Q.Integrate the following function: $\int \sec x (\sec x + \tan x) dx$

Showing the 12 most recent of 19 on this concept.

1. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.$\displaystyle\sum_{n=1}^{\infty} \int_0^1 x^3(1 - x^2)^n\,dx =$ (A) $\dfrac{1}{4}$ (B) $\dfrac{1}{2}$ (C) $1$ (D) $2$

   ›Reveal solutionSolution

   Substituting $u=x^2$ turns the inner integral into a Beta-function value

   $\frac{1}{2(n+1)(n+2)}$; summing that telescoping series over $n\ge1$ gives $1/4$.

   Concept and Intuition

   $\int_0^1u^p(1-u)^q\,du=B(p+1,q+1)=\dfrac{p!\,q!}{(p+q+1)!}$ for non-negative integers

   $p,q$. Once the per-$n$ integral is reduced to this closed form, summing over $n$ often

   telescopes because $\frac{1}{(n+1)(n+2)}=\frac1{n+1}-\frac1{n+2}$.

   Step-by-Step Solution

   1. Substitute $u=x^2$, $du=2x\,dx$. Then $x^3\,dx=x^2\cdot x\,dx=u\cdot\dfrac{du}{2}$.
   2. $\displaystyle\int_0^1x^3(1-x^2)^n\,dx=\dfrac12\int_0^1u(1-u)^n\,du$.
   3. $\displaystyle\int_0^1u(1-u)^n\,du=B(2,n+1)=\dfrac{1!\,n!}{(n+2)!}=\dfrac{1}{(n+1)(n+2)}$.
   4. So the per-$n$ integral equals $\dfrac{1}{2(n+1)(n+2)}$. …

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2. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.If $\int\frac{1}{\cot\frac{x}{2}\cot\frac{x}{3}\cot\frac{x}{6}}dx = A\log\left|\cos\frac{x}{2}\right| + B\log\left|\cos\frac{x}{3}\right| + C\log\left|\cos\frac{x}{6}\right| + k$ then $A + B + C =$ (A) $7$ (B) $11$ (C) $-7$ (D) $1$

   ›Reveal solutionSolution

   Using $\tan3a\tan2a\tan a=\tan3a-\tan2a-\tan a$ (with $a=x/6$) turns the triple-tangent product into individually-integrable tangents, giving $A=-2,B=3,C=6$ so $A+B+C=7$.

   Concept and Intuition

   Products of three tangents whose angles sum in a $3a=2a+a$ pattern often simplify via the tangent-addition identity rearranged into $\tan(A+B)(1-\tan A\tan B)=\tan A+\tan B$, which for $A+B+C=$ a whole angle gives $\tan A\tan B\tan C=\tan(A+B+C)-\tan A-\tan B-\tan C$ type relations — turning an unwieldy product integral into a sum of standard $\int\tan(\cdot)\,dx$ terms.

   Step-by-Step Solution

   1. $\dfrac{1}{\cot\frac x2\cot\frac x3\cot\frac x6}=\tan\dfrac x2\tan\dfrac x3\tan\dfrac x6$.
   2. Let $a=\dfrac x6$; then $\dfrac x3=2a$ and $\dfrac x2=3a$, so the integrand is $\tan3a\tan2a\tan a$.
   3. From $\tan(3a)=\tan(2a+a)=\dfrac{\tan2a+\tan a}{1-\tan2a\tan a}$, cross-multiplying: $\tan3a(1-\tan2a\tan a)=\tan2a+\tan a$, i.e. $\tan3a-\tan2a-\tan a=\tan3a\tan2a\tan a$.
   4. So the integrand equals $\tan\dfrac x2-\tan\dfrac x3-\tan\dfrac x6$.
   5. $\displaystyle\int\tan\frac x2\,dx=-2\log\left|\cos\frac x2\right|$ (chain rule factor $2$), $\displaystyle\int\tan\frac x3\,dx=-3\log\left|\cos\frac x3\right|$, $\displaystyle\int\tan\frac x6\,dx=-6\log\left|\cos\frac x6\right|$. …

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3. AP EAPCET 2025Set eng-2025-05-23-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{x^4-16x^2+2x+8}{x^3-4x^2+2}\,dx=$ (A) $\dfrac{x^2+8x+c}{2}$ (B) $x^2+8x+c$ (C) $x^3-4x+c$ (D) $\dfrac{x^2-8x+c}{2}$

   ›Reveal solutionSolution

   The rational integrand is actually a polynomial in disguise — long division of the numerator by the denominator leaves zero remainder, collapsing the integral to a simple polynomial integral.

   Concept and Intuition

   When a numerator's degree exceeds the denominator's by 1, always try polynomial long division first; if the division is exact, the "hard" rational integral vanishes into an easy polynomial one.

   Step-by-Step Solution

   1. Divide $x^4+0x^3-16x^2+2x+8$ by $x^3-4x^2+0x+2$.
   2. First term of quotient: $x$. Multiply: $x^4-4x^3+0x^2+2x$. Subtract: $4x^3-16x^2+0x+8$.
   3. Next term of quotient: $4$. Multiply: $4x^3-16x^2+0x+8$. Subtract: remainder $=0$.
   4. So the integrand simplifies exactly to $x+4$. …

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4. AP EAPCET 2025Set eng-2025-05-26-AN1 markMCQ
   Q.$\displaystyle\int \sin^3 x \cos^2 x\, dx =$ (A) $\dfrac{\sin^4 x \cos x}{5} - \dfrac{\sin^2 x \cos x}{15} - \dfrac{2\cos x}{15} + c$ (B) $-\dfrac{\sin^4 x \cos x}{5} - \dfrac{\sin^2 x \cos x}{15} + \dfrac{2\cos x}{15} + c$ (C) $\dfrac{\sin^4 x \cos x}{5} - \dfrac{\sin^2 x \cos x}{15} + \dfrac{2x}{15} + c$ (D) $\dfrac{\sin^4 x \cos x}{5} + \dfrac{\sin^2 x \cos x}{3} - \dfrac{2x}{15} + c$

   ›Reveal solutionSolution

   Splitting off one $\sin x$ as $d(\cos x)$ and using $\sin^2x=1-\cos^2x$ reduces the integral to powers of $\cos x$, which re-expressed in $\sin x,\cos x$ mixed form matches option (A).

   Concept and Intuition

   For $\int\sin^{\text{odd}}x\cos^{\text{even}}x\,dx$, peel off one factor of $\sin x$ to pair with $dx$ (since $d(\cos x)=-\sin x\,dx$), turning the rest into a polynomial in $\cos x$ via $\sin^2x=1-\cos^2x$.

   Step-by-Step Solution

   1. $\sin^3x\cos^2x=\sin x(1-\cos^2x)\cos^2x=\sin x\cos^2x-\sin x\cos^4x$.
   2. $\displaystyle\int\sin x\cos^2x\,dx=-\dfrac{\cos^3x}3$, and $\displaystyle\int\sin x\cos^4x\,dx=-\dfrac{\cos^5x}5$.
   3. So the integral is $-\dfrac{\cos^3x}3-\left(-\dfrac{\cos^5x}5\right)=\dfrac{\cos^5x}5-\dfrac{\cos^3x}3+c$.
   4. Rewrite using $\cos^5x=(1-\sin^2x)^2\cos x$ and $\cos^3x=(1-\sin^2x)\cos x$: $\dfrac15(1-2\sin^2x+\sin^4x)\cos x-\dfrac13(1-\sin^2x)\cos x.$ …

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5. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.If $\displaystyle\int \dfrac{x^4 + 1}{x^2 + 1}\,dx = Ax^3 + Bx^2 + Cx + D\tan^{-1}x + E$, then $A + B + C + D =$ (A) $\dfrac{3}{2}$ (B) $\dfrac{4}{3}$ (C) $\dfrac{1}{3}$ (D) $\dfrac{2}{3}$

   ›Reveal solutionSolution

   Polynomial division turns $\frac{x^4+1}{x^2+1}$ into $x^2-1+\frac{2}{x^2+1}$, whose integral matches the given form with $A+B+C+D=\frac43$.

   Concept and Intuition

   When the numerator's degree is higher than the denominator's, divide first (or find a clever algebraic split) to reduce the integral to standard forms — here a polynomial plus a $\tan^{-1}$ term.

   Step-by-Step Solution

   1. Write $x^4+1 = (x^2+1)(x^2-1)+2$ — check: $(x^2+1)(x^2-1)=x^4-1$, plus $2$ gives $x^4+1$. ✓
   2. So $\dfrac{x^4+1}{x^2+1} = x^2-1+\dfrac{2}{x^2+1}$.
   3. Integrate term by term: $\displaystyle\int(x^2-1)dx + \int\frac{2}{x^2+1}dx = \frac{x^3}{3}-x+2\tan^{-1}x+E$. …

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6. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.$\displaystyle\int_0^2 x^2(2-x)^5\,dx =$ (A) $\dfrac{128}{21}$ (B) $\dfrac{64}{7}$ (C) $\dfrac{32}{21}$ (D) $\dfrac{16}{7}$

   ›Reveal solutionSolution

   This tests the standard Beta-integral shortcut $\int_0^a x^m(a-x)^n\,dx=a^{m+n+1}\dfrac{m!\,n!}{(m+n+1)!}$, which avoids expanding $(2-x)^5$ by hand.

   Concept and Intuition

   Any integral of the form $\int_0^a x^m(a-x)^n\,dx$ is a scaled Beta function: substituting $x=at$ turns it into $a^{m+n+1}\int_0^1 t^m(1-t)^n\,dt=a^{m+n+1}B(m+1,n+1)$, and $B(p,q)=\dfrac{(p-1)!(q-1)!}{(p+q-1)!}$ for positive integers. This is far quicker than binomially expanding $(2-x)^5$ and integrating term by term.

   Step-by-Step Solution

   1. Identify $a=2,\ m=2,\ n=5$, so $m+n+1=8$.
   2. Apply the formula: $I=a^{8}\cdot\dfrac{m!\,n!}{8!}=2^8\cdot\dfrac{2!\,5!}{8!}$.
   3. Compute: $2^8=256$, $2!=2$, $5!=120$, $8!=40320$.
   4. $I=256\cdot\dfrac{2\cdot120}{40320}=256\cdot\dfrac{240}{40320}=\dfrac{61440}{40320}$. …

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7. AP EAPCET 2024Set eng-2024-05-21-FN1 markMCQ
   Q.If $729 \displaystyle\int_1^3 \dfrac{1}{x^3(x^2+9)^2}\, dx = a + \log b$, then $a - b =$ (A) $4$ (B) $-\dfrac{4}{5}$ (C) $\dfrac{4}{5}$ (D) $-4$

   ›Reveal solutionSolution

   A partial-fraction decomposition (using $u=x^2$ to exploit the odd symmetry) reduces the integral to elementary logs and rational terms; evaluating between $1$ and $3$ gives $\dfrac{21}{5}-\ln5$, so $a-b=4$.

   Concept and Intuition

   Since the integrand is an odd function of $x$ (odd power of $x$ times an even function of $x$), substituting $u=x^2$ inside a partial-fraction split of $\dfrac{1}{u(u+9)^2}$ handles the repeated quadratic factor cleanly, after which everything integrates to logs and negative powers of $x$.

   Step-by-Step Solution

   1. Write $\dfrac{1}{x^3(x^2+9)^2}=\dfrac1x\cdot\dfrac{1}{x^2(x^2+9)^2}$ and set $u=x^2$: partial fractions give $\dfrac{1}{u(u+9)^2}=\dfrac{1/81}{u}-\dfrac{1/81}{u+9}-\dfrac{1/9}{(u+9)^2}$.
   2. Substituting back and further decomposing $\dfrac{1}{x(x^2+9)}$ and $\dfrac{1}{x(x^2+9)^2}$ (standard partial fractions), one arrives at the antiderivative

   $I(x)=-\frac{1}{162x^2}-\frac{1}{162(x^2+9)}-\frac{2}{729}\log x+\frac{1}{729}\log(x^2+9)+C,$

   which can be verified by direct differentiation to reproduce $\dfrac{1}{x^3(x^2+9)^2}$.

   3. Multiply by $729$: $729\,I(x)=-\dfrac{4.5}{x^2}-\dfrac{4.5}{x^2+9}-2\log x+\log(x^2+9)+C'$. …

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8. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{\sin 7x}{\sin 2x \sin 5x}\,dx =$ (A) $\log(\sin 5x \sin 2x) + c$ (B) $\log \sin 5x + \log \sin 2x + c$ (C) $\dfrac{1}{5}\log \sin 5x + \dfrac{1}{2}\log \sin 2x + c$ (D) $\dfrac{1}{5}\log \sin x + \dfrac{1}{2}\log \sin x + c$

   ›Reveal solutionSolution

   Splitting $\sin7x=\sin(2x+5x)$ via the angle-sum formula turns the fraction into a sum of two cotangents. Answer: option (C).

   Concept and Intuition

   Whenever an integral has $\sin(\text{sum of two angles})$ over the product of the sines of those two angles, expanding the numerator using the angle-addition formula is the key trick — it makes the fraction split into two simple $\cot$ terms.

   Step-by-Step Solution

   1. Note $7x = 2x+5x$, so $\sin7x = \sin2x\cos5x + \cos2x\sin5x$.
   2. Divide by $\sin2x\sin5x$: $\frac{\sin7x}{\sin2x\sin5x} = \frac{\cos5x}{\sin5x} + \frac{\cos2x}{\sin2x} = \cot5x+\cot2x.$
   3. Integrate term by term: $\displaystyle\int \cot5x\,dx = \frac15\log|\sin5x|$, and $\displaystyle\int\cot2x\,dx=\frac12\log|\sin2x|$.
   4. Sum: $\dfrac15\log\sin5x + \dfrac12\log\sin2x + c$.

   Common Mistakes …

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9. AP EAPCET 2024Set eng-2024-05-23-FN1 markMCQ
   Q.If $\int\dfrac{\sqrt[4]{x}}{\sqrt{x}+\sqrt[4]{x}}dx=\dfrac{2}{3}\left[A\sqrt[4]{x^3}+B\sqrt[4]{x^2}+C\sqrt[4]{x}+D\log\left(1+\sqrt[4]{x}\right)\right]+K$ then $\dfrac{2}{3}(A+B+C+D)=$ (A) $2/3$ (B) $-2/3$ (C) $4/3$ (D) $-4/3$

   ›Reveal solutionSolution

   Substitute $t=x^{1/4}$ to turn the integral into a rational function of $t$; polynomial-divide, integrate, and match coefficients.

   Concept and Intuition

   When an integrand only involves $x^{1/2}$ and $x^{1/4}$, substituting $t=x^{1/4}$ (the finest common root) rationalizes everything into a polynomial/rational function of $t$, which is then straightforward to integrate by polynomial long division.

   Step-by-Step Solution

   1. Let $t=x^{1/4}$, so $x=t^4$ and $dx=4t^3\,dt$. Also $x^{1/2}=t^2$.
   2. The integral becomes $\displaystyle\int \frac{t}{t^2+t}\cdot 4t^3\,dt = \int \frac{4t^3}{t+1}\,dt$ (cancelling one factor of $t$ from $t(t+1)$).
   3. Polynomial divide: $t^3 = (t+1)(t^2-t+1) - 1$, so $\dfrac{t^3}{t+1} = t^2-t+1-\dfrac{1}{t+1}$.
   4. So the integral is $4\displaystyle\int\left(t^2-t+1-\frac{1}{t+1}\right)dt = 4\left[\frac{t^3}{3}-\frac{t^2}{2}+t-\log(t+1)\right]+K = \frac43t^3-2t^2+4t-4\log(t+1)+K$. …

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10. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
    Q.If $C_j$ stands for ${}^{n}C_j$, then $\dfrac{C_0}{2}+\dfrac{C_1}{2.2^2}+\dfrac{C_2}{3.2^3}+\ldots+\dfrac{C_n}{(n+1)2^{n+1}}=$ (A) $\dfrac{3^n}{2^{n+1}(n+1)}$ (B) $\dfrac{3^{n+1}}{2^{n+1}(n+1)}$ (C) $\dfrac{3^n}{2^{n}(n+1)}$ (D) $\dfrac{3^{n+1}}{2^{n}(n+1)}$

    ›Reveal solutionSolution

    The sum is a disguised integral of a binomial expansion. By recognizing each term as $\frac{1}{2^{n+1}}\int_0^{1/2} t^k \,dt$ and summing, we get $\frac{1}{2^{n+1}(n+1)}\left(1+\frac12\right)^{n+1} - \frac{1}{2^{n+1}(n+1)} = \frac{3^{n+1}-1}{2^{n+1}(n+1)}$, but the given sum starts at $k=0$ with denominator $(k+1)2^{k+1}$, leading to $\frac{3^{n+1}-1}{2^{n+1}(n+1)}$. However, the options lack the $-1$ term; rechecking shows the sum is actually $\frac{1}{n+1}\left[\left(1+\frac12\right)^{n+1} - 1\right] = \frac{3^{n+1}-2^{n+1}}{2^{n+1}(n+1)}$, which matches option (B) after correcting the index: the sum is $\frac{3^{n+1}}{2^{n+1}(n+1)}$ only if we include a missing term? Wait — careful: the sum runs $k=0$ to $n$, so the integral method gives $\frac{1}{2^{n+1}(n+1)}\left[(1+1/2)^{n+1} - 1\right] = \frac{3^{n+1}-2^{n+1}}{2^{n+1}(n+1)}$. None of the options match that. Let's re-derive properly: The sum is $\sum_{k=0}^n \frac{C_k}{(k+1)2^{k+1}} = \frac{1}{2^{n+1}}\sum_{k=0}^n \frac{2^{n-k}C_k}{k+1}$. Use identity $\frac{C_k}{k+1} = \frac{1}{n+1}C_{k+1}^{n+1}$. Then sum becomes $\frac{1}{2^{n+1}(n+1)}\sum_{k=0}^n 2^{n-k} C_{k+1}^{n+1} = \frac{1}{2^{n+1}(n+1)}\sum_{j=1}^{n+1} 2^{n+1-j} C_j^{n+1} = \frac{1}{2^{n+1}(n+1)}\left[(1+2)^{n+1} - 2^{n+1}\right] = \frac{3^{n+1}-2^{n+1}}{2^{n+1}(n+1)}$. This is not among options. But wait — the problem writes $\frac{C_0}{2} + \frac{C_1}{2\cdot 2^2} + \cdots$; the first term is $C_0/2 = 1/2$. For $n=1$, sum = $1/2 + 1/(2\cdot4)=1/2+1/8=5/8$. Option (B) gives $3^2/(2^2\cdot2)=9/8$, too big. Option (A): $3^1/(2^2\cdot2)=3/8$, too small. So none match? There's a misprint: likely the sum is $\frac{C_0}{1\cdot2^1} + \frac{C_1}{2\cdot2^2} + \cdots + \frac{C_n}{(n+1)2^{n+1}}$, which equals $\frac{3^{n+1}-1}{2^{n+1}(n+1)}$? No. Let's check standard result: $\sum_{k=0}^n \frac{C_k}{k+1} = \frac{2^{n+1}-1}{n+1}$. Here we have extra $2^{-(k+1)}$, so it's $\frac{1}{n+1}\left[(1+\frac12)^{n+1} - 1\right] = \frac{1}{n+1}\left[\frac{3^{n+1}}{2^{n+1}} - 1\right] = \frac{3^{n+1}-2^{n+1}}{2^{n+1}(n+1)}$. Still not matching. Given options, the intended answer is likely (B) $\frac{3^{n+1}}{2^{n+1}(n+1)}$, which would arise if the sum started from $k=0$ to $n$ but with an extra factor? Actually, if the sum were $\sum_{k=0}^n \frac{C_k}{(k+1)2^{k}}$ then it becomes $\frac{3^{n+1}-1}{2^{n}(n+1)}$? Not. Let's trust the standard integral method: $\int_0^{1/2} (1+x)^n dx = \frac{(3/2)^{n+1}-1}{n+1} = \sum_{k=0}^n \frac{C_k}{k+1}(1/2)^{k+1}$. So the sum is exactly $\frac{(3/2)^{n+1}-1}{n+1} = \frac{3^{n+1}-2^{n+1}}{2^{n+1}(n+1)}$. Since that's not an option, the problem likely has a typo and expects (B) as the closest, or they meant the sum without the $-1$ term. Many textbooks give $\sum_{k=0}^n \frac{C_k}{(k+1)2^{k+1}} = \frac{3^{n+1}-2^{n+1}}{2^{n+1}(n+1)}$, but here the options suggest they forgot the $-2^{n+1}$ term. Given the pattern, option (B) is the only one with $3^{n+1}$ and $2^{n+1}(n+1)$ denominator. So we select (B).

    The sum is a binomial expansion integrated termwise from 0 to 1/2. The result simplifies to $\frac{3^{n+1}-2^{n+1}}{2^{n+1}(n+1)}$, but among the given choices, option (B) $\frac{3^{n+1}}{2^{n+1}(n+1)}$ is the intended answer (likely a slight misprint in the problem).

    Concept and Intuition: Integration By Expansion

    When you see a sum of the form $\sum \frac{C_k}{k+1} a^{k+1}$, think of the integral $\int_0^a (1+x)^n \,dx$. Why? Because expanding $(1+x)^n = \sum C_k x^k$ and integrating termwise gives $\sum C_k \frac{a^{k+1}}{k+1}$. Here $a = 1/2$, so the sum is exactly that integral. This turns a messy finite sum into a clean closed form.

    Step-by-step solution

    1. Recognize the pattern

       The general term is $\frac{C_k}{(k+1)2^{k+1}}$, where $C_k = \binom{n}{k}$. This looks like $\frac{1}{2^{k+1}} \cdot \frac{C_k}{k+1}$. The factor $\frac{1}{k+1}$ suggests an integral: $\int_0^{1/2} x^k \,dx = \frac{(1/2)^{k+1}}{k+1}$.

    2. Rewrite the sum as an integral

       $$$$

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11. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
    Q.$\displaystyle\int \dfrac{1-\cos x}{\cos x(1+\cos x)}\,dx=$ (A) $\log|\sec x+\tan x|-2(\csc x-\cot x)+C$ (B) $\log|\sec x+\tan x|-2(\csc x+\cot x)+C$ (C) $\log|\sec x+\tan x|+2(\csc x-\cot x)+C$ (D) $\log|\sec x+\tan x|+2(\csc x+\cot x)+C$

    ›Reveal solutionSolution

    Splitting the integrand by partial fractions in $\cos x$ reduces it to $\sec x-\sec^2(x/2)$, whose integral, rewritten via the half-angle identity $\tan(x/2)=\csc x-\cot x$, matches option (A).

    Concept and Intuition

    Treating $c=\cos x$ as the variable lets us do ordinary partial fractions; then the half-angle substitution $1+\cos x=2\cos^2(x/2)$ turns the resulting term into a standard $\sec^2$ integral.

    Step-by-Step Solution

    1. Let $c=\cos x$. Write $\dfrac{1-c}{c(1+c)}=\dfrac{A}{c}+\dfrac{B}{1+c}$: $1-c=A(1+c)+Bc$. At $c=0$: $A=1$. At $c=-1$: $2=-B\Rightarrow B=-2$.
    2. So the integrand $=\dfrac1{\cos x}-\dfrac{2}{1+\cos x}=\sec x-\dfrac{2}{2\cos^2(x/2)}=\sec x-\sec^2(x/2)$.
    3. $\int\sec x\,dx=\log|\sec x+\tan x|+C$.
    4. $\int\sec^2(x/2)\,dx=2\tan(x/2)$, so $-2\int\sec^2(x/2)\cdot\tfrac12\,dx$... directly: $\int\dfrac{2}{1+\cos x}dx=2\tan(x/2)$. …

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12. AP EAPCET 2023Set eng-2023-05-18-FN1 markMCQ
    Q.$\displaystyle\int \dfrac{dx}{\sin(x-a)\cos(x-b)}=$ (A) $\dfrac{1}{\sin(a-b)}\log\left|\dfrac{\sin(x-a)}{\cos(x-b)}\right|+C$ (B) $\dfrac{1}{\cos(b-a)}\log\left|\dfrac{\sin(x-a)}{\cos(x-b)}\right|+C$ (C) $\dfrac{1}{\cos(b-a)}\left[\log|\sin(x-a)\cos(x-b)|\right]+C$ (D) $\dfrac{1}{\sin(a-b)}\left[\log|\sin(x-a)\cos(x-b)|\right]+C$

    ›Reveal solutionSolution

    This is a classic 'insert a constant angle difference' trick: express $\cos(a-b)$ (a constant) using the compound-angle formula on $(x-a)$ and $(x-b)$, then split the integrand into $\cot(x-a)+\tan(x-b)$, each of which integrates to a log.

    Concept and Intuition

    When an integrand has $\sin(x-a)\cos(x-b)$ in the denominator, the key observation is that $(x-a)-(x-b)=b-a$ is a constant (independent of $x$). This lets us build a compound-angle identity for $\cos(b-a)$ or $\sin(b-a)$ purely in terms of $(x-a)$ and $(x-b)$, and divide through to split the fraction into two integrable pieces ($\cot$ and $\tan$).

    Step-by-Step Solution

    1. Note $(x-a)-(x-b)=b-a$, so $\cos(b-a)=\cos[(x-a)-(x-b)]=\cos(x-a)\cos(x-b)+\sin(x-a)\sin(x-b)$.
    2. Divide both sides by $\sin(x-a)\cos(x-b)$:

    $\dfrac{\cos(b-a)}{\sin(x-a)\cos(x-b)}=\dfrac{\cos(x-a)}{\sin(x-a)}+\dfrac{\sin(x-b)}{\cos(x-b)}=\cot(x-a)+\tan(x-b)$

    3. So $\dfrac{1}{\sin(x-a)\cos(x-b)}=\dfrac{1}{\cos(b-a)}\left[\cot(x-a)+\tan(x-b)\right]$.
    4. Integrate: $\displaystyle\int[\cot(x-a)+\tan(x-b)]\,dx=\log|\sin(x-a)|-\log|\cos(x-b)|=\log\left|\dfrac{\sin(x-a)}{\cos(x-b)}\right|$. …

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