Q.Integrate the following function: $\sin^3 x \cos^3 x$

Showing the 12 most recent of 30 on this concept.

1. 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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2. 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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3. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
   Q.$\int \frac{\sin^8 x - \cos^8 x}{1 - 2\sin^2 x \cos^2 x} dx =$ (A) $\frac{1}{2} \cos 2x + c$ (B) $-\frac{1}{2} \cos 2x + c$ (C) $\frac{-1}{(1 + \tan x)^2} + c$ (D) $-\frac{1}{2} \sin 2x + c$

   ›Reveal solutionSolution

   Factor $\sin^8x-\cos^8x$; the factor $\sin^4x+\cos^4x$ equals the denominator $1-2\sin^2x\cos^2x$, leaving $-\cos2x$, whose integral is $-\tfrac12\sin2x+c$, option (D).

   1. Factor the numerator (difference of squares).

   $\sin^8x-\cos^8x=(\sin^4x-\cos^4x)(\sin^4x+\cos^4x).$

   The first factor:

   $\sin^4x-\cos^4x=(\sin^2x-\cos^2x)(\sin^2x+\cos^2x)=\sin^2x-\cos^2x=-\cos2x.$

   2. Match the second factor to the denominator. $\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-2\sin^2x\cos^2x.$ …

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4. TG EAPCET 2022Set eng-2022-07-19-AN1 markMCQ
   Q.$\int_{0}^{\frac{\pi}{2}} \sin^4 \theta \cos^3 \theta \, d\theta =$ (A) $\dfrac{1}{35}$ (B) $\dfrac{2}{35}$ (C) $\dfrac{4}{35}$ (D) $\dfrac{8}{35}$

   ›Reveal solutionSolution

   We evaluate the definite integral by using a substitution $u = \sin \theta$, which is effective because the power of $\cos \theta$ is odd. The final result is $\boxed{\frac{2}{35}}$.

   When faced with integrals involving products of powers of sine and cosine, like $\int \sin^m \theta \cos^n \theta \, d\theta$, a common and effective strategy is to use a substitution. The choice of substitution depends on whether $m$ or $n$ (or both) are odd.

   The core idea is to "save" one factor of the trigonometric function with the odd power to be part of $du$, and then convert the remaining even power of that function into terms of the other trigonometric function using the Pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$. This makes the entire integrand expressible in terms of the chosen substitution variable.

   In this problem, we have $\int_{0}^{\frac{\pi}{2}} \sin^4 \theta \cos^3 \theta \, d\theta$.

   Here, the power of $\sin \theta$ is $m=4$ (even), and the power of $\cos \theta$ is $n=3$ (odd). Since the power of $\cos \theta$ is odd, we will save one $\cos \theta$ for $du$ and convert the remaining $\cos^2 \theta$ into $\sin^2 \theta$. This suggests that $u = \sin \theta$ will be the appropriate substitution.

   Tip

   For integrals of the form $\int \sin^m x \cos^n x \, dx$:

   • If $n$ is odd, save one $\cos x$ for $du$, convert the remaining $\cos^{n-1} x$ to powers of $\sin x$ using $\cos^2 x = 1 - \sin^2 x$, and substitute $u = \sin x$.
   • If $m$ is odd, save one $\sin x$ for $du$, convert the remaining $\sin^{m-1} x$ to powers of $\cos x$ using $\sin^2 x = 1 - \cos^2 x$, and substitute $u = \cos x$.
   • If both $m$ and $n$ are odd, either substitution works.
   • If both $m$ and $n$ are even, use half-angle identities ($\sin^2 x = \frac{1 - \cos 2x}{2}$, $\cos^2 x = \frac{1 + \cos 2x}{2}$) to reduce the powers.

   Let's apply this strategy step-by-step:

   1. Rewrite the integrand to prepare for substitution. We have $\cos^3 \theta$. Since the power is odd, we separate one factor of $\cos \theta$: $\cos^3 \theta = \cos^2 \theta \cdot \cos \theta$. Now, use the identity $\cos^2 \theta = 1 - \sin^2 \theta$ to express $\cos^2 \theta$ in terms of $\sin \theta$: $\cos^3 \theta = (1 - \sin^2 \theta) \cos \theta$. Substitute this back into the integral:

   $\int_{0}^{\frac{\pi}{2}} \sin^4 \theta (1 - \sin^2 \theta) \cos \theta \, d\theta$

   2. Perform the substitution. Let $u = \sin \theta$. Then, the differential $du$ is $du = \cos \theta \, d\theta$. Since this is a definite integral, we must also change the limits of integration according to our substitution:
      • When $\theta = 0$, $u = \sin(0) = 0$.
      • When $\theta = \frac{\pi}{2}$, $u = \sin\left(\frac{\pi}{2}\right) = 1$. …

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5. 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).

   ---

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

   ---

   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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6. 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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7. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
   Q.$\int_{0}^{\frac{\pi}{2}} \sin^6 x \cos^4 x \, dx =$ (A) $\dfrac{\pi}{256}$ (B) $\dfrac{\pi}{512}$ (C) $\dfrac{3\pi}{512}$ (D) $\dfrac{5\pi}{512}$

   ›Reveal solutionSolution

   This integral of a product of powers of sine and cosine over a quarter-period is a classic Beta function disguised as a trigonometric integral. Using the Beta–Gamma relation, the value simplifies to $\frac{3\pi}{512}$, which corresponds to option (C).

   The key insight is that integrals of the form $\int_0^{\pi/2} \sin^m x \cos^n x \, dx$ are directly expressible in terms of the Beta function $B\!\left(\frac{m+1}{2}, \frac{n+1}{2}\right)$, which in turn is a ratio of Gamma functions. This avoids messy repeated integration by parts or trigonometric reduction formulas.

   1. Recognize the Beta function form The standard identity is:

   $\int_0^{\pi/2} \sin^{p-1} x \cos^{q-1} x \, dx = \frac{1}{2} B\!\left(\frac{p}{2}, \frac{q}{2}\right),$

   but more directly useful here:

   $\int_0^{\pi/2} \sin^{2a-1} x \cos^{2b-1} x \, dx = \frac{1}{2} B(a, b).$

   Our integral has $\sin^6 x = \sin^{2\cdot 3.5?}$ — careful: we need exponents of the form $2a-1$ and $2b-1$.

   Here $\sin^6 x$ means exponent $6 = 2a-1 \Rightarrow a = \frac{7}{2}$.

   And $\cos^4 x$ means exponent $4 = 2b-1 \Rightarrow b = \frac{5}{2}$.

   2. Apply the Beta–Gamma relation

   $\int_0^{\pi/2} \sin^{2a-1} x \cos^{2b-1} x \, dx = \frac{1}{2} B(a, b) = \frac{1}{2} \cdot \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}.$

   Substitute $a = \frac{7}{2}$, $b = \frac{5}{2}$:

   $\int_0^{\pi/2} \sin^6 x \cos^4 x \, dx = \frac{1}{2} \cdot \frac{\Gamma\!\left(\frac{7}{2}\right) \Gamma\!\left(\frac{5}{2}\right)}{\Gamma\!\left(\frac{7}{2} + \frac{5}{2}\right)} = \frac{1}{2} \cdot \frac{\Gamma\!\left(\frac{7}{2}\right) \Gamma\!\left(\frac{5}{2}\right)}{\Gamma(6)}.$

   3. Evaluate the Gamma values Recall $\Gamma(n) = (n-1)!$ for integers, and for half-integers:

   $\Gamma\!\left(\frac{1}{2}\right) = \sqrt{\pi}, \quad \Gamma\!\left(\frac{3}{2}\right) = \frac{1}{2}\sqrt{\pi}, \quad \Gamma\!\left(\frac{5}{2}\right) = \frac{3}{4}\sqrt{\pi}, \quad \Gamma\!\left(\frac{7}{2}\right) = \frac{15}{8}\sqrt{\pi}.$

   Also $\Gamma(6) = 5! = 120$.

   4. Plug in and simplify $$$$

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8. 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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9. TG EAPCET 2022Set eng-2022-07-19-FN1 markMCQ
   Q.

   [!FORMULA] $\int \frac{\sin x \cdot \sec^2 x - \tan x \cdot \sin x + \cos x}{(1 - \cos 2x)} \, dx =$

   (A) $\frac{1}{2} \left[ \sec x - \csc x - \log \left| \tan \left( \frac{x}{2} \right) \tan \left( \frac{\pi}{4} + \frac{x}{2} \right) \right| \right] + c$ (B) $\sec x - \csc x + \log \left| \frac{\tan \left( \frac{x}{2} \right)}{\tan \left( \frac{\pi}{4} + \frac{x}{2} \right)} \right| + c$ (C) $\frac{1}{2} \left[ \sec x - \csc x - \log \left| \frac{\tan \left( \frac{\pi}{4} + \frac{x}{2} \right)}{\tan \left( \frac{x}{2} \right)} \right| \right] + c$ (D) $\sec x + \csc x - \log \left| \tan \left( \frac{x}{2} \right) \right| + c$

   ›Reveal solutionSolution

   Using $1-\cos 2x = 2\sin^2 x$ and splitting term-by-term, the integrand becomes $\tfrac12\big(\sec x\tan x+\csc x\cot x+\csc x-\sec x\big)$, which integrates to $\tfrac12\big[\sec x-\csc x-\log\big|\tfrac{\tan(\pi/4+x/2)}{\tan(x/2)}\big|\big]+c$ — option (C).

   Concept & intuition

   The denominator $1-\cos 2x$ is the giveaway: it equals $2\sin^2 x$. Dividing each numerator term by $2\sin^2 x$ collapses the integrand into a sum of standard, directly-integrable pieces ($\sec x\tan x$, $\csc x\cot x$, $\csc x$, $\sec x$).

   Step-by-step solution

   1. Simplify the denominator.

   $1-\cos 2x = 2\sin^2 x.$

   So the integral is $\displaystyle\int \frac{\sin x\,\sec^2 x-\tan x\,\sin x+\cos x}{2\sin^2 x}\,dx.$

   2. Divide each term by $2\sin^2 x$.

      • $\dfrac{\sin x\,\sec^2 x}{2\sin^2 x}=\dfrac{1}{2\sin x\cos^2 x}$
      • $-\dfrac{\tan x\,\sin x}{2\sin^2 x}=-\dfrac{1}{2\cos x}=-\tfrac12\sec x$
      • $\dfrac{\cos x}{2\sin^2 x}=\tfrac12\csc x\cot x$

   3. Split the first term using $\dfrac{1}{\sin x\cos^2 x}=\dfrac{\sin^2 x+\cos^2 x}{\sin x\cos^2 x}=\dfrac{\sin x}{\cos^2 x}+\dfrac{1}{\sin x}$:

   $\frac{1}{2\sin x\cos^2 x}=\tfrac12\sec x\tan x+\tfrac12\csc x.$

   4. Collect all pieces.

   $\text{integrand}=\tfrac12\big(\sec x\tan x+\csc x\cot x+\csc x-\sec x\big).$

   5. Integrate each standard form. $\int\sec x\tan x\,dx=\sec x,\quad \int\csc x\cot x\,dx=-\csc x,$ …

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10. TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
    Q.If n is a positive integer greater than 1 and $I_n = \int \frac{\sin nx}{\sin x} dx$, then $I_{n+1} - I_{n-1} =$ (A) $\frac{2}{n-1} \cos{(n-1)x}$ (B) $\frac{2}{n-1} \sin{(n-1)x}$ (C) $\frac{2}{n} \cos{nx}$ (D) $\frac{2}{n} \sin{nx}$

    ›Reveal solutionSolution

    The key is to use the sine addition formula to express $\sin((n+1)x) - \sin((n-1)x)$ as $2\cos(nx)\sin x$, which simplifies the integrand to $2\cos(nx)$. The result is $\frac{2}{n}\sin(nx)$, so the correct option is (D).

    We are given $I_n = \int \frac{\sin nx}{\sin x} \, dx$ for $n>1$, and we need $I_{n+1} - I_{n-1}$. Instead of integrating each separately, we combine them under one integral:

    $I_{n+1} - I_{n-1} = \int \frac{\sin((n+1)x) - \sin((n-1)x)}{\sin x} \, dx.$

    The numerator is a difference of sines. Using the identity

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

    with $A = (n+1)x$ and $B = (n-1)x$, we get:

    $\frac{A+B}{2} = \frac{(n+1)x + (n-1)x}{2} = nx, \quad \frac{A-B}{2} = \frac{(n+1)x - (n-1)x}{2} = x.$

    Thus:

    $\sin((n+1)x) - \sin((n-1)x) = 2\cos(nx)\sin x.$

    Now the integrand becomes:

    $\frac{2\cos(nx)\sin x}{\sin x} = 2\cos(nx),$

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

    $I_{n+1} - I_{n-1} = \int 2\cos(nx) \, dx = 2 \cdot \frac{\sin(nx)}{n} + C = \frac{2}{n}\sin(nx) + C.$ …

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11. 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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12. 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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