Q.Evaluate: $\int \sqrt{2ax-x^2}\,dx$

Showing the 12 most recent of 35 on this concept.

1. 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: $\dfrac{d}{du}\left[\dfrac{-u}{a^2\sqrt{u^2-a^2}}\right]=\dfrac{1}{(u^2-a^2)^{3/2}}$ ✓. …

   🔒 Start your 14-day free trial to see the full solution →
2. AP EAPCET 2025Set eng-2025-05-24-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{x}{\sqrt{x^2 - 2x + 5}}\, dx =$ (A) $\sqrt{x^2-2x+5} + \mathrm{Sinh}^{-1}\left(\dfrac{x-1}{2}\right) + c$ (B) $\dfrac{1}{2}\sqrt{x^2-2x+5} + \mathrm{Sin}^{-1}\left(\dfrac{x-1}{2}\right) + c$ (C) $2\sqrt{x^2-2x+5} + \mathrm{Cosh}^{-1}\left(\dfrac{x-1}{2}\right) + c$ (D) $\sqrt{x^2-2x+5} - \mathrm{Cos}^{-1}\left(\dfrac{x-1}{2}\right) + c$

   ›Reveal solutionSolution

   A rational-times-radical integral of the form $\int\frac{x\,dx}{\sqrt{ax^2+bx+c}}$, split by writing the numerator to match the derivative of the radicand. Answer: $\sqrt{x^2-2x+5}+\mathrm{Sinh}^{-1}\!\left(\frac{x-1}{2}\right)+c$.

   Concept and Intuition

   The standard technique for $\int\dfrac{x\,dx}{\sqrt{x^2+bx+c}}$ is to complete the square in the radicand and write $x$ as (half the derivative of the radicand) plus a constant — this splits the integral into an easy "$u/\sqrt{u^2+a^2}$" piece and a standard inverse hyperbolic-sine piece.

   Step-by-Step Solution

   1. Complete the square: $x^2-2x+5=(x-1)^2+4$.
   2. Let $u=x-1\Rightarrow x=u+1,\ dx=du$. The integral becomes $\displaystyle\int\dfrac{u+1}{\sqrt{u^2+4}}\,du$.
   3. Split: $\displaystyle\int\dfrac{u}{\sqrt{u^2+4}}du+\int\dfrac{du}{\sqrt{u^2+4}}$.
   4. First piece: $\displaystyle\int\dfrac{u\,du}{\sqrt{u^2+4}}=\sqrt{u^2+4}+C_1$ (direct substitution $w=u^2+4$).
   5. Second piece: $\displaystyle\int\dfrac{du}{\sqrt{u^2+4}}=\mathrm{Sinh}^{-1}\!\left(\dfrac{u}{2}\right)+C_2$ (standard form $\int\frac{du}{\sqrt{u^2+a^2}}=\mathrm{Sinh}^{-1}(u/a)$). …

   🔒 Start your 14-day free trial to see the full solution →
3. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
   Q.$\int \frac{dx}{\sqrt{7 - 6x - x^2}} =$ (A) $\mathrm{Sinh}^{-1}\left( \frac{x+3}{4} \right) + c$ (B) $\log\left| \frac{x+3}{4} \right| + c$ (C) $\mathrm{Sin}^{-1}\left( \frac{x+3}{4} \right) + c$ (D) $\frac{1}{2}\mathrm{Sin}^{-1}\left( \frac{x+3}{4} \right) + c$

   ›Reveal solutionSolution

   Completing the square under the root reveals the standard form $\int\frac{dx}{\sqrt{a^2-u^2}}=\sin^{-1}(u/a)$. Answer: $\sin^{-1}\left(\frac{x+3}{4}\right)+c$.

   Concept and Intuition

   A quadratic under a square root, when completed to the square, reveals which standard integral form applies: $a^2-(x-h)^2$ gives an arcsine, while $(x-h)^2+a^2$ or $(x-h)^2-a^2$ give hyperbolic-inverse/log forms.

   Step-by-Step Solution

   1. $7-6x-x^2 = -(x^2+6x-7) = -[(x+3)^2-9-7] = -(x+3)^2+16 = 16-(x+3)^2$.
   2. So the integral is $\displaystyle\int\frac{dx}{\sqrt{4^2-(x+3)^2}}$.
   3. Using $\int\frac{du}{\sqrt{a^2-u^2}}=\sin^{-1}\left(\frac ua\right)+c$ with $u=x+3,\ a=4$: …

   🔒 Start your 14-day free trial to see the full solution →
4. 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, …

   🔒 Start your 14-day free trial to see the full solution →
5. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.$\displaystyle\int \sqrt{x^2+x+1}\,dx$ (A) $\dfrac{(2x+1)}{4}\sqrt{x^2+x+1}+\dfrac{3}{8}\mathrm{Sinh}^{-1}\left(\dfrac{2x+1}{\sqrt{3}}\right)+c$ (B) $\dfrac{x+1}{4}\sqrt{x^2+x+1}+\dfrac{3}{8}\mathrm{Sinh}^{-1}\left(\dfrac{2x+1}{\sqrt{3}}\right)+c$ (C) $\dfrac{x+1}{4}\sqrt{x^2+x+1}-\dfrac{3}{8}\mathrm{Sinh}^{-1}\left(\dfrac{2x+1}{\sqrt{3}}\right)+c$ (D) $\dfrac{(2x+1)}{4}\sqrt{x^2+x+1}-\dfrac{3}{8}\mathrm{Sinh}^{-1}\left(\dfrac{2x+1}{\sqrt{3}}\right)+c$

   ›Reveal solutionSolution

   Completing the square turns $\sqrt{x^2+x+1}$ into the standard $\sqrt{u^2+a^2}$ form, whose known integral gives option (A).

   Concept and Intuition

   The standard result $\displaystyle\int\sqrt{u^2+a^2}\,du=\frac u2\sqrt{u^2+a^2}+\frac{a^2}2\,\mathrm{Sinh}^{-1}\!\left(\frac ua\right)+c$ applies to any quadratic under a square root once it's written as a perfect square plus a constant.

   Step-by-Step Solution

   1. Complete the square: $x^2+x+1=\left(x+\dfrac12\right)^2+\dfrac34$. Let $u=x+\dfrac12$, $a^2=\dfrac34$ (so $a=\dfrac{\sqrt3}2$).
   2. Apply the formula: $\displaystyle\int\sqrt{u^2+a^2}\,du=\frac u2\sqrt{u^2+a^2}+\frac{a^2}2\mathrm{Sinh}^{-1}\!\left(\frac ua\right)+c$.
   3. $\dfrac u2 = \dfrac{x+\frac12}{2}=\dfrac{2x+1}{4}$, and $\sqrt{u^2+a^2}=\sqrt{x^2+x+1}$.
   4. $\dfrac{a^2}2=\dfrac{3/4}{2}=\dfrac38$, and $\dfrac ua = \dfrac{x+\frac12}{\sqrt3/2}=\dfrac{2x+1}{\sqrt3}$. …

   🔒 Start your 14-day free trial to see the full solution →
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 $t=\sqrt{x-2}$: …

   🔒 Start your 14-day free trial to see the full solution →
7. 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 $\sqrt{t^2-1}=\sqrt{\sin2x}$: …

   🔒 Start your 14-day free trial to see the full solution →
8. 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 $\dfrac12\sinh^{-1}\left(\dfrac{u}{a}\right)=\dfrac12\sinh^{-1}\left(\dfrac{2x+1}{\sqrt3}\right)$. …

   🔒 Start your 14-day free trial to see the full solution →
9. 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 $\dfrac{\tan x+1}{\sqrt{2\tan x}}$. …

   🔒 Start your 14-day free trial to see the full solution →
10. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
    Q.If $\displaystyle\int \dfrac{2x+5}{\sqrt{7-6x-x^2}} dx = A\sqrt{7-6x-x^2} + B\sin^{-1}\left(\dfrac{x+3}{4}\right) + c$ then the ordered pair $(A,B)=$ (A) $(-2,-1)$ (B) $(2,-1)$ (C) $(-2,1)$ (D) $(2,1)$

    ›Reveal solutionSolution

    Splitting $2x+5$ into a multiple of the derivative of $7-6x-x^2$ plus a constant reduces the integral to a standard $\sqrt{}$ term plus an $\sin^{-1}$ term, giving $(A,B)=(-2,-1)$.

    Concept and Intuition

    For $\int \dfrac{px+q}{\sqrt{ax^2+bx+c}}dx$, always split the numerator as (multiple of the derivative of the quadratic under the root) + (constant), because $\int\frac{f'(x)}{\sqrt{f(x)}}dx=2\sqrt{f(x)}$ handles the first part exactly, leaving a pure $1/\sqrt{\text{quadratic}}$ integral for the second part (an inverse-sine form after completing the square).

    Step-by-Step Solution

    1. Let $f(x)=7-6x-x^2$. Then $f'(x)=-6-2x$.
    2. Write $2x+5 = \lambda(-6-2x) + \mu$. Matching coefficients of $x$: $2=-2\lambda\Rightarrow\lambda=-1$. Matching constants: $5=-6\lambda+\mu = 6+\mu\Rightarrow\mu=-1$.
    3. So $2x+5 = -1\cdot(-6-2x) - 1$.
    4. $\displaystyle\int\frac{2x+5}{\sqrt{f(x)}}dx = -1\displaystyle\int\frac{f'(x)}{\sqrt{f(x)}}dx - \int\frac{dx}{\sqrt{f(x)}} = -1\cdot 2\sqrt{f(x)} - \int\frac{dx}{\sqrt{f(x)}}$.
    5. Complete the square: $7-6x-x^2 = -(x^2+6x-7) = -((x+3)^2-16) = 16-(x+3)^2$. …

    🔒 Start your 14-day free trial to see the full solution →
11. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
    Q.$\displaystyle\int \dfrac{2-\sin x}{2\cos x+3} \, dx =$ (A) $\dfrac{2}{\sqrt{5}}\mathrm{Tan}^{-1}\left(\dfrac{1}{\sqrt{3}}\tan\dfrac{x}{2}\right) - \log\sqrt{2\cos x+3}+c$ (B) $\dfrac{4}{\sqrt{5}}\mathrm{Tan}^{-1}\left(\dfrac{1}{\sqrt{5}}\tan\dfrac{x}{2}\right) + \log\sqrt{2\cos x+3}+c$ (C) $\dfrac{3}{\sqrt{5}}\mathrm{Tan}^{-1}\left(\dfrac{1}{\sqrt{5}}\tan\dfrac{x}{2}\right) + \log\sqrt{2\cos x-3}+c$ (D) $\dfrac{1}{\sqrt{5}}\mathrm{Tan}^{-1}\left(\dfrac{1}{\sqrt{5}}\tan\dfrac{x}{3}\right) - \log\sqrt{2\cos x-3}+c$

    ›Reveal solutionSolution

    Decompose the numerator into a multiple of the denominator's derivative plus a constant; the log part and arctan part combine to option (B).

    Concept and Intuition

    For integrals of the form $\displaystyle\int\dfrac{p+q\sin x}{a+b\cos x}dx$, write the numerator as $\lambda\cdot(\text{derivative of denominator}) + \mu$ (a pure constant), so the integral splits into a straightforward logarithmic piece and a standard $\displaystyle\int\dfrac{dx}{a+b\cos x}$ piece (solved via the Weierstrass/half-angle substitution).

    Step-by-Step Solution

    1. Let $D(x)=2\cos x+3$, so $D'(x)=-2\sin x$.
    2. Write $2-\sin x = \alpha D'(x) + \lambda = -2\alpha\sin x + \lambda$. Matching $\sin x$ coefficients: $-2\alpha=-1\Rightarrow\alpha=\dfrac12$. Matching constants: $\lambda=2$.
    3. So $\displaystyle\int\dfrac{2-\sin x}{2\cos x+3}dx = \dfrac12\int\dfrac{D'(x)}{D(x)}dx + 2\int\dfrac{dx}{2\cos x+3} = \dfrac12\log|2\cos x+3| + 2I$, where $I=\displaystyle\int\dfrac{dx}{2\cos x+3}$.
    4. Standard result (with $a=3,\ b=2$, $a>b$): $I=\dfrac{2}{\sqrt{a^2-b^2}}\tan^{-1}\!\left(\sqrt{\dfrac{a-b}{a+b}}\tan\dfrac x2\right)=\dfrac{2}{\sqrt5}\tan^{-1}\!\left(\dfrac1{\sqrt5}\tan\dfrac x2\right)$ (since $a^2-b^2=5$ and $\sqrt{(a-b)/(a+b)}=\sqrt{1/5}$).
    5. So $2I=\dfrac4{\sqrt5}\tan^{-1}\!\left(\dfrac1{\sqrt5}\tan\dfrac x2\right)$. …

    🔒 Start your 14-day free trial to see the full solution →
12. 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 …

    🔒 Start your 14-day free trial to see the full solution →