Q.Integrate the following function: $\sin^2 (2x + 5)$

Concept and Intuition

The reduction formula for powers of sine comes from integrating by parts, peeling off

one power of $\sin x$ at a time and expressing the boundary term via $\sin^{n-1}x\cos x$;

it relates $I_n$ to $I_{n-2}$, which is exactly why $I_6-\tfrac56I_4$ isolates only the

"boundary" piece.

Step-by-Step Solution

1. Recall (or derive by parts, taking $u=\sin^{n-1}x$, $dv=\sin x\,dx$) the reduction formula: $I_n=\displaystyle\int\sin^nx\,dx=-\dfrac{\sin^{n-1}x\cos x}{n}+\dfrac{n-1}{n}I_{n-2}$.
2. Set $n=6$: $I_6=-\dfrac{\sin^5x\cos x}{6}+\dfrac{5}{6}I_4$.
3. Rearranging: $I_6-\dfrac56I_4=-\dfrac{\sin^5x\cos x}{6}$.

Common Mistakes

• Misremembering the reduction formula's sign or the power of $\cos x$ in the boundary term (it's $\cos^1x$, not $\cos^2x$, for this formula).
• Using $n=4$ instead of $n=6$ when substituting into the formula.

ANSWER: A

Concept and Intuition

Cubes of sine/cosine at special complementary angles ($10°,50°,70°$ are related since $10°+50°+70°$ pattern echoes $30°,150°,210°$ under tripling) are best handled by converting $\sin^3\theta$ into a linear combination of $\sin\theta$ and $\sin3\theta$. The tripled angles $30°,150°,210°$ have known simple sine values, which is exactly why $10°,50°,70°$ were chosen.

Step-by-Step Solution

1. Use the identity $\sin^3\theta=\dfrac{3\sin\theta-\sin3\theta}{4}$.
2. $\sin^3 10°=\dfrac{3\sin10°-\sin30°}{4}=\dfrac{3\sin10°-\frac12}{4}$
3. $\sin^3 50°=\dfrac{3\sin50°-\sin150°}{4}=\dfrac{3\sin50°-\frac12}{4}$
4. $\sin^3 70°=\dfrac{3\sin70°-\sin210°}{4}=\dfrac{3\sin70°-(-\frac12)}{4}=\dfrac{3\sin70°+\frac12}{4}$
5. Sum: $\sin^310°+\sin^350°-\sin^370°=\dfrac{3(\sin10°+\sin50°-\sin70°) -\frac12-\frac12-\frac12}{4}=\dfrac{3(\sin10°+\sin50°-\sin70°)-\frac32}{4}$.
6. Now $\sin10°+\sin50°=2\sin30°\cos20°=2\left(\frac12\right)\cos20°=\cos20°$.
7. Since $\sin70°=\cos20°$, we get $\sin10°+\sin50°-\sin70°=\cos20°-\cos20°=0$.
8. So the sum $=\dfrac{3(0)-\frac32}{4}=\dfrac{-\frac32}{4}=-\dfrac38$.

Common Mistakes

• Forgetting the sign of $\sin210°=-\dfrac12$ (it's in the third quadrant) when applying the triple-angle formula to $\sin^3 70°$.
• Trying to compute each cube numerically without simplification — the exact identity route avoids rounding errors that could mislead you between close options.

ANSWER: A

Concept and Intuition

Angles $5\pi/8$ and $7\pi/8$ are supplementary-related to $3\pi/8$ and $\pi/8$ respectively ($\sin(\pi-x)=\sin x$), collapsing the four-term sum to just two distinct terms, one of which is naturally a cosine (co-function) of the other.

Step-by-Step Solution

1. $\sin\dfrac{5\pi}8 = \sin\left(\pi-\dfrac{5\pi}8\right)=\sin\dfrac{3\pi}8$, and $\sin\dfrac{7\pi}8=\sin\left(\pi-\dfrac{7\pi}8\right)=\sin\dfrac\pi8$.
2. So the sum becomes $\sin^4\dfrac\pi8+\sin^4\dfrac{3\pi}8+\sin^4\dfrac{3\pi}8+\sin^4\dfrac\pi8 = 2\sin^4\dfrac\pi8+2\sin^4\dfrac{3\pi}8$.
3. Also $\dfrac{3\pi}8=\dfrac\pi2-\dfrac\pi8$, so $\sin\dfrac{3\pi}8=\cos\dfrac\pi8$, giving $\sin^4\dfrac{3\pi}8=\cos^4\dfrac\pi8$.
4. Sum $=2\left(\sin^4\dfrac\pi8+\cos^4\dfrac\pi8\right)$.
5. Use the identity $\sin^4\theta+\cos^4\theta = (\sin^2\theta+\cos^2\theta)^2-2\sin^2\theta\cos^2\theta = 1-\dfrac12\sin^2(2\theta)$.
6. With $\theta=\pi/8$: $2\theta=\pi/4$, $\sin(\pi/4)=\dfrac{\sqrt2}2$, so $\sin^2(\pi/4)=\dfrac12$.
7. $\sin^4\dfrac\pi8+\cos^4\dfrac\pi8 = 1-\dfrac12\cdot\dfrac12 = \dfrac34$.
8. Total sum $=2\times\dfrac34=\dfrac32$.

Common Mistakes

• Trying to compute each $\sin^4$ term individually using half-angle formulas — the supplementary-angle symmetry and co-function identity make this far faster.

ANSWER: C

Concept and Intuition

The key simplifications are $\sin^4x+\cos^4x = 1-\tfrac12\sin^2(2x)$ and $\sin^6x+\cos^6x = 1-\tfrac34\sin^2(2x)$ (both standard identities derivable from $(\sin^2x+\cos^2x)^2$ and $^3$ expansions). Once each $f_n$ is written in terms of $u=\sin^2(2x)$, we can just combine algebraically and watch the $u$ terms cancel.

Step-by-Step Solution

1. $f_1(x)=\dfrac{1}{2}[\sin^2x+\cos^2x]=\dfrac12(1)=\dfrac12$.
2. $\sin^4x+\cos^4x = (\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x = 1-\tfrac12\sin^2(2x)$. With $u=\sin^2(2x)$: $f_2(x)=\dfrac14\left(1-\dfrac{u}{2}\right)=\dfrac14-\dfrac{u}{8}$.
3. $\sin^6x+\cos^6x = 1-3\sin^2x\cos^2x = 1-\tfrac34\sin^2(2x) = 1-\tfrac34u$. So $f_3(x)=\dfrac16\left(1-\dfrac{3u}{4}\right)=\dfrac16-\dfrac{u}{8}$.
4. $f_1+f_2-f_3 = \dfrac12+\left(\dfrac14-\dfrac{u}{8}\right)-\left(\dfrac16-\dfrac{u}{8}\right) = \dfrac12+\dfrac14-\dfrac16$ (the $u/8$ terms cancel exactly).
5. Common denominator 12: $\dfrac{6}{12}+\dfrac{3}{12}-\dfrac{2}{12}=\dfrac{7}{12}$.

Common Mistakes

• Misremembering the reduction formulas for $\sin^6x+\cos^6x$ (using the wrong coefficient $3/4$ vs $1/2$).
• Not noticing the $u$-terms cancel, and instead trying to solve for a specific $x$ unnecessarily.

ANSWER: D

Concept and Intuition

The five angles $\frac{\pi}{20},\frac{3\pi}{20},\frac{5\pi}{20},\frac{7\pi}{20},\frac{9\pi}{20}$ are symmetric about $\frac{\pi}{2}\left(=\frac{10\pi}{20}\right)$ in pairs: $\left(\frac{\pi}{20},\frac{9\pi}{20}\right)$ and $\left(\frac{3\pi}{20},\frac{7\pi}{20}\right)$ each sum to $\frac{\pi}{2}$, while $\frac{5\pi}{20}=\frac{\pi}{4}$ is the unpaired middle term. Using $\sin\left(\frac{\pi}{2}-\theta\right)=\cos\theta$, each pair of $\sin^2$ terms becomes $\sin^2\theta+\cos^2\theta=1$.

Step-by-Step Solution

1. Write out the terms: $k=0:\sin^2\frac{\pi}{20}$, $k=1:\sin^2\frac{3\pi}{20}$, $k=2:\sin^2\frac{5\pi}{20}=\sin^2\frac{\pi}{4}$, $k=3:\sin^2\frac{7\pi}{20}$, $k=4:\sin^2\frac{9\pi}{20}$.
2. Since $\frac{9\pi}{20}=\frac{\pi}{2}-\frac{\pi}{20}$: $\sin^2\frac{9\pi}{20}=\cos^2\frac{\pi}{20}$, so $\sin^2\frac{\pi}{20}+\sin^2\frac{9\pi}{20}=1$.
3. Since $\frac{7\pi}{20}=\frac{\pi}{2}-\frac{3\pi}{20}$: $\sin^2\frac{7\pi}{20}=\cos^2\frac{3\pi}{20}$, so $\sin^2\frac{3\pi}{20}+\sin^2\frac{7\pi}{20}=1$.
4. Middle term: $\sin^2\frac{\pi}{4}=\left(\frac{1}{\sqrt2}\right)^2=\frac12$.
5. Total sum $=1+1+\frac12=\frac52$.

Common Mistakes

• Trying to compute each $\sin^2$ numerically instead of spotting the complementary-angle pairing.
• Mispairing the terms (wrong indices) when checking which angles sum to $\pi/2$.

ANSWER: B

Concept and Intuition

Any even power of $\cos\theta$ or $\sin\theta$ can always be written as a sum of cosines of multiple angles — this is the reverse of the multiple-angle formula, and it is exactly how you reduce a power to a form you can integrate or match term-by-term. The trick is to apply $\cos^2 x = \dfrac{1+\cos2x}{2}$ twice, once on $\theta$ and then again on the resulting $\cos^2 2\theta$ term.

Step-by-Step Solution

1. Start with $\cos^2\theta = \dfrac{1+\cos2\theta}{2}$.
2. Square both sides: $\cos^4\theta = \dfrac{(1+\cos2\theta)^2}{4} = \dfrac{1+2\cos2\theta+\cos^22\theta}{4}$.
3. Reduce $\cos^22\theta$ using the same identity with $x=2\theta$: $\cos^22\theta = \dfrac{1+\cos4\theta}{2}$.
4. Substitute: $\cos^4\theta = \dfrac{1}{4}\left(1+2\cos2\theta+\dfrac{1+\cos4\theta}{2}\right) = \dfrac{1}{4}\left(\dfrac{2+4\cos2\theta+1+\cos4\theta}{2}\right) = \dfrac{3+4\cos2\theta+\cos4\theta}{8}$.
5. So $\cos^4\theta = \dfrac18\cos4\theta + \dfrac12\cos2\theta + \dfrac38$, giving $a=\dfrac18$, $b=\dfrac12$, $c=\dfrac38$.

Common Mistakes

• Forgetting to reduce $\cos^22\theta$ again and stopping at a quadratic-in-$\cos2\theta$ form.
• Mixing up which coefficient multiplies $\cos4\theta$ versus $\cos2\theta$ when reading off $(a,b,c)$.

ANSWER: B

Concept and Intuition

Any product of $\sin^4\theta$ and $\cos^2\theta$ (degree 6 in sines/cosines) can be expressed as a combination of $\cos0\theta, \cos2\theta, \cos4\theta, \cos6\theta$ only — there's no way to generate a $\cos8\theta$ term from a degree-6 trigonometric polynomial. So all coefficients from $n=4$ onward are exactly zero.

Step-by-Step Solution

1. Write $\sin^2\theta = \dfrac{1-\cos2\theta}{2}$, so $\sin^4\theta = \dfrac{(1-\cos2\theta)^2}{4} = \dfrac{1-2\cos2\theta+\cos^22\theta}{4}$.
2. Write $\cos^2\theta = \dfrac{1+\cos2\theta}{2}$.
3. Product: $\sin^4\theta\cos^2\theta = \dfrac18(1-2\cos2\theta+\cos^22\theta)(1+\cos2\theta)$.
4. Expanding: $= \dfrac18\left[1 - \cos2\theta - \cos^22\theta + \cos^32\theta\right]$.
5. Using $\cos^2\phi=\dfrac{1+\cos2\phi}{2}$ and $\cos^3\phi=\dfrac{3\cos\phi+\cos3\phi}{4}$ with $\phi=2\theta$: $\cos^22\theta = \dfrac{1+\cos4\theta}{2}$, $\cos^32\theta = \dfrac{3\cos2\theta+\cos6\theta}{4}$.
6. Substituting and simplifying: $\sin^4\theta\cos^2\theta = \dfrac1{16} - \dfrac1{32}\cos2\theta - \dfrac1{16}\cos4\theta + \dfrac1{32}\cos6\theta$.
7. So $a_0=\tfrac1{16}$ ($n=0$), $a_2=-\tfrac1{32}$ ($n=1$), $a_4=-\tfrac1{16}$ ($n=2$), $a_6=\tfrac1{32}$ ($n=3$), and $a_{2n}=0$ for all $n\ge4$.
8. The least $n$ with $a_{2n}=0$ is therefore $n=4$.

Common Mistakes

• Forgetting to convert $\cos^3(2\theta)$ correctly into a sum of cosines (using the wrong triple-angle expansion).
• Miscounting the index $n$ (confusing the coefficient of $\cos6\theta$, which is $n=3$, with the first vanishing one at $n=4$).

ANSWER: D

Concept and Intuition

Products of $\sin^2$ and $\cos^2$ of the same angle are classic candidates for the double-angle identity $\sin2\theta=2\sin\theta\cos\theta$, and then the half-angle-style identity $\sin^2 x = \frac{1-\cos2x}{2}$ collapses everything into a single cosine of a quadrupled angle.

Step-by-Step Solution

1. $\sin\theta\cos\theta = \dfrac{1}{2}\sin2\theta$, so $\sin^2\theta\cos^2\theta = \dfrac{1}{4}\sin^2 2\theta$.
2. Apply $\sin^2 x = \dfrac{1-\cos2x}{2}$ with $x=2\theta$: $\sin^2 2\theta = \dfrac{1-\cos4\theta}{2}$.
3. Combine: $\dfrac{1}{4}\cdot\dfrac{1-\cos4\theta}{2} = \dfrac{1-\cos4\theta}{8}$.

Common Mistakes

• Stopping after the first substitution and forgetting the extra factor of $\frac12$ from the second identity.
• Confusing $\sin^2(2\theta)$ with $\sin(2\theta)^2=\sin(4\theta^2)$ (a nonsense simplification).

ANSWER: D

Concept and Intuition

Power-reduction formulas let us rewrite even powers of sine/cosine as linear combinations of cosines of multiple angles, useful for integration and simplification. Apply the formula $\cos^2\theta = \dfrac{1+\cos2\theta}{2}$ twice.

Step-by-Step Solution

1. $\cos^4x = (\cos^2x)^2 = \left(\dfrac{1+\cos2x}{2}\right)^2 = \dfrac{1+2\cos2x+\cos^22x}{4}$.
2. Reduce $\cos^22x$ using the same identity with angle $2x$: $\cos^22x = \dfrac{1+\cos4x}{2}$.
3. Substitute: $\cos^4x = \dfrac{1+2\cos2x+\frac{1+\cos4x}{2}}{4} = \dfrac{1}{4}+\dfrac{1}{2}\cos2x+\dfrac{1}{8}+\dfrac{1}{8}\cos4x$.
4. Combine constants: $\dfrac14+\dfrac18=\dfrac38$, giving $\cos^4x = \dfrac38+\dfrac12\cos2x+\dfrac18\cos4x$.

Common Mistakes

• Arithmetic slip when combining the constant terms $\tfrac14+\tfrac18$.
• Sign error, producing the chemistry-formula-like variant with a minus sign (option B), which is actually the expansion for $\sin^4x$, not $\cos^4x$.

ANSWER: A