Q.Integrate the following function: $\int \frac{dx}{x^2 + 2x + 2}$ equals (A) $x \tan^{-1} (x + 1) + C$ (B) $\tan^{-1} (x + 1) + C$ (C) $(x + 1) \tan^{-1} x + C$ (D) $\tan^{-1} x + C$

Showing the 12 most recent of 35 on this concept.

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

   ›Reveal solutionSolution

   Complete the square in the denominator and apply the standard $\int \frac{dx}{x^2+a^2}=\frac1a\tan^{-1}(x/a)$ form. Answer: option (B).

   Concept and Intuition

   Any irreducible quadratic $ax^2+bx+c$ in a denominator under a simple rational integrand can be handled by completing the square to reduce it to the standard $u^2+a^2$ form, whose antiderivative is a scaled arctangent.

   Step-by-Step Solution

   1. Complete the square: $1+x+x^2 = \left(x+\dfrac12\right)^2 + \dfrac34 = \left(x+\dfrac12\right)^2+\left(\dfrac{\sqrt3}{2}\right)^2$.
   2. So the integral is $\displaystyle 2\int \frac{dx}{(x+\frac12)^2+(\frac{\sqrt3}{2})^2}$.
   3. Using $\int \frac{du}{u^2+a^2}=\frac1a\tan^{-1}(u/a)$ with $u=x+\frac12$, $a=\frac{\sqrt3}{2}$:

   $2\cdot\frac{1}{\sqrt3/2}\tan^{-1}\!\left(\frac{x+1/2}{\sqrt3/2}\right) = \frac{4}{\sqrt3}\tan^{-1}\!\left(\frac{2x+1}{\sqrt3}\right)+c.$

   Common Mistakes …

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2. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
   Q.If $\int \frac{x^2+1}{x^4+1}\, dx = f(x) + c$, then $f(x) =$ ________ (A) $\frac{1}{\sqrt{2}} \mathrm{Tan}^{-1}\left( \frac{x^2+1}{\sqrt{2}x} \right)$ (B) $\frac{1}{\sqrt{2}} \mathrm{Tan}^{-1}\left( \frac{x^2-1}{\sqrt{2}x} \right)$ (C) $\frac{1}{\sqrt{2}} \mathrm{Tan}^{-1}\left( \frac{1-x^2}{\sqrt{2}x} \right)$ (D) $\frac{1}{\sqrt{2}} \mathrm{Tan}^{-1}\left( \frac{1+x^4}{\sqrt{2}x} \right)$

   ›Reveal solutionSolution

   The classic "divide by $x^2$, substitute $u=x-1/x$" trick converts this rational integral into an arctangent. Answer: $\frac1{\sqrt2}\tan^{-1}\left(\frac{x^2-1}{\sqrt2 x}\right)$.

   Concept and Intuition

   Integrals of the form $\int\frac{x^2\pm1}{x^4+1}dx$ are classically solved by dividing through by $x^2$ and recognizing $x^2+\frac1{x^2} = \left(x\mp\frac1x\right)^2\pm2$, which lets $u=x\mp\frac1x$ turn it into $\int\frac{du}{u^2\pm2}$.

   Step-by-Step Solution

   1. Divide numerator and denominator by $x^2$: $\dfrac{x^2+1}{x^4+1}=\dfrac{1+1/x^2}{x^2+1/x^2}$.
   2. Let $u=x-\dfrac1x\Rightarrow du=\left(1+\dfrac1{x^2}\right)dx$.
   3. Note $u^2=x^2-2+\dfrac1{x^2}\Rightarrow x^2+\dfrac1{x^2}=u^2+2$.
   4. So the integral becomes $\displaystyle\int\frac{du}{u^2+2}=\frac1{\sqrt2}\tan^{-1}\left(\frac{u}{\sqrt2}\right)+c$. …

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3. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{1}{1+x+x^2}\, dx =$ (A) $\dfrac{2}{\sqrt{3}}\log\left(\dfrac{2x+1+\sqrt{3}}{2x-1-\sqrt{3}}\right) + c$ (B) $\dfrac{1}{\sqrt{3}}\log\left(\dfrac{2x+1-\sqrt{3}}{2x+1+\sqrt{3}}\right) + c$ (C) $\dfrac{2}{\sqrt{3}}\tan^{-1}\left(\dfrac{2x+1}{\sqrt{3}}\right) + c$ (D) $\dfrac{2}{\sqrt{5}}\tan^{-1}\left(\dfrac{2x+1}{\sqrt{5}}\right) + c$

   ›Reveal solutionSolution

   Completing the square turns the quadratic denominator into a sum of squares, a standard $\int\frac{dx}{u^2+a^2}$ form.

   Concept and Intuition

   Any $\int\dfrac{dx}{ax^2+bx+c}$ with no real roots in the denominator reduces, by completing the square, to the standard arctan integral $\int\dfrac{du}{u^2+a^2}=\dfrac1a\tan^{-1}\dfrac ua+c$.

   Step-by-Step Solution

   1. $1+x+x^2=\left(x+\dfrac12\right)^2+1-\dfrac14=\left(x+\dfrac12\right)^2+\dfrac34$.
   2. So the integral is $\displaystyle\int\dfrac{dx}{\left(x+\tfrac12\right)^2+\left(\tfrac{\sqrt3}2\right)^2}$.
   3. Using $\int\dfrac{du}{u^2+a^2}=\dfrac1a\tan^{-1}\dfrac ua+c$ with $u=x+\tfrac12$, $a=\dfrac{\sqrt3}2$: $=\dfrac{1}{\sqrt3/2}\tan^{-1}\left(\dfrac{x+1/2}{\sqrt3/2}\right)+c=\dfrac2{\sqrt3}\tan^{-1}\left(\dfrac{2x+1}{\sqrt3}\right)+c.$ …

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4. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.$\displaystyle\int (\sqrt{\tan x} + \sqrt{\cot x})\, dx =$ (A) $2\,\mathrm{Tan}^{-1}\left(\dfrac{\tan x - 1}{\sqrt{\tan x}}\right) + c$ (B) $\mathrm{Tan}^{-1}\left(\dfrac{\tan x - 2}{2\sqrt{\tan x}}\right) + c$ (C) $\sqrt{2}\,\mathrm{Tan}^{-1}\left(\dfrac{\tan x - 1}{\sqrt{2\tan x}}\right) + c$ (D) $\sqrt{2}\,\mathrm{Tan}^{-1}\left(\dfrac{\tan x + 1}{\sqrt{2\tan x}}\right) + c$

   ›Reveal solutionSolution

   The integral $\int(\sqrt{\tan x}+\sqrt{\cot x})dx$ is a classic "sum of square-root trig" integral that reduces to an arctangent form; verified here by direct differentiation, giving option (C).

   Concept and Intuition

   $\sqrt{\tan x}+\sqrt{\cot x}$ combines to $\dfrac{\tan x + 1}{\sqrt{\tan x}}$, and integrals of this shape are standard results that produce an inverse-tangent (or occasionally inverse-sine) antiderivative involving $\sqrt{2\tan x}$. Rather than re-deriving the substitution from scratch, the fastest reliable check with multiple-choice options is to differentiate each candidate and see which one reproduces the integrand exactly.

   Step-by-Step Solution

   1. Rewrite the integrand: $\sqrt{\tan x}+\sqrt{\cot x}=\sqrt{\tan x}+\dfrac{1}{\sqrt{\tan x}}=\dfrac{\tan x+1}{\sqrt{\tan x}}$.
   2. Test option (C): let $u=\dfrac{\tan x-1}{\sqrt{2\tan x}}$, and check $\dfrac{d}{dx}\left[\sqrt2\,\mathrm{Tan}^{-1}(u)\right]=\sqrt2\cdot\dfrac{u'}{1+u^2}$.
   3. Compute $1+u^2 = 1+\dfrac{(\tan x-1)^2}{2\tan x}=\dfrac{2\tan x+(\tan x-1)^2}{2\tan x}=\dfrac{\tan^2x+1}{2\tan x}=\dfrac{\sec^2x}{2\tan x}$.
   4. Compute $u'$ (quotient rule on $u=(\tan x-1)(2\tan x)^{-1/2}$):

   $u'=\sec^2x(2\tan x)^{-3/2}\big[(2\tan x)-( \tan x-1)\big] = \sec^2x(2\tan x)^{-3/2}(\tan x+1).$

   5. Combine: $\dfrac{u'}{1+u^2}=\sec^2x(2\tan x)^{-3/2}(\tan x+1)\cdot\dfrac{2\tan x}{\sec^2x}=\dfrac{2\tan x(\tan x+1)}{(2\tan x)^{3/2}}=\dfrac{\tan x+1}{\sqrt{2\tan x}}\cdot\dfrac{1}{1}$ (after simplifying $(2\tan x)^{3/2}/(2\tan x)=\sqrt{2\tan x}$), giving $\dfrac{\tan x+1}{\sqrt{2\tan x}}$. …

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5. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.If $\displaystyle\int \dfrac{x^2 - x + 2}{x^2 + x + 2}\,dx = x - \log(f(x)) + \dfrac{2}{\sqrt{7}}\tan^{-1}(g(x)) + c$, then $f(-1) + \sqrt{7}\,g(-1) =$ (A) $1$ (B) $0$ (C) $-1$ (D) $2$

   ›Reveal solutionSolution

   Splitting the numerator to isolate the derivative of the denominator identifies $f(x)=x^2+x+2$ and $g(x)=\frac{2x+1}{\sqrt7}$, giving $f(-1)+\sqrt7 g(-1)=1$.

   Concept and Intuition

   For a rational function whose numerator and denominator are both quadratics with the same leading coefficient, subtracting the denominator from the numerator strips off the constant term, leaving a simpler linear-over-quadratic integral that splits into a logarithm part (from the derivative of the denominator) and an arctan part (from completing the square).

   Step-by-Step Solution

   1. Note $x^2+x+2 - 2x = x^2-x+2$, so $\dfrac{x^2-x+2}{x^2+x+2} = 1 - \dfrac{2x}{x^2+x+2}$.
   2. $\displaystyle\int 1\,dx = x$.
   3. For $\displaystyle\int\frac{2x}{x^2+x+2}dx$, write $2x = (2x+1)-1$:

   • $\displaystyle\int\frac{2x+1}{x^2+x+2}dx = \log(x^2+x+2)$ (numerator is exactly the derivative of the denominator).
   • $\displaystyle\int\frac{-1}{x^2+x+2}dx = -\int\frac{dx}{(x+\frac12)^2+\frac74} = -\frac{2}{\sqrt7}\tan^{-1}\!\left(\frac{2x+1}{\sqrt7}\right)$.

   4. So $\displaystyle\int\frac{2x}{x^2+x+2}dx = \log(x^2+x+2) - \frac{2}{\sqrt7}\tan^{-1}\!\left(\frac{2x+1}{\sqrt7}\right)$. …

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6. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{\sqrt{x-2}}{2x+4}\, dx =$ (A) $\sqrt{x-2} - \dfrac{1}{2}\mathrm{Tan}^{-1}\left(\dfrac{\sqrt{x-2}}{2}\right) + c$ (B) $\sqrt{x-2} - 2\,\mathrm{Tan}^{-1}\left(\dfrac{\sqrt{x-2}}{2}\right) + c$ (C) $\sqrt{x-2} + 2\,\mathrm{Tan}^{-1}\left(\dfrac{\sqrt{x-2}}{2}\right) + c$ (D) $\sqrt{x-2} + \dfrac{1}{2}\mathrm{Tan}^{-1}\left(\dfrac{\sqrt{x-2}}{2}\right) + c$

   ›Reveal solutionSolution

   A rationalizing substitution $x-2=t^2$ converts the integral into a simple $\int\left(1-\dfrac{4}{t^2+4}\right)dt$, giving $\sqrt{x-2}-2\,\mathrm{Tan}^{-1}\!\left(\dfrac{\sqrt{x-2}}{2}\right)+c$.

   Concept and Intuition

   Whenever an integral has a single square root of a linear expression (here $\sqrt{x-2}$), the substitution $x-2=t^2$ (so $t=\sqrt{x-2}$) removes the square root entirely and typically converts the integral into a rational function of $t$, which is then handled by the standard $\int\dfrac{dt}{t^2+a^2}$ arctan formula.

   Step-by-Step Solution

   1. Simplify the denominator first: $2x+4=2(x+2)$, so the integral is $\dfrac12\displaystyle\int\frac{\sqrt{x-2}}{x+2}dx$.
   2. Substitute $x-2=t^2 \Rightarrow x = t^2+2$, $dx=2t\,dt$, and $x+2 = t^2+4$.
   3. The integral becomes

   $\frac12\int\frac{t}{t^2+4}\cdot 2t\,dt = \int\frac{t^2}{t^2+4}dt.$

   4. Split the rational function: $\dfrac{t^2}{t^2+4}=1-\dfrac{4}{t^2+4}$.
   5. Integrate termwise: $\displaystyle\int\left(1-\frac{4}{t^2+4}\right)dt = t - 4\cdot\frac12\,\mathrm{Tan}^{-1}\!\left(\frac t2\right)+c = t-2\,\mathrm{Tan}^{-1}\!\left(\frac t2\right)+c$.
   6. Substitute back $t=\sqrt{x-2}$: …

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7. AP EAPCET 2024Set eng-2024-05-18-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{x+1}{\sqrt{x^2+x+1}}\,dx=$ (A) $\dfrac{1}{2}\sqrt{x^2+x+1}+\dfrac{1}{2}\cosh^{-1}\left(\dfrac{x+2}{\sqrt{3}}\right)+c$ (B) $\dfrac{1}{2}\sqrt{x^2+x+1}+\dfrac{2}{\sqrt{3}}\tan^{-1}\left(\dfrac{2x+1}{\sqrt{3}}\right)+c$ (C) $\sqrt{x^2+x+1}+\dfrac{2}{\sqrt{3}}\log|x^2+x+1|+c$ (D) $\sqrt{x^2+x+1}+\dfrac{1}{2}\sinh^{-1}\left(\dfrac{2x+1}{\sqrt{3}}\right)+c$

   ›Reveal solutionSolution

   Splitting the numerator into a multiple of the derivative of the radicand plus a constant is the standard technique for $\int\dfrac{px+q}{\sqrt{ax^2+bx+c}}dx$; here it gives $\sqrt{x^2+x+1}+\dfrac12\sinh^{-1}\left(\dfrac{2x+1}{\sqrt3}\right)+c$.

   Concept and Intuition

   For $\int\dfrac{\text{linear}}{\sqrt{\text{quadratic}}}dx$, write the linear numerator as $A\cdot(\text{derivative of quadratic})+B$. The $A$-part becomes a simple power-rule integral (since it's exactly $u^{-1/2}du$), and the $B$-part reduces to the standard $\int\dfrac{dx}{\sqrt{x^2+a^2}}=\sinh^{-1}(x/a)+c$ form after completing the square.

   Step-by-Step Solution

   1. Write $x+1=\dfrac12(2x+1)+\dfrac12$.
   2. First part: $\dfrac12\displaystyle\int\dfrac{2x+1}{\sqrt{x^2+x+1}}dx$. Let $u=x^2+x+1$, $du=(2x+1)dx$: this is $\dfrac12\int u^{-1/2}du=\dfrac12\cdot2u^{1/2}=\sqrt{x^2+x+1}$.
   3. Second part: $\dfrac12\displaystyle\int\dfrac{dx}{\sqrt{x^2+x+1}}=\dfrac12\int\dfrac{dx}{\sqrt{(x+1/2)^2+3/4}}$.
   4. This is of the form $\dfrac12\int\dfrac{du}{\sqrt{u^2+a^2}}$ with $u=x+1/2$, $a=\sqrt3/2$, giving $\dfrac12\sinh^{-1}\left(\dfrac{u}{a}\right)=\dfrac12\sinh^{-1}\left(\dfrac{2x+1}{\sqrt3}\right)$. …

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8. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{dx}{(2ax+x^2)^{3/2}}=$ (A) $\dfrac{1}{a^2}\left(\dfrac{x+a}{\sqrt{2ax+x^2}}\right)+C$ (B) $\dfrac{1}{a^2}\left(\dfrac{x-a}{\sqrt{2ax+x^2}}\right)+C$ (C) $\dfrac{-1}{a^2}\left(\dfrac{x-a}{\sqrt{2ax+x^2}}\right)+C$ (D) $\dfrac{-1}{a^2}\left(\dfrac{x+a}{\sqrt{2ax+x^2}}\right)+C$

   ›Reveal solutionSolution

   Completing the square converts the integral into the standard form $\int du/(u^2-a^2)^{3/2}$, which has a known closed form; back-substituting gives option (D).

   Concept and Intuition

   Many integrals of the form $\int dx/(\text{quadratic})^{3/2}$ become standard once the quadratic is completed to a perfect-square-minus-constant form; the resulting substitution $u=x+a$ reduces it to a memorized/derivable antiderivative.

   Step-by-Step Solution

   1. $2ax+x^2=x^2+2ax+a^2-a^2=(x+a)^2-a^2$. Let $u=x+a$, $du=dx$.
   2. The integral becomes $\displaystyle\int\frac{du}{(u^2-a^2)^{3/2}}$.
   3. Standard result (verifiable by differentiation): $\displaystyle\int\frac{du}{(u^2-a^2)^{3/2}}=\frac{-u}{a^2\sqrt{u^2-a^2}}+C$. Check: $\dfrac{d}{du}\left[\dfrac{-u}{a^2\sqrt{u^2-a^2}}\right]=\dfrac{1}{(u^2-a^2)^{3/2}}$ ✓. …

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9. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.$\int\frac{(3x-2)\tan(\sqrt{9x^2-12x+1})}{\sqrt{9x^2-12x+1}}dx =$ (A) $\frac{1}{3}\sec^2\sqrt{9x^2-12x+1} + c$ (B) $\frac{1}{3}\sec^2 x + c$ (C) $\frac{1}{2}\log\left|\sec\sqrt{9x^2-12x+1}\right| + c$ (D) $\frac{1}{3}\log\left|\sec\sqrt{9x^2-12x+1}\right| + c$

   ›Reveal solutionSolution

   A double substitution — first $u=9x^2-12x+1$, then $w=\sqrt u$ — turns the integral into a plain $\int\tan w\,dw$, giving $\frac13\log|\sec\sqrt{9x^2-12x+1}|+c$.

   Concept and Intuition

   When $(3x-2)$ (half the derivative of $9x^2-12x+1$) sits outside a function of $\sqrt{9x^2-12x+1}$, a chained substitution — first for the quadratic, then for its square root — collapses the whole expression to a single-variable standard integral.

   Step-by-Step Solution

   1. Let $u=9x^2-12x+1$. Then $du=(18x-12)\,dx=6(3x-2)\,dx$, so $(3x-2)\,dx=\dfrac{du}{6}$.
   2. Integral becomes $\displaystyle\int\frac{\tan\sqrt u}{\sqrt u}\cdot\frac{du}{6}=\frac16\int\frac{\tan\sqrt u}{\sqrt u}\,du$.
   3. Let $w=\sqrt u$, so $dw=\dfrac{du}{2\sqrt u}$, i.e. $\dfrac{du}{\sqrt u}=2\,dw$.
   4. Integral becomes $\dfrac16\int\tan w\cdot2\,dw=\dfrac13\int\tan w\,dw=\dfrac13\left(-\log|\cos w|\right)+c=\dfrac13\log|\sec w|+c$. …

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10. AP EAPCET 2021Set eng-2021-08-23-FN1 markMCQ
    Q.If $\displaystyle\int \dfrac{dx}{\cos^4 x + \sin^4 x} = \dfrac{1}{\sqrt{2}} \tan^{-1}[g(x)] + c$, then $g(x)$ equals ______ (A) $\dfrac{\tan x - \cot x}{\sqrt{2}}$ (B) $\dfrac{\tan x + \cot x}{\sqrt{2}}$ (C) $\dfrac{\sin x - \cos x}{\sqrt{2}}$ (D) $\dfrac{\sin x + \cos x}{\sqrt{2}}$

    ›Reveal solutionSolution

    A standard trick for $\int dx/(\cos^4x+\sin^4x)$: convert to tangent, then use the $z=u-1/u$ substitution that turns a symmetric quartic denominator into a simple quadratic.

    Concept and Intuition

    Expressions like $u^4+1$ in a denominator, paired with a numerator $1+u^2$, are tailor-made for the substitution $z=u-1/u$ (or $z=u+1/u$), because $u^2+1/u^2 = (u\mp 1/u)^2 \pm 2$ collapses the quartic structure into a quadratic in $z$.

    Step-by-Step Solution

    1. Divide numerator and denominator by $\cos^4x$: $\dfrac{dx}{\cos^4x+\sin^4x} = \dfrac{\sec^4x\,dx}{1+\tan^4x}$.
    2. Write $\sec^4x = (1+\tan^2x)\sec^2x$, so the integral is $\displaystyle\int \dfrac{(1+\tan^2x)\sec^2x\,dx}{1+\tan^4x}$.
    3. Let $u=\tan x$, $du=\sec^2x\,dx$: integral becomes $\displaystyle\int \dfrac{(1+u^2)\,du}{1+u^4}$.
    4. Divide numerator and denominator by $u^2$: $\displaystyle\int \dfrac{(1+1/u^2)\,du}{u^2+1/u^2}$.
    5. Let $z = u - \dfrac1u$, so $dz = \left(1+\dfrac{1}{u^2}\right)du$, and $u^2+\dfrac{1}{u^2} = z^2+2$. …

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11. AP EAPCET 2026Set eng-2026-05-18-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

    Two successive substitutions ($t=\sqrt x$, then $w=\sqrt{(1-t)/(1+t)}$) collapse the integral to $\int -2\,dw$; the answer is $-2\sqrt{\dfrac{1-\sqrt x}{1+\sqrt x}}+c$.

    Concept and Intuition

    The presence of $\sqrt x$ inside $(1+\sqrt x)$ and inside $\sqrt{x-x^2}=\sqrt x\sqrt{1-x}$ both suggest first substituting $t=\sqrt x$. What remains — a rational-times-square-root expression in $t$ symmetric under $t\to -t$ in a $(1\pm t)$ sense — is the classic cue for the substitution $w^2=\dfrac{1-t}{1+t}$, which rationalizes everything at once.

    Step-by-Step Solution

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

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

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

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

    3. Let $w=\sqrt{\dfrac{1-t}{1+t}}$, so $t=\dfrac{1-w^2}{1+w^2}$ and $dt=\dfrac{-4w}{(1+w^2)^2}\,dw$. One finds

    $1+t=\frac{2}{1+w^2},\qquad 1-t=\frac{2w^2}{1+w^2},$

    so

    $(1+t)^{3/2}(1-t)^{1/2}=\frac{4w}{(1+w^2)^2}.$

    4. Substituting, …

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12. AP EAPCET 2021Set eng-2021-08-23-AN1 markMCQ
    Q.$\displaystyle\int \dfrac{\sin x - \cos x}{\sqrt{\sin 2x}}\,dx =$ (A) $-\log|\sin x - \cos x + \sqrt{\sin 2x}| + c$ (B) $-\log|\sin x + \cos x - \sqrt{\sin 2x}| + c$ (C) $-\log|\sin x + \cos x + \sqrt{\sin 2x}| + c$ (D) $-\log|\sin x - \cos x - \sqrt{\sin 2x}| + c$

    ›Reveal solutionSolution

    The key substitution is $t=\sin x+\cos x$, which turns $\sin 2x$ into $t^2-1$ and the numerator into $-dt$, reducing the integral to a standard $\int \dfrac{dt}{\sqrt{t^2-1}}$ form.

    Concept and Intuition

    Expressions like $\sin x \pm \cos x$ paired with $\sin 2x$ are a classic signal to substitute $t=\sin x\pm\cos x$, because $(\sin x+\cos x)^2 = 1+\sin2x$ and $(\sin x-\cos x)^2=1-\sin2x$ — this converts everything to a single variable.

    Step-by-Step Solution

    1. Let $t=\sin x+\cos x$. Then $dt = (\cos x-\sin x)\,dx = -(\sin x-\cos x)\,dx$.
    2. Also $t^2 = \sin^2x+\cos^2x+2\sin x\cos x = 1+\sin2x$, so $\sin2x = t^2-1$, i.e. $\sqrt{\sin2x}=\sqrt{t^2-1}$.
    3. Rewrite the integral:

    $\int \frac{\sin x-\cos x}{\sqrt{\sin2x}}\,dx = \int \frac{-dt}{\sqrt{t^2-1}} = -\log\left|t+\sqrt{t^2-1}\right|+c.$

    4. Substitute back $t=\sin x+\cos x$ and $\sqrt{t^2-1}=\sqrt{\sin2x}$: …

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