Q.Integrate the following function: $\frac{\sin^{-1}x}{\sqrt{1 - x^2}}$

Showing the 12 most recent of 51 on this concept.

1. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{(1 + \sqrt{x})\sqrt{x - x^2}} =$ (A) $-2\sqrt{\dfrac{1 + \sqrt{x}}{1 - \sqrt{x}}} + c$ (B) $-\sqrt{\dfrac{1 - \sqrt{x}}{1 + \sqrt{x}}} + c$ (C) $-2\sqrt{\dfrac{1 - \sqrt{x}}{1 + \sqrt{x}}} + c$ (D) $2\sqrt{\dfrac{1 + \sqrt{x}}{1 - \sqrt{x}}} + c$

   ›Reveal solutionSolution

   Substituting $t=\sqrt x$ turns the surd-heavy integrand into $\int \frac{2\,dt}{(1+t)^{3/2}(1-t)^{1/2}}$, whose antiderivative is exactly $-2\sqrt{\frac{1-t}{1+t}}$.

   Concept and Intuition

   When an integrand mixes $\sqrt x$ and $\sqrt{x-x^2}=\sqrt{x}\sqrt{1-x}$, substituting $t=\sqrt x$ clears every square root of $x$ at once, converting the whole thing into a rational-power integral in $t$ that matches the derivative of $\sqrt{\frac{1-t}{1+t}}$ — a standard "recognise the derivative" pattern worth memorising for CET-style problems.

   Step-by-Step Solution

   1. Write $\sqrt{x-x^2}=\sqrt{x(1-x)}=\sqrt x\sqrt{1-x}$, so the integral is

   $I=\int\frac{dx}{(1+\sqrt x)\sqrt x\sqrt{1-x}}.$

   2. Let $t=\sqrt x$, so $x=t^2$, $dx=2t\,dt$:

   $I=\int\frac{2t\,dt}{(1+t)\cdot t\cdot\sqrt{1-t^2}}=\int\frac{2\,dt}{(1+t)\sqrt{(1-t)(1+t)}}=\int\frac{2\,dt}{(1+t)^{3/2}(1-t)^{1/2}}.$

   3. Let $y=\sqrt{\dfrac{1-t}{1+t}}$. Differentiating $y^2=\dfrac{1-t}{1+t}$: $2y\,y'=\frac{-(1+t)-(1-t)}{(1+t)^2}=\frac{-2}{(1+t)^2}\ \Rightarrow\ y'=\frac{-1}{y(1+t)^2}=\frac{-1}{(1+t)^{3/2}(1-t)^{1/2}}.$ …

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2. 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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3. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
   Q.If $\displaystyle\int \sqrt{x}(1-x^3)^{\frac{-1}{2}}dx=\dfrac{2}{3}g(f(x))+c$, then (A) $f(x)=\sqrt{x},\ g(x)=\sin^{-1}x$ (B) $f(x)=x^{3/2},\ g(x)=\sin^{-1}x$ (C) $f(x)=x^{3/2},\ g(x)=\cos^{-1}x$ (D) $f(x)=\sqrt{x},\ g(x)=\cos^{-1}x$

   ›Reveal solutionSolution

   A substitution $u=x^{3/2}$ turns the integral into the standard $\int du/\sqrt{1-u^2}$ form, giving $f(x)=x^{3/2}$ and $g=\sin^{-1}$.

   Concept and Intuition

   The presence of $\sqrt x\,dx$ alongside $x^3=(x^{3/2})^2$ inside a square root strongly signals the substitution $u=x^{3/2}$ (its derivative is proportional to $\sqrt x$, exactly what's needed to absorb the leftover $\sqrt x\,dx$). Once substituted, the integral collapses to the standard arcsine form.

   Step-by-Step Solution

   1. Let $u=x^{3/2}$. Then $du=\dfrac32 x^{1/2}\,dx=\dfrac32\sqrt x\,dx$, so $\sqrt x\,dx=\dfrac23\,du$.
   2. Also, $u^2=x^3$, so $1-x^3=1-u^2$.
   3. Substitute into the integral: $\displaystyle\int\sqrt x\,(1-x^3)^{-1/2}\,dx=\int\dfrac23\cdot\dfrac{du}{\sqrt{1-u^2}}=\dfrac23\int\dfrac{du}{\sqrt{1-u^2}}$.
   4. This is the standard form: $\displaystyle\int\dfrac{du}{\sqrt{1-u^2}}=\sin^{-1}u+c$.
   5. So the integral $=\dfrac23\sin^{-1}(u)+c=\dfrac23\sin^{-1}\left(x^{3/2}\right)+c$. …

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4. 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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5. 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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6. 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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7. 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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8. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{1}{x^5\sqrt[5]{x^5+1}}\,dx=$ (A) $\dfrac{4}{\sqrt[5]{x^5+1}}+c$ (B) $4x^4(x^5+1)^{4/5}+c$ (C) $-\dfrac{(x^5+1)^{4/5}}{4x^4}+c$ (D) $-\dfrac{(x^5+1)^{4/5}}{4x^5}+c$

   ›Reveal solutionSolution

   Rewriting the integrand to expose $1+x^{-5}$ as the natural substitution variable solves this cleanly; the answer is $-\dfrac{(x^5+1)^{4/5}}{4x^4}+c$.

   Concept and Intuition

   When an integral mixes a power of $x$ with a root of a polynomial in $x$, factoring out the highest power of $x$ from inside the root often converts the expression into a function of $1/x$ (or $x^{-5}$ here), whose derivative is already present elsewhere in the integrand — a clean substitution.

   Step-by-Step Solution

   1. $(x^5+1)^{-1/5}=\left(x^5\left(1+x^{-5}\right)\right)^{-1/5}=x^{-1}(1+x^{-5})^{-1/5}$.
   2. So the integrand $x^{-5}(x^5+1)^{-1/5}=x^{-6}(1+x^{-5})^{-1/5}$.
   3. Let $t=1+x^{-5}$, so $dt=-5x^{-6}dx\Rightarrow x^{-6}dx=-\dfrac{dt}{5}$.
   4. Integral $=\displaystyle\int t^{-1/5}\left(-\dfrac{dt}{5}\right)=-\dfrac15\cdot\dfrac{t^{4/5}}{4/5}+c=-\dfrac14t^{4/5}+c$. …

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9. 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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10. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
    Q.$\displaystyle\int \dfrac{dx}{(x^5+1)^{6/5}} =$ (A) $\dfrac{1}{\sqrt[5]{x^5+1}} + c$ (B) $\dfrac{\sqrt[5]{x^5+1}}{x} + c$ (C) $\dfrac{x}{\sqrt[5]{x^5+1}} + c$ (D) $\sqrt[5]{x^5+1} + c$

    ›Reveal solutionSolution

    Recognizing the integrand as the derivative of $\dfrac{x}{\sqrt[5]{x^5+1}}$ (verified by direct differentiation) gives the antiderivative immediately.

    Concept and Intuition

    For integrals of the form $\int\frac{dx}{(x^n+1)^{(n+1)/n}}$, a useful trick is to guess that the antiderivative looks like $\frac{x}{(x^n+1)^{1/n}}$ (a ratio designed to make the product-rule differentiation collapse nicely), and then verify by differentiating it — if it reproduces the integrand exactly, we're done. This is often faster than a substitution for this particular family.

    Step-by-Step Solution

    1. Guess the antiderivative $g(x) = \dfrac{x}{(x^5+1)^{1/5}} = x(x^5+1)^{-1/5}$.
    2. Differentiate using the product rule: $g'(x) = (x^5+1)^{-1/5} + x\cdot\left(-\dfrac15\right)(x^5+1)^{-6/5}\cdot5x^4$.
    3. Simplify the second term: $x\cdot\left(-\dfrac15\right)(5x^4)(x^5+1)^{-6/5} = -x^5(x^5+1)^{-6/5}$.
    4. So $g'(x) = (x^5+1)^{-1/5} - x^5(x^5+1)^{-6/5}$.
    5. Factor out $(x^5+1)^{-6/5}$: $g'(x) = (x^5+1)^{-6/5}\left[(x^5+1) - x^5\right] = (x^5+1)^{-6/5}\times 1 = (x^5+1)^{-6/5}$. …

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11. AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
    Q.$0 < x < 1$, $\int \dfrac{dx}{\sqrt{x^2 - x^5}} = \dfrac{1}{3}\log|f(x)| + C$, then $f(1/2) =$ (A) $\dfrac{\sqrt{8} - \sqrt{7}}{\sqrt{8} + \sqrt{7}}$ (B) $\dfrac{\sqrt{8} + \sqrt{7}}{\sqrt{8} - \sqrt{7}}$ (C) $2(\sqrt{8} - \sqrt{7})$ (D) $2(\sqrt{8} - \sqrt{7})^2$

    ›Reveal solutionSolution

    Substituting $t=x^3$ then $1-t=w^2$ integrates $\int dx/(x\sqrt{1-x^3})$ cleanly to $\tfrac13\log\left|\tfrac{1-\sqrt{1-x^3}}{1+\sqrt{1-x^3}}\right|$, and evaluating at $x=1/2$ gives $\dfrac{\sqrt8-\sqrt7}{\sqrt8+\sqrt7}$.

    Concept and Intuition

    The key simplification is $x^2-x^5=x^2(1-x^3)$, since $0<x<1$ makes $x>0$ so $\sqrt{x^2(1-x^3)}=x\sqrt{1-x^3}$. From there, the substitution $t=x^3$ turns the integral into the very standard form $\int\dfrac{dt}{t\sqrt{1-t}}$, solvable by a further substitution $1-t=w^2$.

    Step-by-Step Solution

    1. Rewrite: $\displaystyle I=\int\frac{dx}{\sqrt{x^2-x^5}}=\int\frac{dx}{x\sqrt{1-x^3}}$ (using $x>0$).
    2. Let $t=x^3\Rightarrow dt=3x^2dx\Rightarrow dx=\dfrac{dt}{3x^2}$. Then $I=\displaystyle\int\frac{dt}{3x^2\cdot x\sqrt{1-t}}=\int\frac{dt}{3x^3\sqrt{1-t}}=\frac13\int\frac{dt}{t\sqrt{1-t}}$ (since $x^3=t$).
    3. Let $1-t=w^2\Rightarrow t=1-w^2,\ dt=-2w\,dw$: $\displaystyle\int\frac{dt}{t\sqrt{1-t}}=\int\frac{-2w\,dw}{(1-w^2)w}=-2\int\frac{dw}{1-w^2}=-\log\left|\frac{1+w}{1-w}\right|=\log\left|\frac{1-w}{1+w}\right|$.
    4. So $I=\dfrac13\log\left|\dfrac{1-w}{1+w}\right|+C$ where $w=\sqrt{1-t}=\sqrt{1-x^3}$, i.e. $f(x)=\dfrac{1-\sqrt{1-x^3}}{1+\sqrt{1-x^3}}$.
    5. Verify by differentiating (chain rule through $w$) that this reproduces $\dfrac{1}{x\sqrt{1-x^3}}$ — confirmed. …

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12. 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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