Q.Integrate the following function: $\cos^4 2x$

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: $\cos^4x = \dfrac{1+2\cos2x+\frac{1+\cos4x}{2}}{4} = \dfrac{1}{4}+\dfrac{1}{2}\cos2x+\dfrac{1}{8}+\dfrac{1}{8}\cos4x$. …

Concept and Intuition

Angles like $\frac{5\pi}{8}=\pi-\frac{3\pi}{8}$ have cosines that are negatives of $\cos\frac{3\pi}{8}$, but raised to an even power (4th), the sign disappears. Also $\cos\frac{3\pi}{8}=\sin\frac{\pi}{8}$ (complementary angles), so everything reduces to $\sin^4\theta+\cos^4\theta$ for $\theta=\pi/8$, which has the neat identity $1-2\sin^2\theta\cos^2\theta$.

Step-by-Step Solution

1. $\cos\frac{5\pi}{8}=\cos\left(\pi-\frac{3\pi}{8}\right)=-\cos\frac{3\pi}{8}$, so $\cos^4\frac{5\pi}{8}=\cos^4\frac{3\pi}{8}$.
2. $\cos\frac{7\pi}{8}=\cos\left(\pi-\frac{\pi}{8}\right)=-\cos\frac{\pi}{8}$, so $\cos^4\frac{7\pi}{8}=\cos^4\frac{\pi}{8}$.
3. Sum $=2\cos^4\frac{\pi}{8}+2\cos^4\frac{3\pi}{8}$.
4. $\cos\frac{3\pi}{8}=\cos\left(\frac{\pi}{2}-\frac{\pi}{8}\right)=\sin\frac{\pi}{8}$, so this is $2\left(\cos^4\frac{\pi}{8}+\sin^4\frac{\pi}{8}\right)$. …

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 $x=2\theta$: $\sin^2 2\theta = \dfrac{1-\cos4\theta}{2}$. …

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 $x=2\theta$: $\cos^22\theta = \dfrac{1+\cos4\theta}{2}$. …

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 $=2\left(\sin^4\dfrac\pi8+\cos^4\dfrac\pi8\right)$. …

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 $n=6$: $I_6=-\dfrac{\sin^5x\cos x}{6}+\dfrac{5}{6}I_4$. …

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 $\phi=2\theta$: $\cos^22\theta = \dfrac{1+\cos4\theta}{2}$, $\cos^32\theta = \dfrac{3\cos2\theta+\cos6\theta}{4}$. …

Concept and Intuition

When a sum of $\sin^2$ runs over angles symmetric about $45^\circ$ in equal steps, each pair $(\theta, 90^\circ-\theta)$ contributes exactly 1, turning a long sum into simple counting.

Step-by-Step Solution

1. The terms are $\sin^25^\circ,\sin^210^\circ,\dots,\sin^290^\circ$ — an arithmetic sequence of angles with step $5^\circ$, giving $18$ terms.
2. Pair each $\theta$ with $90^\circ-\theta$: $(5,85),(10,80),(15,75),(20,70),(25,65),(30,60),(35,55),(40,50)$ — 8 pairs (16 terms), each summing to $\sin^2\theta+\sin^2(90^\circ-\theta)=\sin^2\theta+\cos^2\theta=1$. …

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 $f_3(x)=\dfrac16\left(1-\dfrac{3u}{4}\right)=\dfrac16-\dfrac{u}{8}$. …

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 $\sin^2\frac{\pi}{20}+\sin^2\frac{9\pi}{20}=1$. …

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 $\sin10°+\sin50°=2\sin30°\cos20°=2\left(\frac12\right)\cos20°=\cos20°$. …