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

Step 1 — The symmetry property to use

For an interval symmetric about the origin,

Here $a = \pi$, so the parity of the integrand settles everything.

Step 2 — Determine the parity of each factor

• $1 - x^2$: replacing $x \to -x$ gives $1 - x^2$ ⇒ even
• $\sin x$: $\sin(-x) = -\sin x$ ⇒ odd
• $\cos^2 x$: $\cos^2(-x) = (\cos x)^2 = \cos^2 x$ ⇒ even

Step 3 — Combine

$f(-x) = \underbrace{(1 - x^2)}_{\text{even}}\cdot \underbrace{(-\sin x)}_{\text{odd}}\cdot \underbrace{\cos^2 x}_{\text{even}} = -\,(1-x^2)\sin x\cos^2 x = -f(x)$ …

Concept: simplify the trigonometric integrand before integrating.

Denominator: cot x - tan x = cos x/sin x - sin x/cos x = (cos^2 x - sin^2 x)/(sin x cos x) = cos 2x / ((1/2) sin 2x) = 2 cos 2x / sin 2x = 2 cot 2x.

Numerator: cos 4x + 1 = 2 cos^2 2x.

So the integrand becomes …

Step 0 — The limits.

The printed limits render as $\pm\tfrac12$, but with those limits the value would be $\sin(0.5)\approx 0.479$, which is not among the options. The intended (and standard KCET) integral is over the symmetric interval $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ — the $\pi$ has been lost in typesetting. Every option is consistent with that reading.

Step 1 — Set up the king's-property trick.

Replace $x\to -x$ (limits are symmetric, so the value is unchanged):

$I=\int_{-\pi/2}^{\pi/2}\frac{\cos(-x)}{1+e^{-x}}\,dx=\int_{-\pi/2}^{\pi/2}\frac{\cos x\;e^{x}}{1+e^{x}}\,dx,$ …

Step 1 — Turn the radicand into a perfect square.

For any angle $\theta$,

This is the standard trick for a $\sqrt{1\pm\sin\theta}$ integral: a square root is only integrable in closed form once the radicand is a square.

Step 2 — Apply it with $\theta=\dfrac{x}{2}$ (so $\dfrac{\theta}{2}=\dfrac{x}{4}$).

Step 3 — Remove the modulus (this is where such problems usually go wrong).

As $x$ runs over $[0,2\pi]$, the argument $\dfrac{x}{4}$ runs over $\left[0,\dfrac{\pi}{2}\right]$. In the first quadrant both $\sin\frac{x}{4}\ge 0$ and $\cos\frac{x}{4}\ge 0$, so the bracket never changes sign and

$\left|\sin\tfrac{x}{4}+\cos\tfrac{x}{4}\right|=\sin\tfrac{x}{4}+\cos\tfrac{x}{4}.$ …

Step 1 — Convert to degrees to see the structure.

Notice at once that $75^\circ + 15^\circ = 90^\circ$ — the angles are complementary. That is the hook the question is built on.

Step 2 — Use the complementary (co-function) identity.

So the required product becomes a sine × cosine of the same angle:

Step 3 — Apply the double-angle identity.

Since $\sin 2\theta = 2\sin\theta\cos\theta$, we have $\sin\theta\cos\theta = \tfrac12 \sin 2\theta$. With $\theta = 75^\circ$:

Step 4 — Evaluate.

$\sin 150^\circ = \sin(180^\circ - 30^\circ) = \sin 30^\circ = \frac12 .$ …

Step 1 — The key identity. From $\cos\phi=2\cos^2\frac{\phi}{2}-1$,

Step 2 — Innermost radical. The stem's innermost term is $2+2\cos 8\theta$ (the three nested radicals require the inner angle $8\theta$; with a bare $\theta$ the expression would reduce to $2\cos\frac{\theta}{8}$, which is not among the options):

Step 3 — Second radical.

Step 4 — Outermost radical. …