Q.$\int_{0}^{\pi} x\log\sin x\,dx$

Concept and Intuition

When integration limits $a,b$ satisfy $a+b=\pi$ (or any constant $L$) and the "weight function" $f(x)$ is invariant under $x\to a+b-x$, the King's-rule trick converts $\int x f(x)\,dx$ into $\frac{a+b}{2}\int f(x)\,dx$. That directly gives $k$.

Step-by-Step Solution

1. Let $I=\displaystyle\int_{\pi/4}^{3\pi/4}\frac{x\sin x}{1+3\cos 2x}dx$ and $J=\displaystyle\int_{\pi/4}^{3\pi/4}\frac{\sin x}{1+3\cos 2x}dx$, so $I=kJ$.
2. Substitute $x\to \pi-x$ in $I$ (limits swap and reverse, net unchanged since $\pi/4+3\pi/4=\pi$):

So $I=\displaystyle\int_{\pi/4}^{3\pi/4}\frac{(\pi-x)\sin x}{1+3\cos 2x}dx=\pi J-I$.

3. Hence $2I=\pi J \Rightarrow I=\dfrac{\pi}{2}J$, so $k=\dfrac{\pi}{2}$.

4. Now compute $\displaystyle\int_0^{k}\sin^{\pi/k}x\,dx=\int_0^{\pi/2}\sin^{2}x\,dx$ (since $\pi/k=\pi/(\pi/2)=2$). …

Concept and Intuition

Whenever an integrand has the shape $x\cdot f(x)$ over $[0,\pi]$ and $f(\pi-x)=f(x)$, the substitution $x\to\pi-x$ shows $\int_0^\pi xf(x)dx = \int_0^\pi(\pi-x)f(x)dx$, so adding gives $2\int_0^\pi xf(x)dx = \pi\int_0^\pi f(x)dx$, i.e. the "King's rule". Here $f(x)=\sin^5x\cos^6x$ satisfies this because the even power on $\cos x$ absorbs the sign flip from $\cos(\pi-x)=-\cos x$.

Step-by-Step Solution

1. Let $f(x)=\sin^5x\cos^6x$. Check: $f(\pi-x) = \sin^5(\pi-x)\cos^6(\pi-x) = \sin^5x\cdot(-\cos x)^6 = \sin^5x\cos^6x = f(x)$ (even power kills the sign).
2. By King's rule: $\displaystyle\int_0^\pi x f(x)\,dx = \frac{\pi}{2}\int_0^\pi f(x)\,dx$.
3. $f(x)$ is also symmetric about $x=\pi/2$ in the same way, so $\displaystyle\int_0^\pi f(x)\,dx = 2\int_0^{\pi/2}\sin^5x\cos^6x\,dx$.
4. Use the Wallis-type formula for $m$ odd (here $m=5$), $n$ even (here $n=6$):

Here $(m-1)!!=4!!=8$, $(n-1)!!=5!!=15$, $(m+n)!!=11!!=10395$.

$\int_0^{\pi/2}\sin^5x\cos^6x\,dx = \frac{8\cdot15}{10395} = \frac{120}{10395} = \frac{8}{693}.$ …

Concept and Intuition

Whenever the integrand is $x$ times a function of $\sin x$ and even powers of $\cos x$ (so that replacing $x\to\pi-x$ leaves the non-$x$ part unchanged), the King's Rule $\int_0^\pi x\,g(x)dx=\dfrac\pi2\int_0^\pi g(x)dx$ removes the awkward factor of $x$ entirely, turning it into an ordinary trigonometric integral.

Step-by-Step Solution

1. Let $g(x)=\dfrac{\sin x}{4\cos^2x+3\sin^2x}$. Check $g(\pi-x)$: $\sin(\pi-x)=\sin x$, $\cos^2(\pi-x)=\cos^2x$, $\sin^2(\pi-x)=\sin^2x$, so $g(\pi-x)=g(x)$.
2. By the King's Rule, $I=\displaystyle\int_0^\pi x\,g(x)\,dx=\frac\pi2\int_0^\pi g(x)\,dx=\frac\pi2 J$.
3. Simplify the denominator: $4\cos^2x+3\sin^2x=3(\sin^2x+\cos^2x)+\cos^2x=3+\cos^2x$.
4. $J=\displaystyle\int_0^\pi\frac{\sin x}{3+\cos^2x}dx$. Let $u=\cos x$, $du=-\sin x\,dx$; limits $x=0\to u=1$, $x=\pi\to u=-1$. …

Concept and Intuition

The key tool is the property $\displaystyle\int_a^b g(x)\,dx = \int_a^b g(a+b-x)\,dx$ (substituting $x\to a+b-x$ reflects the interval onto itself). Applying this to $g(x)=\dfrac{f(x)}{f(x)+f(a+b-x)}$:

Adding these two expressions for $I$: $2I = \int_a^b 1\,dx = b-a$, so $I=\dfrac{b-a}{2}$ — true for any function $f$ for which the integral makes sense, which is exactly what R states.

Step-by-Step Solution

1. Verify R is a genuinely true, general statement (shown above): $\displaystyle\int_a^b\frac{f(x)}{f(x)+f(a+b-x)}dx=\frac{b-a}2$.
2. In the Assertion, $a=\pi/6$, $b=\pi/3$, so $a+b=\pi/2$, matching R's structure with $f(x)=(\sin x)^{\sqrt2}$.
3. Compute: $\dfrac{b-a}2 = \dfrac{\pi/3-\pi/6}{2} = \dfrac{\pi/6}{2}=\dfrac{\pi}{12}$. …

Concept and Intuition

This is a "King's rule" style integral: whenever the integrand can be paired with its reflection so the pair sums to a constant, the integral over a symmetric range collapses to (constant)$\times$(range)/2 — no actual antiderivative is needed.

Step-by-Step Solution

1. Let $f(x)=\dfrac{2023^{\sin^2x}}{2023^{\sin^2x}+2023^{\cos^2x}}$. Since $\sin^2x,\cos^2x$ both have period $\pi$, $f$ has period $\pi$.
2. Check $f(\pi/2-x)$: $\sin^2(\pi/2-x)=\cos^2x$, $\cos^2(\pi/2-x)=\sin^2x$, so $f(\pi/2-x)=\dfrac{2023^{\cos^2x}}{2023^{\cos^2x}+2023^{\sin^2x}}=1-f(x)$.
3. So $\displaystyle\int_0^{\pi/2}f(x)dx=\int_0^{\pi/2}f(\pi/2-x)dx$ (substitution), and adding both expressions: $2\int_0^{\pi/2}f\,dx=\int_0^{\pi/2}1\,dx=\pi/2 \Rightarrow \int_0^{\pi/2}f=\pi/4$.
4. Similarly, substituting $x\to\pi-x$ on $[\pi/2,\pi]$ shows $\int_{\pi/2}^{\pi}f\,dx=\int_0^{\pi/2}f\,dx=\pi/4$. So $\int_0^{\pi}f\,dx=\pi/2$. …

Concept and Intuition

The messy bracket is actually a cubic in $\sin x$ that factors cleanly, and once the $1/\sin x$ cancels with a factor of $\sin x$ from the bracket, the integral reduces to standard $\int_0^\pi x\sin^2x\,dx$, $\int_0^\pi x\sin x\,dx$, and $\int_0^\pi x\,dx$ — all classic results.

Step-by-Step Solution

1. Simplify the bracket: $3\cos^2x+2\sin x+\sin^3x-3 = -3(1-\cos^2x)+2\sin x+\sin^3x = -3\sin^2x+2\sin x+\sin^3x = \sin^3x-3\sin^2x+2\sin x$.
2. Factor: $\sin x(\sin^2x-3\sin x+2) = \sin x(\sin x-1)(\sin x-2)$.
3. So the integrand $\dfrac{x}{\sin x}\times\sin x(\sin x-1)(\sin x-2) = x(\sin x-1)(\sin x-2) = x(\sin^2x-3\sin x+2)$.
4. Integral $I = \displaystyle\int_0^\pi x\sin^2x\,dx - 3\int_0^\pi x\sin x\,dx + 2\int_0^\pi x\,dx$.
5. $\int_0^\pi x\,dx = \pi^2/2$. $\int_0^\pi x\sin x\,dx = [-x\cos x+\sin x]_0^\pi = \pi$. …

Concept and Intuition

Over the interval $[0,\pi/2]$, the substitution $x\to\frac{\pi}{2}-x$ swaps $\sin x\leftrightarrow\cos x$ and hence $\tan x\leftrightarrow\cot x$, while leaving the limits of integration unchanged (just reversed and re-reversed). This is the standard reason $\int_0^{\pi/2}\tan^n x\,dx=\int_0^{\pi/2}\cot^n x\,dx$ for any $n$.

Step-by-Step Solution

1. Consider $I=\displaystyle\int_0^{\pi/2}\cot^n(x)\,dx$.
2. Substitute $x=\frac{\pi}{2}-u$, so $dx=-du$. When $x=0,u=\frac{\pi}{2}$; when $x=\frac{\pi}{2},u=0$.
3. $\cot(x)=\cot\left(\frac{\pi}{2}-u\right)=\tan(u)$.
4. So $I=\displaystyle\int_{\pi/2}^{0}\tan^n(u)\,(-du)=\displaystyle\int_0^{\pi/2}\tan^n(u)\,du$.
5. This is exactly $\displaystyle\int_0^{\pi/2}\tan^n(x)\,dx$ (renaming $u\to x$). …

Concept and Intuition

Both integrals are standard results worth memorizing (or re-deriving via King's rule $x\to a-x$). The second one uses the substitution $x\to\pi/4-x$, which turns $1+\tan x$ into $\dfrac{2}{1+\tan x}$, producing a self-referential equation for the integral.

Step-by-Step Solution

1. $\displaystyle\int_0^\pi\log(\sin x)\,dx=2\int_0^{\pi/2}\log(\sin x)\,dx$ (symmetry about $\pi/2$), and the classical value of $\int_0^{\pi/2}\log(\sin x)\,dx=-\dfrac{\pi}{2}\log2$.
2. So $\displaystyle\int_0^\pi\log(\sin x)\,dx=-\pi\log2=8k\Rightarrow k=-\dfrac{\pi\log2}{8}$.
3. For $I=\displaystyle\int_0^{\pi/4}\log(1+\tan x)\,dx$, substitute $x\to\pi/4-x$: $\tan(\pi/4-x)=\dfrac{1-\tan x}{1+\tan x}$, so $1+\tan(\pi/4-x)=\dfrac{2}{1+\tan x}$. …

Concept and Intuition

The substitution $x\to(a+b-x)$ (here $a=0,b=1$, so $x\to1-x$) is the standard trick for definite integrals over a symmetric interval — it often reveals that two seemingly different integrals are actually equal.

Step-by-Step Solution

1. Start with $I=\displaystyle\int_0^1 x^m(1-x)^n\,dx$.
2. Substitute $x=1-u$, so $dx=-du$; when $x=0,u=1$; when $x=1,u=0$.
3. $I=\displaystyle\int_1^0(1-u)^m u^n(-du)=\int_0^1(1-u)^m u^n\,du$.
4. Renaming the dummy variable $u\to x$: $I=\displaystyle\int_0^1 x^n(1-x)^m\,dx$. …