}} = \dfrac{1}{x^5\left(1+\frac{1}{x^4}\right)^{3/4}} = x^{-5}\left(1+x^{-4}\right)^{-3/4}$.
   3. Let $t = 1+x^{-4}$. Then $dt = -4x^{-5}dx$, so $x^{-5}dx = -\dfrac{dt}{4}$. …

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6. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{x^3 \tan^{-1} x^4}{1 + x^8}\,dx =$ (A) $\dfrac{(\tan^{-1}(x^4))^2}{8} + c$ (B) $\dfrac{(\tan^{-1}(x^4))^3}{3} + c$ (C) $\dfrac{(\tan^{-1}(x^4))^2}{4} + c$ (D) $\dfrac{(\tan^{-1}(x^4))^2}{2} + c$

   ›Reveal solutionSolution

   A double substitution (first $u=x^4$, then $v=\tan^{-1}u$) turns this into a trivial $\int v\,dv$. Answer: $\dfrac{(\tan^{-1}x^4)^2}{8}+c$.

   Concept and Intuition

   The presence of $x^3\,dx$ alongside $x^4$ inside the arctan and $x^8=(x^4)^2$ in the denominator is a strong signal to substitute $u=x^4$ first. After that, the structure $\tan^{-1}u \cdot \dfrac{du}{1+u^2}$ is exactly of the form "function times its own derivative," solved by a second substitution.

   Step-by-Step Solution

   1. Let $u=x^4$, so $du=4x^3\,dx \Rightarrow x^3\,dx = \dfrac{du}{4}$.
   2. The integral becomes $\displaystyle\int \frac{\tan^{-1}u}{1+u^2}\cdot\frac{du}{4} = \frac14\int \frac{\tan^{-1}u}{1+u^2}\,du$. …

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7. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{\cos^4 x}{\left(\sin^2 x + \sin^{-3} x \cos^5 x\right)^{3}}\;dx =$ (A) $\dfrac{1}{5}\left(1 + \cot^5 x\right)^{-2} + C$ (B) $\dfrac{1}{10}\left(1 + \cot^2 x\right)^{-5} + C$ (C) $\dfrac{1}{10}\left(1 + \cot^5 x\right)^{-2} + C$ (D) $\dfrac{1}{5}\left(1 + \cot^5 x\right)^{-5} + C$

   ›Reveal solutionSolution

   Factoring $\sin^2x$ out of the denominator turns it into $\sin^2x(1+\cot^5x)$, and the substitution $t=1+\cot^5x$ makes the whole integral a simple power-rule integration, giving $\dfrac{1}{10}(1+\cot^5x)^{-2}+C$.

   Concept and Intuition

   Integrals with mixed powers of $\sin x$ and $\cos x$ in odd/negative combinations often simplify beautifully once you factor out a common power to expose a $(1+\cot^n x)$ or $(1+\tan^n x)$ structure — this is exactly the kind of expression whose derivative (via chain rule) reproduces $\cot^{n-1}x\csc^2x$, matching what's left over in the integrand.

   Step-by-Step Solution

   1. Denominator: $\sin^2x+\sin^{-3}x\cos^5x$. Factor out $\sin^2x$: $=\sin^2x\left[1+\dfrac{\cos^5x}{\sin^5x}\right]=\sin^2x\,(1+\cot^5x)$.
   2. So the full denominator cubed: $\left[\sin^2x(1+\cot^5x)\right]^3=\sin^6x\,(1+\cot^5x)^3$.
   3. Integrand: $\dfrac{\cos^4x}{\sin^6x(1+\cot^5x)^3} = \dfrac{\cos^4x}{\sin^4x}\cdot\dfrac{1}{\sin^2x}\cdot(1+\cot^5x)^{-3}=\cot^4x\,\csc^2x\,(1+\cot^5x)^{-3}$.
   4. Substitute $t=1+\cot^5x$. Then $\dfrac{dt}{dx}=5\cot^4x\cdot(-\csc^2x)=-5\cot^4x\csc^2x$, so $\cot^4x\csc^2x\,dx=-\dfrac{dt}{5}$. …

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8. 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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9. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.$\int\frac{x^4 - 1}{x^2\sqrt{x^4 + x^2 + 1}}dx =$ (A) $\frac{2\sqrt{x^4+x^2+1}}{x} + c$ (B) $\frac{\sqrt{x^4+x^2+1}}{x} + c$ (C) $\frac{\sqrt{x^4+x^2+1}}{2x} + c$ (D) $\frac{4\sqrt{x^4+x^2+1}}{x} + c$

   ›Reveal solutionSolution

   Differentiating the candidate $\frac{\sqrt{x^4+x^2+1}}{x}$ reproduces the given integrand exactly, confirming it as the antiderivative.

   Concept and Intuition

   When an integrand looks like it could come from a quotient rule (a square root over a power of $x$), it is often faster to differentiate a plausible candidate of that shape and check, rather than search for a substitution from scratch.

   Step-by-Step Solution

   1. Try $g(x)=\dfrac{\sqrt{x^4+x^2+1}}{x}=\dfrac{N}{x}$ where $N=\sqrt{x^4+x^2+1}$.
   2. $N'=\dfrac{4x^3+2x}{2\sqrt{x^4+x^2+1}}=\dfrac{x(2x^2+1)}{N}$.
   3. Quotient rule: $g'(x)=\dfrac{N'x-N}{x^2}=\dfrac{\frac{x^2(2x^2+1)}{N}-N}{x^2}=\dfrac{x^2(2x^2+1)-N^2}{Nx^2}$.
   4. $N^2=x^4+x^2+1$, so the numerator is $x^2(2x^2+1)-(x^4+x^2+1)=2x^4+x^2-x^4-x^2-1=x^4-1$.
   5. So $g'(x)=\dfrac{x^4-1}{x^2\sqrt{x^4+x^2+1}}$ — exactly the given integrand. …

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10. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
    Q.$\displaystyle\int \dfrac{\sin^6 x}{\cos^8 x}\,dx =$ (A) $\tan 7x + c$ (B) $\dfrac{\tan^7 x}{7} + c$ (C) $\dfrac{\tan 7x}{7} + c$ (D) $\sec^7 x$

    ›Reveal solutionSolution

    $\int \tan^6x\sec^2x\,dx = \dfrac{\tan^7x}{7}+c$ via $u=\tan x$.

    Concept and Intuition

    $\sin^6x/\cos^8x = \tan^6x\cdot\sec^2x$, and $\sec^2x\,dx$ is exactly $d(\tan x)$ — a direct power-rule substitution.

    Step-by-Step Solution

    1. $\dfrac{\sin^6x}{\cos^8x} = \tan^6x\cdot\dfrac{1}{\cos^2x} = \tan^6x\sec^2x$.
    2. Let $u=\tan x \Rightarrow du=\sec^2x\,dx$.
    3. $\int u^6\,du = \dfrac{u^7}{7}+c = \dfrac{\tan^7x}{7}+c$.

    Common Mistakes …

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11. 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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12. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
    Q.Find the value of $k$ if $\displaystyle\int \cos^k(x)\sin(x)\,dx = \dfrac{-1}{4}\cos^4(x) + c$ (A) $4$ (B) $3$ (C) $2$ (D) $1$

    ›Reveal solutionSolution

    Differentiate the given antiderivative and match powers of $\cos x$ to find $k=3$.

    Concept and Intuition

    When an integral is given in the form $\int \cos^k(x)\sin(x)\,dx = F(x)+c$, the fastest way to find $k$ is not to integrate but to differentiate the claimed answer $F(x)$ — the derivative must reproduce the original integrand exactly.

    Step-by-Step Solution

    1. Differentiate $F(x) = -\dfrac14\cos^4x$:

    $F'(x) = -\dfrac14\cdot 4\cos^3x\cdot(-\sin x) = \cos^3x\sin x.$

    2. This must equal the integrand $\cos^k x\sin x$.
    3. Comparing powers of $\cos x$: $k=3$. …

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