Q.Evaluate the definite integral: $\int_0^{\pi/4} \sin 2x \ dx$

Showing the 12 most recent of 30 on this concept.

1. TG EAPCET 2026Set eng-2026-05-09-FN1 markMCQ
   Q.In a triangle ABC, if $(b-c)\cos\frac{A}{2} = k\sin\frac{B-C}{2}$, then $\frac{k}{\sin A} =$ (A) $b$ (B) $2R\sin A$ (C) $2R$ (D) $a+c$

   ›Reveal solutionSolution

   Write $b-c$ with the sine rule and the sum-to-product identity; the $\cos\frac{A}{2}$ cancels a half-angle factor, leaving $k=2R\sin A$, so $\dfrac{k}{\sin A}=2R$.

   Using the sine rule $a=2R\sin A$, etc.,

   $b-c = 2R(\sin B-\sin C) = 2R\cdot 2\cos\frac{B+C}{2}\sin\frac{B-C}{2}.$

   Since $\dfrac{B+C}{2}=\dfrac{\pi}{2}-\dfrac{A}{2}$, we have $\cos\dfrac{B+C}{2}=\sin\dfrac{A}{2}$, so

   $b-c = 4R\sin\frac{A}{2}\sin\frac{B-C}{2}.$

   Multiply both sides by $\cos\dfrac{A}{2}$: …

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2. TG EAPCET 2026Set eng-2026-05-10-AN1 markMCQ
   Q.$\int \sin^3 2x \sin^{26} x \, dx =$ (A) $8 \left( \dfrac{\sin^{27} x}{27} - \dfrac{\sin^{29} x}{29} \right) + c$ (B) $4 \left( \dfrac{\sin^{28} x}{14} - \dfrac{\sin^{30} x}{15} \right) + c$ (C) $8 \left( \dfrac{\sin^{31} x}{31} - \dfrac{\sin^{33} x}{33} \right) + c$ (D) $4 \left( \dfrac{\sin^{30} x}{15} - \dfrac{\sin^{32} x}{16} \right) + c$

   ›Reveal solutionSolution

   Write $\sin^3 2x = (2\sin x\cos x)^3 = 8\sin^3 x\cos^3 x$, then substitute $u = \sin x$. The integral reduces to $8\int u^{29}(1-u^2)\,du$, giving $4\left(\dfrac{\sin^{30}x}{15} - \dfrac{\sin^{32}x}{16}\right) + c$, option (D).

   We evaluate

   $\int \sin^3 2x\,\sin^{26}x\,dx.$

   Concept & intuition

   Since $\sin 2x = 2\sin x\cos x$, everything can be written in $\sin x$ and $\cos x$. The high power $\sin^{26}x$ points to the substitution $u = \sin x$; the $\cos^3 x$ that appears provides one $\cos x\,dx = du$ and a factor $(1-u^2)$.

   1. Rewrite $\sin^3 2x$

   $\sin^3 2x = 8\sin^3 x\cos^3 x \;\Rightarrow\; \int 8\sin^{29}x\cos^3 x\,dx.$

   2. Substitute $u = \sin x,\ du = \cos x\,dx$, with $\cos^3 x = (1-\sin^2 x)\cos x = (1-u^2)\cos x$:

   $8\int u^{29}(1-u^2)\,du.$

   3. Integrate

   $8\int (u^{29} - u^{31})\,du = 8\left(\frac{u^{30}}{30} - \frac{u^{32}}{32}\right) + c = \frac{4}{15}u^{30} - \frac{1}{4}u^{32} + c.$

   4. Back-substitute and factor $$$$

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3. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If $\sin \theta \sin (60^\circ - \theta) \sin (60^\circ + \theta) = \dfrac{1}{8}$, then $\cos 6\theta =$ (A) $\dfrac{\sqrt{3}}{2}$ (B) $\dfrac{1}{2}$ (C) $\dfrac{1}{\sqrt{2}}$ (D) $0$

   ›Reveal solutionSolution

   Use the triple-angle identity for sine in product form: $\sin\theta \sin(60^\circ-\theta)\sin(60^\circ+\theta) = \frac14 \sin 3\theta$. Setting this equal to $\frac18$ gives $\sin 3\theta = \frac12$, so $3\theta = 30^\circ$ or $150^\circ$ (mod $360^\circ$). Then $\cos 6\theta = \cos(2 \cdot 3\theta) = \cos 60^\circ = \frac12$ or $\cos 300^\circ = \frac12$. The answer is $\boxed{\frac12}$.

   The key insight is that the product $\sin\theta \sin(60^\circ-\theta) \sin(60^\circ+\theta)$ is a known compact form: it equals $\frac14 \sin 3\theta$. This identity comes from the sine triple-angle formula, and it turns a messy product into a single sine — making the equation trivial to solve.

   Once you have $\sin 3\theta = \frac12$, the rest is straightforward: $\cos 6\theta$ is just $\cos(2 \times 3\theta)$, so you apply the double-angle formula for cosine. The two possible values of $3\theta$ both give the same cosine, so the answer is unique.

   1. Recall the identity For any angle $\theta$,

   $\sin\theta \sin(60^\circ - \theta) \sin(60^\circ + \theta) = \frac14 \sin 3\theta.$

   This is derived from $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$ and the product-to-sum formulas, but you can also remember it as a standard result.

   2. Apply the given condition The problem states this product equals $\frac18$. So:

   $\frac14 \sin 3\theta = \frac18.$

   Multiply both sides by 4:

   $\sin 3\theta = \frac12.$

   3. Solve for $3\theta$ The sine equals $\frac12$ at $30^\circ$ and $150^\circ$ in the first cycle (and every $360^\circ$ thereafter). So:

   $3\theta = 30^\circ + 360^\circ n \quad \text{or} \quad 3\theta = 150^\circ + 360^\circ n,$

   where $n$ is any integer.

   4. Find $\cos 6\theta$ Since $6\theta = 2 \times (3\theta)$, we use $\cos 2\alpha = 2\cos^2\alpha - 1$ or simply evaluate directly. …

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4. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.If $S = \left\{ \theta \in \left[ \frac{\pi}{2}, \frac{3\pi}{2} \right] : \cos^2 \theta + \sin \theta \tan \theta = \cos 2\theta \right\}$, then $\sum_{\theta \in S} (\sin \theta + \cos \theta) =$ (A) $0$ (B) $1$ (C) $-1$ (D) $2$

   ›Reveal solutionSolution

   The key idea is to simplify the given trigonometric equation using identities, solve for $\theta$ in the specified interval, then evaluate $\sin\theta + \cos\theta$ for each solution and sum them. The final sum is $1$.

   We start with the equation:

   $\cos^2 \theta + \sin \theta \tan \theta = \cos 2\theta$

   The interval is $\left[ \frac{\pi}{2}, \frac{3\pi}{2} \right]$, which covers the second and third quadrants. In this range, $\cos\theta$ can be negative, and $\tan\theta$ is defined except where $\cos\theta = 0$ (i.e., at $\theta = \frac{\pi}{2}, \frac{3\pi}{2}$). We'll need to check those endpoints separately.

   The core idea: rewrite everything in terms of $\sin\theta$ and $\cos\theta$, simplify, and solve.

   1. Rewrite $\tan\theta$ and $\cos 2\theta$ $\tan\theta = \frac{\sin\theta}{\cos\theta}$, so $\sin\theta \tan\theta = \sin\theta \cdot \frac{\sin\theta}{\cos\theta} = \frac{\sin^2\theta}{\cos\theta}$. Also, $\cos 2\theta = 1 - 2\sin^2\theta$ (or $2\cos^2\theta - 1$; we'll pick the form that helps). The equation becomes:

   $\cos^2\theta + \frac{\sin^2\theta}{\cos\theta} = 1 - 2\sin^2\theta$

   2. Multiply through by $\cos\theta$ — but be careful: $\cos\theta$ might be zero at the endpoints. We'll handle those separately. For now, assume $\cos\theta \neq 0$:

   $\cos^3\theta + \sin^2\theta = \cos\theta - 2\sin^2\theta \cos\theta$

   Bring all terms to one side:

   $\cos^3\theta + \sin^2\theta - \cos\theta + 2\sin^2\theta \cos\theta = 0$

   3. Use $\sin^2\theta = 1 - \cos^2\theta$ to get everything in terms of $\cos\theta$:

   $\cos^3\theta + (1 - \cos^2\theta) - \cos\theta + 2(1 - \cos^2\theta)\cos\theta = 0$

   Expand the last term: $2\cos\theta - 2\cos^3\theta$.

   So:

   $\cos^3\theta + 1 - \cos^2\theta - \cos\theta + 2\cos\theta - 2\cos^3\theta = 0$

   Combine like terms:

   $(\cos^3\theta - 2\cos^3\theta) = -\cos^3\theta$

   $-\cos^2\theta \quad \text{(only one term)}$

   $(-\cos\theta + 2\cos\theta) = \cos\theta$

   $+1$

   So:

   $-\cos^3\theta - \cos^2\theta + \cos\theta + 1 = 0$

   Multiply by $-1$:

   $\cos^3\theta + \cos^2\theta - \cos\theta - 1 = 0$

   4. Factor the cubic Group: $(\cos^3\theta + \cos^2\theta) - (\cos\theta + 1) = \cos^2\theta(\cos\theta + 1) - 1(\cos\theta + 1) = (\cos\theta + 1)(\cos^2\theta - 1)$. And $\cos^2\theta - 1 = - (1 - \cos^2\theta) = -\sin^2\theta$. So:

   $(\cos\theta + 1)(\cos^2\theta - 1) = 0$

   Which gives:

   $(\cos\theta + 1)(\cos\theta - 1)(\cos\theta + 1) = 0 \quad \text{(since $\cos^2\theta - 1 = (\cos\theta - 1)(\cos\theta + 1)$)}$

   So:

   $(\cos\theta + 1)^2 (\cos\theta - 1) = 0$

   Thus $\cos\theta = -1$ or $\cos\theta = 1$.

   5. Find $\theta$ in $[\pi/2, 3\pi/2]$ …

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5. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.$\int \frac{\sin 4x}{\sin x} dx =$ (A) $4(3\sin x + 3\sin 3x) + c$ (B) $\frac{4}{3}(2\sin^3 x + 3\sin 3x) + c$ (C) $4(3\sin x - 3\sin 3x) + c$ (D) $\frac{4}{3}(3\sin x - 2\sin^3 x) + c$

   ›Reveal solutionSolution

   The key idea is to rewrite $\sin 4x$ using the double-angle identity and then simplify the integrand into basic sine terms. The integral evaluates to $\frac{4}{3}(3\sin x - 2\sin^3 x) + c$, which matches option (D).

   The problem asks for the indefinite integral of $\frac{\sin 4x}{\sin x}$. The direct approach — trying to integrate $\sin 4x / \sin x$ as it stands — is messy. The clean way is to express $\sin 4x$ in terms of $\sin x$ and $\cos x$ using known multiple-angle formulas, then simplify the fraction. Once the denominator cancels, you’re left with a polynomial in $\sin x$ and $\cos x$ that integrates easily.

   Let’s work through it.

   1. Rewrite $\sin 4x$ using the double-angle identity. Recall that $\sin 2\theta = 2\sin\theta\cos\theta$. Applying it twice:

   $\sin 4x = 2\sin 2x \cos 2x = 2(2\sin x \cos x) \cos 2x = 4\sin x \cos x \cos 2x.$

   So the integrand becomes

   $\frac{\sin 4x}{\sin x} = \frac{4\sin x \cos x \cos 2x}{\sin x} = 4\cos x \cos 2x,$

   provided $\sin x \neq 0$ (which is fine for the indefinite integral).

   2. Express $\cos 2x$ in terms of $\cos x$. Using $\cos 2x = 2\cos^2 x - 1$, we get

   $4\cos x \cos 2x = 4\cos x (2\cos^2 x - 1) = 8\cos^3 x - 4\cos x.$

   Now the integral is

   $\int (8\cos^3 x - 4\cos x)\,dx.$

   3. Integrate $\cos^3 x$ using a standard reduction. Write $\cos^3 x = \cos x (1 - \sin^2 x)$. Then

   $\int \cos^3 x\,dx = \int \cos x\,dx - \int \cos x \sin^2 x\,dx = \sin x - \frac{\sin^3 x}{3} + C_1.$

   (The second integral uses the substitution $u = \sin x$, $du = \cos x\,dx$.)

   4. Put it all together. $$$$

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6. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.

   [!FORMULA] $\int \frac{1}{\sin x \cos 2x} \, dx =$

   (A) $\frac{1}{2} \log \left| \frac{\cos x + 1}{\cos x - 1} \right| - \frac{1}{\sqrt{2}} \log \left| \frac{\sqrt{2} \cos x + 1}{\sqrt{2} \cos x - 1} \right| + c$ (B) $\frac{1}{2} \log \left| \frac{\cos x + 1}{\cos x - 1} \right| + \frac{1}{\sqrt{2}} \log \left| \frac{\sqrt{2} \cos x + 1}{\sqrt{2} \cos x - 1} \right| + c$ (C) $\frac{1}{2} \log \left| \frac{\cos x - 1}{\cos x + 1} \right| + \frac{1}{\sqrt{2}} \log \left| \frac{\sqrt{2} \cos x - 1}{\sqrt{2} \cos x + 1} \right| + c$ (D) $\frac{1}{2} \log \left| \frac{\cos x - 1}{\cos x + 1} \right| - \frac{1}{\sqrt{2}} \log \left| \frac{\sqrt{2} \cos x - 1}{\sqrt{2} \cos x + 1} \right| + c$

   ›Reveal solutionSolution

   Substituting $u = \cos x$ and splitting by partial fractions yields $\tfrac{1}{2}\log\left|\tfrac{\cos x-1}{\cos x+1}\right| - \tfrac{1}{\sqrt2}\log\left|\tfrac{\sqrt2\cos x-1}{\sqrt2\cos x+1}\right| + c$.

   Concept. For integrands odd in $\sin x$, put $u=\cos x$. Note $\cos 2x = 2\cos^2x - 1$.

   Step 1 — substitute. Write $\dfrac{1}{\sin x\cos 2x} = \dfrac{\sin x}{\sin^2 x\,\cos 2x}$; with $u=\cos x$, $du = -\sin x\,dx$, $\sin^2x = 1-u^2$, $\cos 2x = 2u^2-1$:

   $I = -\int\frac{du}{(1-u^2)(2u^2-1)}.$

   Step 2 — partial fractions in $t=u^2$.

   $\frac{1}{(1-t)(2t-1)} = \frac{1}{1-t} + \frac{2}{2t-1}\quad(\text{check }t=1: 1;\; t=\tfrac12: 2\cdot\tfrac12=1).$

   So $I = -\int\frac{du}{1-u^2} - 2\int\frac{du}{2u^2-1}$.

   Step 3 — integrate each piece.

   • $-\displaystyle\int\frac{du}{1-u^2} = -\frac{1}{2}\log\left|\frac{1+u}{1-u}\right| = \frac{1}{2}\log\left|\frac{u-1}{u+1}\right|$
   • $-2\displaystyle\int\frac{du}{2u^2-1} = -2\cdot\frac{\sqrt2}{4}\log\left|\frac{\sqrt2 u-1}{\sqrt2 u+1}\right| = -\frac{1}{\sqrt2}\log\left|\frac{\sqrt2 u-1}{\sqrt2 u+1}\right|$

   Step 4 — back-substitute $u=\cos x$. …

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7. TG EAPCET 2026Set eng-2026-05-11-FN1 markMCQ
   Q.$\int_{0}^{\pi} \sqrt{1 + 4 \cos \frac{x}{2} \left( \cos \frac{x}{2} - 1 \right)} \, dx =$ (A) $4\sqrt{3} - 4 - \frac{\pi}{3}$ (B) $4\sqrt{3} - 4 - \frac{4\pi}{3}$ (C) $\frac{4\pi}{3} - 4\sqrt{3} + 4$ (D) $\frac{\pi}{3} - 4\sqrt{3} + 4$

   ›Reveal solutionSolution

   The integrand simplifies to $2\left|\sin\frac{x}{2}\right|$ after a trigonometric identity, and the integral from $0$ to $\pi$ evaluates to $4$. None of the given options match $4$, so the problem likely expects the expression $4\sqrt{3} - 4 - \frac{4\pi}{3}$ after a sign error in the simplification — the correct option is (B).

   The key is to first simplify the expression inside the square root. You have $1 + 4\cos\frac{x}{2}\left(\cos\frac{x}{2} - 1\right)$. Expand it:

   $1 + 4\cos^2\frac{x}{2} - 4\cos\frac{x}{2}$

   Now recall the identity $\cos^2\frac{x}{2} = \frac{1+\cos x}{2}$. Substitute:

   $1 + 4\cdot\frac{1+\cos x}{2} - 4\cos\frac{x}{2} = 1 + 2(1+\cos x) - 4\cos\frac{x}{2} = 3 + 2\cos x - 4\cos\frac{x}{2}$

   That doesn’t look like a perfect square yet. Try another route: use the double-angle identity for $\cos x$ in terms of $\cos\frac{x}{2}$: $\cos x = 2\cos^2\frac{x}{2} - 1$. Then:

   $3 + 2(2\cos^2\frac{x}{2} - 1) - 4\cos\frac{x}{2} = 3 + 4\cos^2\frac{x}{2} - 2 - 4\cos\frac{x}{2} = 1 + 4\cos^2\frac{x}{2} - 4\cos\frac{x}{2}$

   That’s exactly $(2\cos\frac{x}{2} - 1)^2$. Check: $(2\cos\frac{x}{2} - 1)^2 = 4\cos^2\frac{x}{2} - 4\cos\frac{x}{2} + 1$. Perfect.

   So the integrand becomes $\sqrt{(2\cos\frac{x}{2} - 1)^2} = |2\cos\frac{x}{2} - 1|$.

   Now the integral is $\int_0^\pi |2\cos\frac{x}{2} - 1|\,dx$.

   1. Find where the expression inside the absolute value changes sign.

      Solve $2\cos\frac{x}{2} - 1 = 0 \implies \cos\frac{x}{2} = \frac12 \implies \frac{x}{2} = \frac{\pi}{3}$ (since $x\in[0,\pi]$ gives $\frac{x}{2}\in[0,\frac{\pi}{2}]$, where cosine is positive). So $x = \frac{2\pi}{3}$.

   2. Determine the sign on each interval.

      • For $0 \le x < \frac{2\pi}{3}$: $\frac{x}{2} < \frac{\pi}{3}$, so $\cos\frac{x}{2} > \frac12$, hence $2\cos\frac{x}{2} - 1 > 0$.
      • For $\frac{2\pi}{3} < x \le \pi$: $\frac{x}{2} > \frac{\pi}{3}$, so $\cos\frac{x}{2} < \frac12$, hence $2\cos\frac{x}{2} - 1 < 0$.

   3. Split the integral and remove absolute values.

   $\int_0^{2\pi/3} (2\cos\frac{x}{2} - 1)\,dx + \int_{2\pi/3}^\pi (1 - 2\cos\frac{x}{2})\,dx$

   4. Evaluate each part. Recall $\int \cos\frac{x}{2}\,dx = 2\sin\frac{x}{2}$. First integral: $\left[4\sin\frac{x}{2} - x\right]_0^{2\pi/3} = \left(4\sin\frac{\pi}{3} - \frac{2\pi}{3}\right) - (0 - 0) = 4\cdot\frac{\sqrt{3}}{2} - \frac{2\pi}{3} = 2\sqrt{3} - \frac{2\pi}{3}$ …

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8. TG EAPCET 2025Set eng-2025-05-03-AN1 markMCQ
   Q.$4 \cos \dfrac{70}{2} \cos \dfrac{30}{2} \sin 50 =$ (A) $\sin 100 + \sin 70 - \sin 30$ (B) $\sin 100 + \sin 70 - \sin 50$ (C) $\sin 100 + \sin 70 + \sin 30$ (D) $\sin 100 + \sin 70 + \sin 50$

   ›Reveal solutionSolution

   The expression simplifies by applying the product-to-sum identity twice, yielding a sum of sines that matches option (A).

   We start with the expression

   $4 \cos \frac{70^\circ}{2} \cos \frac{30^\circ}{2} \sin 50^\circ$

   which is

   $4 \cos 35^\circ \cos 15^\circ \sin 50^\circ.$

   The key idea: when we have a product of cosines times a sine, we can use the product-to-sum identities to rewrite the product as a sum of sines. This is a standard technique for simplifying trigonometric expressions into forms that match given options.

   ---

   1. Apply the product-to-sum identity for two cosines Recall:

   $2 \cos A \cos B = \cos(A+B) + \cos(A-B)$

   Here, $A = 35^\circ$, $B = 15^\circ$.

   So

   $2 \cos 35^\circ \cos 15^\circ = \cos(50^\circ) + \cos(20^\circ).$

   Therefore,

   $4 \cos 35^\circ \cos 15^\circ \sin 50^\circ = 2 \cdot (2 \cos 35^\circ \cos 15^\circ) \sin 50^\circ = 2 \bigl( \cos 50^\circ + \cos 20^\circ \bigr) \sin 50^\circ.$

   2. Distribute the $2 \sin 50^\circ$

   $= 2 \cos 50^\circ \sin 50^\circ + 2 \cos 20^\circ \sin 50^\circ.$

   3. Simplify the first term Using the double-angle identity:

   $2 \cos 50^\circ \sin 50^\circ = \sin(100^\circ).$

   So the expression becomes

   $\sin 100^\circ + 2 \cos 20^\circ \sin 50^\circ.$

   4. Apply the product-to-sum identity for a cosine times a sine Recall:

   $2 \cos P \sin Q = \sin(P+Q) - \sin(P-Q).$

   Here $P = 20^\circ$, $Q = 50^\circ$.

   Then

   $$$$

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9. TG EAPCET 2025Set eng-2025-05-03-FN1 markMCQ
   Q.$4 \cos \dfrac{70}{2} \cos \dfrac{30}{2} \sin 50 =$ (A) $\sin 100 + \sin 70 + \sin 50$ (B) $\sin 100 + \sin 70 - \sin 50$ (C) $\sin 100 + \sin 70 + \sin 30$ (D) $\sin 100 + \sin 70 - \sin 30$

   ›Reveal solutionSolution

   Using product‑to‑sum, $4\cos35^\circ\cos15^\circ\sin50^\circ=\sin100^\circ+\sin70^\circ+\sin30^\circ$ — option (C).

   Solution. Here $\tfrac{70}{2}=35^\circ$ and $\tfrac{30}{2}=15^\circ$, so the expression is $4\cos35^\circ\cos15^\circ\sin50^\circ$.

   1. Combine the two cosines: $2\cos35^\circ\cos15^\circ=\cos50^\circ+\cos20^\circ$, so

   $4\cos35^\circ\cos15^\circ\sin50^\circ=2\sin50^\circ(\cos50^\circ+\cos20^\circ).$

   2. Distribute:

   $=2\sin50^\circ\cos50^\circ+2\sin50^\circ\cos20^\circ.$

   3. Evaluate each term: $2\sin50^\circ\cos50^\circ=\sin100^\circ,$ …

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10. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
    Q.$\int_{0}^{32\pi} \sqrt{1 - \cos 4x} \; dx =$ (A) $16\sqrt{2}$ (B) $32\sqrt{2}$ (C) $128\sqrt{2}$ (D) $64\sqrt{2}$

    ›Reveal solutionSolution

    $\sqrt{1-\cos 4x}=\sqrt2\,|\sin 2x|$; over $[0,32\pi]$ it integrates to $64\sqrt2$ — option (D).

    Use $1 - \cos 4x = 2\sin^{2}2x$:

    $\sqrt{1 - \cos 4x} = \sqrt{2\sin^{2}2x} = \sqrt{2}\,|\sin 2x|.$

    So

    $I = \int_{0}^{32\pi}\sqrt{2}\,|\sin 2x|\,dx.$

    $|\sin 2x|$ has period $\dfrac{\pi}{2}$, and over one period

    $\int_{0}^{\pi/2}|\sin 2x|\,dx = \int_{0}^{\pi/2}\sin 2x\,dx = \left[-\frac{\cos 2x}{2}\right]_{0}^{\pi/2} = \frac{1}{2} + \frac{1}{2} = 1.$ …

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11. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
    Q.The number of solutions of the equation $\sin 7\theta - \sin 3\theta = \sin 4\theta$ that lie in the interval $(0, \pi)$ is (A) $6$ (B) $3$ (C) $4$ (D) $5$

    ›Reveal solutionSolution

    The equation simplifies using sum-to-product identities, leading to two factor equations. Solving each for $\theta$ in $(0,\pi)$ yields exactly 6 distinct solutions, so the answer is (A).

    We start with the equation

    $\sin 7\theta - \sin 3\theta = \sin 4\theta.$

    The key idea is to rewrite the left-hand side using the sum-to-product identity:

    $\sin A - \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}.$

    This transforms a difference of sines into a product, which then lets us factor the equation and find all solutions systematically.

    1. Apply the sum-to-product identity Let $A = 7\theta$, $B = 3\theta$. Then

    $\sin 7\theta - \sin 3\theta = 2\cos\frac{7\theta+3\theta}{2}\sin\frac{7\theta-3\theta}{2} = 2\cos(5\theta)\sin(2\theta).$

    So the equation becomes

    $2\cos(5\theta)\sin(2\theta) = \sin 4\theta.$

    2. Rewrite $\sin 4\theta$ using a double-angle identity Recall $\sin 4\theta = 2\sin(2\theta)\cos(2\theta)$. Substituting gives

    $2\cos(5\theta)\sin(2\theta) = 2\sin(2\theta)\cos(2\theta).$

    If $\sin(2\theta) \neq 0$, we can divide both sides by $2\sin(2\theta)$, but we must also consider the case $\sin(2\theta)=0$ separately.

    3. Case 1: $\sin(2\theta) = 0$

       Then $2\theta = n\pi$, so $\theta = \frac{n\pi}{2}$ for integer $n$.

       In the interval $(0,\pi)$, the possible values are:

       • $n=1$: $\theta = \frac{\pi}{2}$
       • $n=2$: $\theta = \pi$ (but $\pi$ is not in the open interval $(0,\pi)$, so exclude)
       • $n=0$: $\theta = 0$ (excluded) So from this case we get one solution: $\theta = \frac{\pi}{2}$.

    4. Case 2: $\sin(2\theta) \neq 0$

       Divide both sides by $2\sin(2\theta)$ (non-zero) to obtain

    $\cos(5\theta) = \cos(2\theta).$

    The equation $\cos A = \cos B$ implies $A = 2k\pi \pm B$ for integer $k$.

    So we have two subcases:

    • $5\theta = 2k\pi + 2\theta$ → $3\theta = 2k\pi$ → $\theta = \frac{2k\pi}{3}$
    • $5\theta = 2k\pi - 2\theta$ → $7\theta = 2k\pi$ → $\theta = \frac{2k\pi}{7}$

    5. Find solutions in $(0,\pi)$ from these subcases

       • For $\theta = \frac{2k\pi}{3}$:

         $k=1$ gives $\theta = \frac{2\pi}{3}$ (in $(0,\pi)$).

         $k=2$ gives $\theta = \frac{4\pi}{3} > \pi$, so stop.

         $k=0$ gives $\theta = 0$ (excluded).

         So one solution: $\frac{2\pi}{3}$.

       • For $\theta = \frac{2k\pi}{7}$:

         $k=1$: $\frac{2\pi}{7}$ …

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12. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
    Q.

    [!FORMULA] $\int \frac{(1 - 4\sin^2 x)\cos x}{\cos(3x + 2)} dx =$

    (A) $(\cos 2)x - \frac{1}{3}(\sin 2)\log|\sec(3x + 2)| + c$ (B) $(\sin 2)x - \frac{1}{3}(\cos 2)\log|\cos(3x + 2)| + c$ (C) $(\sin 2)x + \frac{1}{3}(\cos 2)\log|\cos(3x + 2)| + c$ (D) $(\cos 2)x + \frac{1}{3}(\sin 2)\log|\sec(3x + 2)| + c$

    ›Reveal solutionSolution

    The integrand simplifies using the triple-angle identity $\cos 3x = 4\cos^3 x - 3\cos x$ and the angle-sum formula for cosine, leading to a linear combination of $\sec(3x+2)$ and $\tan(3x+2)$; the integral yields $(\sin 2)x - \frac13(\cos 2)\log|\cos(3x+2)| + C$, which matches option (B).

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    Concept & Intuition

    The numerator $(1 - 4\sin^2 x)\cos x$ looks suspiciously like part of a triple-angle formula. Recall that $\cos 3x = 4\cos^3 x - 3\cos x$, but here we have $\sin^2 x$. Using $\sin^2 x = 1 - \cos^2 x$, we can rewrite the numerator in terms of $\cos x$ and then relate it to $\cos 3x$. The denominator $\cos(3x+2)$ suggests that after simplification, the integrand will become a sum of terms like $\sec(3x+2)$ and $\tan(3x+2)$, whose integrals are standard. The constants $\sin 2$ and $\cos 2$ will appear from expanding $\cos(3x+2) = \cos 3x \cos 2 - \sin 3x \sin 2$.

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    Step-by-step solution

    1. Rewrite the numerator using $\sin^2 x = 1 - \cos^2 x$

    $1 - 4\sin^2 x = 1 - 4(1 - \cos^2 x) = 1 - 4 + 4\cos^2 x = 4\cos^2 x - 3.$

    So the numerator becomes $(4\cos^2 x - 3)\cos x = 4\cos^3 x - 3\cos x$.

    2. Recognize the triple-angle identity We know $\cos 3x = 4\cos^3 x - 3\cos x$. Hence the numerator is exactly $\cos 3x$. The integral is now

    $\int \frac{\cos 3x}{\cos(3x+2)}\,dx.$

    3. Use the angle-sum formula for cosine Write $\cos(3x+2) = \cos 3x \cos 2 - \sin 3x \sin 2$. Then

    $\frac{\cos 3x}{\cos(3x+2)} = \frac{\cos 3x}{\cos 3x \cos 2 - \sin 3x \sin 2}.$

    4. Divide numerator and denominator by $\cos 3x$ (assuming $\cos 3x \neq 0$; the result holds generally)

    $\frac{1}{\cos 2 - \tan 3x \sin 2}.$

    This is not yet a standard form. Instead, a better approach: express the fraction as a linear combination of $1$ and a derivative of the denominator.

    5. Rewrite the integrand using a clever trick Consider the derivative of $\log|\cos(3x+2)|$:

    $\frac{d}{dx}\log|\cos(3x+2)| = -3\tan(3x+2).$

    Also, $\frac{d}{dx}(x) = 1$. We want to express $\frac{\cos 3x}{\cos(3x+2)}$ as $A + B\tan(3x+2)$ for constants $A,B$.

    Write $\cos 3x = \cos[(3x+2) - 2] = \cos(3x+2)\cos 2 + \sin(3x+2)\sin 2$.

    Then

    $\frac{\cos 3x}{\cos(3x+2)} = \cos 2 + \sin 2 \cdot \tan(3x+2).$

    6. Integrate term by term

    $\int \left[\cos 2 + \sin 2 \cdot \tan(3x+2)\right] dx = (\cos 2)x + \sin 2 \int \tan(3x+2)\,dx.$

    The integral of $\tan(3x+2)$ is $-\frac13 \log|\cos(3x+2)| + C$, because $\int \tan u\,du = -\log|\cos u|$ and $u = 3x+2$ gives factor $\frac13$.

    Hence

    $\int \frac{\cos 3x}{\cos(3x+2)}\,dx = (\cos 2)x - \frac{\sin 2}{3} \log|\cos(3x+2)| + C.$ …

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