Q.Integrate the following function: $\frac{\sin x}{(1 + \cos x)^2}$

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.$\displaystyle\int \dfrac{\cos^3 x}{(1 + \sin x)^4}\, dx =$ (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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4. 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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5. 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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6. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.If $n \ge 2$ is a natural number and $0 < \theta < \dfrac{\pi}{2}$, then $\displaystyle\int \dfrac{(\cos^n\theta - \cos\theta)^{1/n}}{\cos^{n+1}\theta}\sin\theta\, d\theta =$ (A) $\dfrac{n}{n-1}\left(\cos^{(1-n)}\theta - 1\right)^2 + c$ (B) $\dfrac{n}{(n+1)(1-n)}\left(\cos^{(1-n)}\theta - 1\right)^{1+\frac{1}{n}} + c$ (C) $\dfrac{1}{n-1}\left(\cos^{(n-1)}\theta - 1\right)^2 + c$ (D) $\dfrac{n}{1-n^2}\left(1 - \cos^{(1-n)}\theta\right)^{(n+1)/n}$

   ›Reveal solutionSolution

   Factor out $\cos^n\theta$ from inside the radical to isolate a clean power of $\cos\theta$, then substitute $w=\cos^{1-n}\theta$ to reduce the whole integral to $\int w^{1/n}dw$.

   Concept and Intuition

   The key move is factoring $\cos^n\theta-\cos\theta=\cos^n\theta\left(1-\cos^{1-n}\theta\right)$ so the $n$-th root pulls a clean $\cos\theta$ outside, cancelling nicely against the $\cos^{n+1}\theta$ in the denominator and leaving a single power of $w=\cos^{1-n}\theta$ whose differential exactly matches $\sin\theta\,d\theta/\cos^n\theta$ in the integrand.

   Step-by-Step Solution

   1. Factor: $\cos^n\theta-\cos\theta=\cos^n\theta\left(1-\cos^{1-n}\theta\right)$, so

   $(\cos^n\theta-\cos\theta)^{1/n}=\cos\theta\left(1-\cos^{1-n}\theta\right)^{1/n}.$

   2. Divide by $\cos^{n+1}\theta$:

   $\frac{(\cos^n\theta-\cos\theta)^{1/n}}{\cos^{n+1}\theta}=\frac{\left(1-\cos^{1-n}\theta\right)^{1/n}}{\cos^n\theta}.$

   3. Let $w=\cos^{1-n}\theta$. Then $\dfrac{dw}{d\theta}=(1-n)\cos^{-n}\theta\cdot(-\sin\theta)=(n-1)\sin\theta\cos^{-n}\theta$, so $\dfrac{\sin\theta\,d\theta}{\cos^n\theta}=\dfrac{dw}{n-1}$.
   4. The whole integrand times $d\theta$ becomes $(1-w)^{1/n}\cdot\dfrac{dw}{n-1}$ — wait, more directly $w^{1/n}$ is the factor $\left(1-\cos^{1-n}\theta\right)^{1/n}$ once we track $1-\cos^{1-n}\theta$ as the base; carrying the substitution through consistently: …

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7. 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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8. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{\left(1+\sqrt{x}\right)^{2022}} =$ (A) $\dfrac{2}{(1+\sqrt{x})^{2021}}\left[\dfrac{1+\sqrt{x}}{2020} - \dfrac{1}{2021}\right] + C$ (B) $\dfrac{2}{(1+\sqrt{x})^{2022}}\left[\dfrac{1+\sqrt{x}}{2020} - \dfrac{\sqrt{x}}{2021}\right] + C$ (C) $\dfrac{2}{(1+\sqrt{x})}\left[\dfrac{(1+\sqrt{x})^{2022}}{2022} - \dfrac{(1+\sqrt{x})^{2021}}{2021}\right] + C$ (D) $\dfrac{1}{(1+\sqrt{x})^2}\left[\dfrac{1}{(1+\sqrt{x})^{1010}} - \dfrac{1}{(1+\sqrt{x})^{1011}}\right] + C$

   ›Reveal solutionSolution

   Substituting $t=1+\sqrt x$ turns the integral into a simple power-rule integral in $t$; back-substituting and factoring reproduces option (A)'s bracketed form.

   Concept and Intuition

   Whenever an integrand is a function purely of $1+\sqrt x$, the substitution $t=1+\sqrt x$ (so $\sqrt x = t-1$, $x=(t-1)^2$) turns the messy radical expression into a clean power of $t$, and the pieces of $dx$ that are left over ($2(t-1)\,dt$) combine with the $t^{-2022}$ factor to give a difference of two pure power terms — each integrable by the ordinary power rule.

   Step-by-Step Solution

   1. Let $t=1+\sqrt x$. Then $\sqrt x=t-1$, $x=(t-1)^2$, and $dx=2(t-1)\,dt$.
   2. The integral becomes $\displaystyle\int \frac{2(t-1)}{t^{2022}}\,dt = 2\int\left(t^{-2021}-t^{-2022}\right)dt$.
   3. Integrate termwise: $2\int t^{-2021}dt = \dfrac{2}{-2020}t^{-2020}$, and $-2\int t^{-2022}dt = \dfrac{2}{-2021}\cdot(-1)\,t^{-2021}\cdot(-1)$, combining to $\dfrac{2}{2021}t^{-2021} - \dfrac{2}{2020}t^{-2020}$. …

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9. 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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10. 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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11. 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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12. 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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