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

We start with the given:

$\sin\alpha + \cos\alpha = m$.

We want $\sin^6\alpha + \cos^6\alpha$.

Concept and intuition:

The expression $\sin^6\alpha + \cos^6\alpha$ is a sum of sixth powers. A classic trick is to rewrite it as $(\sin^2\alpha)^3 + (\cos^2\alpha)^3$, then use the sum of cubes factorization:

$a^3+b^3 = (a+b)^3 - 3ab(a+b)$.

Here $a=\sin^2\alpha$, $b=\cos^2\alpha$, so $a+b=1$ (since $\sin^2\alpha+\cos^2\alpha=1$). That reduces the problem to finding $\sin^2\alpha\cos^2\alpha$, which we can get from $m$ because $m^2 = 1 + 2\sin\alpha\cos\alpha$.

Let’s work it out step by step.

1. Express the sixth power sum using cubes.

Let $u = \sin^2\alpha$, $v = \cos^2\alpha$. Then $u+v=1$, and

So we need $uv = \sin^2\alpha \cos^2\alpha$.

2. Find $\sin\alpha\cos\alpha$ from $m$. Square $m$:

Hence

3. Compute $\sin^2\alpha\cos^2\alpha$. $$$$

The angles in the sine product run from $\pi/14$ to $13\pi/14$ in steps of $2\pi/14$, covering every odd multiple of $\pi/14$. Using $\sin(\pi-\theta)=\sin\theta$:

• $\sin(13\pi/14) = \sin(\pi/14)$
• $\sin(11\pi/14) = \sin(3\pi/14)$
• $\sin(9\pi/14) = \sin(5\pi/14)$
• $\sin(7\pi/14) = \sin(\pi/2) = 1$

So the seven factors pair into three repeated pairs plus the middle term:

Step-by-step solution

1. Convert sines to cosines using $\sin\theta = \cos(\pi/2-\theta)$:

So $P = \cos\tfrac{3\pi}{7}\cos\tfrac{2\pi}{7}\cos\tfrac{\pi}{7}$.

2. Relate $\cos(3\pi/7)$ to $\cos(4\pi/7)$ using $\cos(\pi-x)=-\cos x$, with $3\pi/7 = \pi - 4\pi/7$:

so

3. Evaluate the classic product $\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}$ using repeated application of $\sin(2\theta)=2\sin\theta\cos\theta$: $\cos\tfrac{\pi}{7}\cos\tfrac{2\pi}{7}\cos\tfrac{4\pi}{7} = \frac{\sin(8\pi/7)}{8\sin(\pi/7)} = \frac{-\sin(\pi/7)}{8\sin(\pi/7)} = -\frac{1}{8}.$ …

Since $\cos\dfrac{3\pi}{14}=\cos\!\left(\dfrac{\pi}{2}-\dfrac{2\pi}{7}\right)=\sin\dfrac{2\pi}{7}$ and $\cos\dfrac{5\pi}{14}=\cos\!\left(\dfrac{\pi}{2}-\dfrac{\pi}{7}\right)=\sin\dfrac{\pi}{7}$, and using the standard product $\cos\dfrac{\pi}{7}\cos\dfrac{2\pi}{7}\cos\dfrac{3\pi}{7}=\dfrac18$:

$\text{Product}=\dfrac18\,\sin\dfrac{\pi}{7}\sin\dfrac{2\pi}{7}=\dfrac{1}{16}\left(\cos\dfrac{\pi}{7}-\cos\dfrac{3\pi}{7}\right)\approx 0.0424.$ …

The key insight is that the angles $\frac{\pi}{8}, \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{7\pi}{8}$ are symmetric about $\frac{\pi}{2}$. This symmetry lets us pair terms and use the identity $\sin^4 x + \cos^4 x = 1 - \frac12 \sin^2 2x$, which simplifies the sum dramatically without needing to compute each fourth power separately.

1. Recognize the symmetry Notice that $\frac{5\pi}{8} = \pi - \frac{3\pi}{8}$ and $\frac{7\pi}{8} = \pi - \frac{\pi}{8}$. Since $\cos(\pi - \alpha) = -\cos\alpha$ and $\sin(\pi - \alpha) = \sin\alpha$, their fourth powers are the same: $\cos^4\frac{5\pi}{8} = \cos^4\frac{3\pi}{8}$ and $\sin^4\frac{7\pi}{8} = \sin^4\frac{\pi}{8}$. So the sum becomes:

2. Pair the terms Group them as:

3. Apply the identity For any angle $x$, we have $\sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac12 \sin^2 2x$. So:

Similarly, for $x = \frac{3\pi}{8}$, note that $2x = \frac{3\pi}{4}$, and $\sin\frac{3\pi}{4} = \frac{\sqrt{2}}{2}$ as well. So:

4. Sum the two pairs

5. Set up the equation for $\theta$ …

We are given $A+B+C = 4S$. The expression to simplify is

The trick is to notice that each term involves $2S$ minus one of $A, B, C$. Using the given relation, we can express $2S$ in terms of the half-angles, making the sum-to-product identities very clean.

1. Rewrite each angle using $S$ From $A+B+C = 4S$, we have $S = \frac{A+B+C}{4}$. Then

Similarly,

And $2S = \frac{A+B+C}{2}$.

2. Pair the first two sines and apply sum-to-product Consider $\sin(2S-A) + \sin(2S-B)$. Using

with $X = \frac{-A+B+C}{2}$ and $Y = \frac{A-B+C}{2}$:

So

3. Add the third sine and subtract $\sin 2S$ Now we have

Write $\sin(2S-C) = \sin\frac{A+B-C}{2}$ and $\sin 2S = \sin\frac{A+B+C}{2}$.

Pair these two:

Use $\sin P - \sin Q = 2\cos\frac{P+Q}{2}\sin\frac{P-Q}{2}$:

So

4. Combine everything The whole expression becomes $2\sin\frac{C}{2}\cos\frac{B-A}{2} - 2\cos\frac{A+B}{2}\sin\frac{C}{2} = 2\sin\frac{C}{2}\left[\cos\frac{B-A}{2} - \cos\frac{A+B}{2}\right].$ …

The key insight is that the condition $A+B+C=4S$ lets us express every angle in the trigonometric sum in terms of $S$ and the original angles. For instance, $2S-A = \frac{A+B+C}{2} - A = \frac{B+C-A}{2}$, and similarly for the others. This transforms the problem into one where we can systematically apply sum-to-product formulas.

1. Rewrite each term using $S$. Since $4S = A+B+C$, we have $2S = \frac{A+B+C}{2}$. Then:

And $2S$ itself is $\frac{A+B+C}{2}$.

So the given expression becomes:

2. Group the first two cosines and apply sum-to-product. For any $X$ and $Y$, $\cos X + \cos Y = 2\cos\frac{X+Y}{2}\cos\frac{X-Y}{2}$. Here $X = \frac{B+C-A}{2}$ and $Y = \frac{A+C-B}{2}$. Their sum: $\frac{B+C-A + A+C-B}{2} = \frac{2C}{2} = C$, so $\frac{X+Y}{2} = \frac{C}{2}$. Their difference: $\frac{B+C-A - (A+C-B)}{2} = \frac{B+C-A-A-C+B}{2} = \frac{2B-2A}{2} = B-A$, so $\frac{X-Y}{2} = \frac{B-A}{2}$. Hence:

3. Group the last two cosines (with a minus sign) and apply sum-to-product. We have $-\cos\frac{A+B-C}{2} - \cos\frac{A+B+C}{2} = -\left(\cos\frac{A+B-C}{2} + \cos\frac{A+B+C}{2}\right)$. For the sum inside: $X = \frac{A+B-C}{2}$, $Y = \frac{A+B+C}{2}$. Their sum: $\frac{A+B-C + A+B+C}{2} = \frac{2A+2B}{2} = A+B$, so $\frac{X+Y}{2} = \frac{A+B}{2}$. Their difference: $\frac{A+B-C - (A+B+C)}{2} = \frac{A+B-C-A-B-C}{2} = \frac{-2C}{2} = -C$, so $\frac{X-Y}{2} = -\frac{C}{2}$. Since cosine is even, $\cos\frac{X-Y}{2} = \cos\frac{C}{2}$. Thus:

Therefore:

4. Combine the two parts. The whole expression is now:

Factor $2\cos\frac{C}{2}$:

$2\cos\frac{C}{2}\left(\cos\frac{B-A}{2} - \cos\frac{A+B}{2}\right)$ …

The key insight here is that when $A+B+C$ is a fixed angle, many trigonometric expressions collapse into constants or simple functions of that sum. Instead of expanding everything blindly, we can use identities that express sums of cosines and products of sines in terms of the sum of angles.

We are given $A+B+C = \frac{3\pi}{2}$. Notice that $\cos(A+B+C) = \cos\frac{3\pi}{2} = 0$. So option (D) becomes $2 - 0 = 2$, and option (B) becomes $0$, option (C) becomes $-1$, option (A) becomes $-(-1)=1$. So the answer is likely a constant — let’s find which one.

1. Rewrite the product $4\sin A \sin B \sin C$

   Use the identity: $2\sin A \sin B = \cos(A-B) - \cos(A+B)$.

   So $4\sin A \sin B \sin C = 2\sin C \cdot [2\sin A \sin B] = 2\sin C [\cos(A-B) - \cos(A+B)]$.

   That gives $2\sin C \cos(A-B) - 2\sin C \cos(A+B)$.

2. Simplify $2\sin C \cos(A+B)$

   Since $A+B = \frac{3\pi}{2} - C$, we have $\cos(A+B) = \cos\left(\frac{3\pi}{2} - C\right) = -\sin C$.

   So $2\sin C \cos(A+B) = 2\sin C (-\sin C) = -2\sin^2 C$.

3. Simplify $2\sin C \cos(A-B)$

   Use the product-to-sum identity: $2\sin C \cos(A-B) = \sin(C + A - B) + \sin(C - A + B)$.

   Now $C + A - B = (A+B+C) - 2B = \frac{3\pi}{2} - 2B$, and $C - A + B = (A+B+C) - 2A = \frac{3\pi}{2} - 2A$.

   So $2\sin C \cos(A-B) = \sin\left(\frac{3\pi}{2} - 2B\right) + \sin\left(\frac{3\pi}{2} - 2A\right)$.

   Since $\sin\left(\frac{3\pi}{2} - \theta\right) = -\cos\theta$, this becomes $-\cos 2B - \cos 2A$.

4. Combine the product part

   So $4\sin A \sin B \sin C = (-\cos 2A - \cos 2B) - (-2\sin^2 C) = -\cos 2A - \cos 2B + 2\sin^2 C$.

5. Add $\cos 2A + \cos 2B + \cos 2C$

   The full expression is: …

Since $A+B+C=2S$, we have $2S-A-B=C$ and $2S-C=A+B$.

Expand each product with $\sin P\cos Q=\tfrac12\big[\sin(P+Q)+\sin(P-Q)\big]$: