Q.Integrate the following function: $\int \frac{dx}{\sqrt{9x - 4x^2}}$ equals (A) $\frac{1}{9} \sin^{-1} \left(\frac{9x - 8}{8}\right) + C$ (B) $\frac{1}{2} \sin^{-1} \left(\frac{8x - 9}{9}\right) + C$ (C) $\frac{1}{3} \sin^{-1} \left(\frac{9x - 8}{8}\right) + C$ (D) $\frac{1}{2} \sin^{-1} \left(\frac{9x - 8}{9}\right) + C$

Showing the 12 most recent of 35 on this concept.

1. 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 →
2. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.$\int\frac{1}{9\cos^2x - 24\sin x\cos x + 16\sin^2 x}dx =$ (A) $\frac{\cos x}{4(3\cos x - 4\sin x)} + c$ (B) $\frac{\sin x}{4(3\cos x - 4\sin x)} + c$ (C) $\frac{\cos x}{3\cos x - 4\sin x} + c$ (D) $\frac{\sin x}{3\cos x - 4\sin x} + c$

   ›Reveal solutionSolution

   Recognising the denominator as the perfect square $(3\cos x-4\sin x)^2$ and substituting $u=3-4\tan x$ reduces the integral to a simple power rule, giving $\frac{\cos x}{4(3\cos x-4\sin x)}+c$.

   Concept and Intuition

   When a trigonometric denominator has the form $a^2\cos^2x - 2ab\sin x\cos x + b^2\sin^2x$, check whether it is a perfect square $(a\cos x - b\sin x)^2$ first — this instantly turns a messy-looking integral into $\int\sec^2(\cdot)/(\text{linear in }\tan x)^2\,dx$, solvable by substitution.

   Step-by-Step Solution

   1. Observe $9\cos^2x-24\sin x\cos x+16\sin^2x=(3\cos x-4\sin x)^2$ (matches $a^2-2ab+b^2$ with $a=3\cos x,\ b=4\sin x$).
   2. So the integral is $\displaystyle\int\frac{dx}{(3\cos x-4\sin x)^2}$.
   3. Divide numerator and denominator by $\cos^2x$: $=\displaystyle\int\frac{\sec^2x\,dx}{(3-4\tan x)^2}$.
   4. Let $u=3-4\tan x$, so $du=-4\sec^2x\,dx$, i.e. $\sec^2x\,dx=-\tfrac14du$. …

   🔒 Start your 14-day free trial to see the full solution →
3. 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 →
4. 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 →
5. 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 →
6. 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 →
7. 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 →
8. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.$\displaystyle\int \dfrac{\sqrt{x^4 + x^{-4} + 2}}{x^3}\, dx =$ (A) $\log|x| - \dfrac{1}{4x^4} + C$ (B) $\log|x| + \dfrac{1}{4x^4} + C$ (C) $\log|x| - \dfrac{4}{x^4} + C$ (D) $\log|x| + \dfrac{4}{x^4} + C$

   ›Reveal solutionSolution

   The expression under the square root is a perfect square, simplifying the integral to a sum of two elementary power terms.

   Concept and Intuition

   Recognize $x^4+x^{-4}+2$ as $(x^2+x^{-2})^2$ using the identity $a^2+b^2+2ab=(a+b)^2$ with $a=x^2,b=x^{-2}$.

   Step-by-Step Solution

   1. $(x^2+x^{-2})^2 = x^4+2+x^{-4}$ — matches the expression under the square root exactly.
   2. $\sqrt{x^4+x^{-4}+2} = |x^2+x^{-2}| = x^2+\dfrac{1}{x^2}$ (always positive for real $x\ne0$).
   3. Divide by $x^3$: $\dfrac{x^2+1/x^2}{x^3} = \dfrac1x+\dfrac{1}{x^5}$. …

   🔒 Start your 14-day free trial to see the full solution →
9. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
   Q.$\displaystyle\int \left(\sqrt{x + \sqrt{12x - 36}} + \sqrt{x - \sqrt{12x - 36}}\right) dx =$ (A) $2\sqrt{3}x + C, \ \forall x$ (B) $\dfrac{4(x - 3)^{3/2}}{3} + C, \ \forall x$ (C) $\begin{cases} \dfrac{4}{3}(x - 3)^{3/2} + C, & x > 6 \\ 2\sqrt{3}x + C, & 3 \leq x \leq 6 \end{cases}$ (D) $\begin{cases} \dfrac{4}{3}(x - 3)^{3/2} + C, & 3 \leq x \leq 6 \\ 2\sqrt{3}x + C, & x > 6 \end{cases}$

   ›Reveal solutionSolution

   Substituting $t=\sqrt{x-3}$ turns both nested radicals into perfect squares $(t\pm\sqrt3)^2$; the integrand collapses to a constant $2\sqrt3$ on $[3,6]$ and to $2\sqrt{x-3}$ for $x>6$, giving the piecewise antiderivative in (C).

   Concept and Intuition

   Nested radicals of the form $\sqrt{x\pm\sqrt{\text{linear in }x}}$ often simplify to perfect squares under a substitution that removes the inner square root. Here the key is recognizing $12x-36$ is a perfect multiple of $(x-3)$.

   Step-by-Step Solution

   1. Let $t=\sqrt{x-3}\ge0$ (valid for $x\ge3$), so $x=t^2+3$ and $12x-36=12t^2\Rightarrow \sqrt{12x-36}=2\sqrt3\,t$.
   2. $x+\sqrt{12x-36} = t^2+3+2\sqrt3\,t = (t+\sqrt3)^2 \Rightarrow \sqrt{x+\sqrt{12x-36}} = t+\sqrt3$ (always non-negative).
   3. $x-\sqrt{12x-36} = t^2+3-2\sqrt3\,t=(t-\sqrt3)^2 \Rightarrow \sqrt{x-\sqrt{12x-36}} = |t-\sqrt3|$.
   4. Sum $= (t+\sqrt3)+|t-\sqrt3|$. If $t\ge\sqrt3$ (i.e. $x-3\ge3 \Rightarrow x\ge6$): sum $=2t=2\sqrt{x-3}$. If $0\le t<\sqrt3$ (i.e. $3\le x<6$): sum $=(t+\sqrt3)+(\sqrt3-t)=2\sqrt3$. …

   🔒 Start your 14-day free trial to see the full solution →
10. 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 →
11. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
    Q.$\displaystyle\int \dfrac{x^2 - 1}{x^3\sqrt{2x^4 - 2x^2 + 1}}\,dx$ (A) $\dfrac{1 + 2x^2 + 2x^4}{2x^2} + c$ (B) $\dfrac{(1 + 2x^2 + 2x^4)^{1/2}}{2x^2} + c$ (C) $\dfrac{1 - 2x^2 + 2x^4}{2x^2} + c$ (D) $\dfrac{(1 - 2x^2 + 2x^4)^{1/2}}{2x^2} + c$

    ›Reveal solutionSolution

    Verifying the antiderivative by differentiating each candidate is faster and safer than guessing a substitution; the derivative of option (D) reproduces the integrand exactly. Answer: option (D).

    Concept and Intuition

    When an integral has an awkward-looking algebraic form under a square root, and the options are all algebraic expressions (not transcendental), the fastest rigorous check is to differentiate the candidate answers and see which one reproduces the integrand — this avoids errors in choosing a substitution.

    Step-by-Step Solution

    1. Let $N(x) = 1-2x^2+2x^4$ (note this equals $2x^4-2x^2+1$, the expression under the root in the integrand).
    2. Try $f(x) = \dfrac{N^{1/2}}{2x^2} = \dfrac12 N^{1/2}x^{-2}$.
    3. Differentiate: $f'(x) = \dfrac{N'}{4N^{1/2}x^2} - \dfrac{N^{1/2}}{x^3}$, where $N'=-4x+8x^3=4x(2x^2-1)$.
    4. First term becomes $\dfrac{(2x^2-1)}{xN^{1/2}}$. Combine both terms over the common denominator $x^3N^{1/2}$:

    $f'(x)=\frac{(2x^2-1)x^2 - N}{x^3N^{1/2}}.$

    5. Numerator: $(2x^2-1)x^2-N = (2x^4-x^2)-(1-2x^2+2x^4) = x^2-1$. …

    🔒 Start your 14-day free trial to see the full solution →
12. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
    Q.If $\int\sqrt{\dfrac{2}{1+\sin x}}dx=2\log|A(x)-B(x)|+C$ and $0\le x\le\dfrac{\pi}{2}$ then $B(\pi/4)=$ (A) $\dfrac{1}{\sqrt{2+3\sqrt{3}}}$ (B) $\dfrac{1}{\sqrt{3+2\sqrt{2}}}$ (C) $\dfrac{-1}{\sqrt{3+2\sqrt{2}}}$ (D) $\dfrac{2}{\sqrt{2+\sqrt{2}}}$

    ›Reveal solutionSolution

    A half-angle substitution turns $\sqrt{2/(1+\sin x)}$ into a secant, whose standard integral gives the $\log|\sec-\tan|$ form; then evaluate at $x=\pi/4$ and simplify the surd.

    Concept and Intuition

    $1+\sin x$ always simplifies via the identity $1+\sin x=1+\cos(\pi/2-x)=2\cos^2\left(\dfrac{\pi}{4}-\dfrac{x}{2}\right)$, turning an awkward square root into a clean secant function that has a standard log-integral.

    Step-by-Step Solution

    1. $1+\sin x=2\cos^2\left(\dfrac\pi4-\dfrac x2\right)$, so $\sqrt{\dfrac{2}{1+\sin x}}=\dfrac{1}{\left|\cos\left(\frac\pi4-\frac x2\right)\right|}=\sec\left(\dfrac\pi4-\dfrac x2\right)$ for $0\le x\le\pi/2$ (the argument stays in $[0,\pi/4]$, cosine positive).
    2. Let $u=\dfrac\pi4-\dfrac x2$, so $du=-\dfrac12 dx$, i.e. $dx=-2\,du$.
    3. $\int\sec u\,(-2\,du)=-2\log|\sec u+\tan u|+C=2\log\left|\dfrac{1}{\sec u+\tan u}\right|+C=2\log|\sec u-\tan u|+C$ (using $\sec^2u-\tan^2u=1$).
    4. So $A(x)=\sec\left(\dfrac\pi4-\dfrac x2\right)$, $B(x)=\tan\left(\dfrac\pi4-\dfrac x2\right)$. …

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