Integration

The key idea is to exploit the symmetry of $\sin x$ about $x=\frac{\pi }{2}$.

Establishing the identity.

Let $I={\int }_0^{\pi }x\mathrm{f}(\sin x)\mathrm{d}x$. Apply the substitution $x=\pi -t$, so $\mathrm{d}x=-\mathrm{d}t$. When $x=0$, $t=\pi$; when $x=\pi$, $t=0$. The integral becomes$$I={\int }_{\pi }^0(\pi -t)\mathrm{f}(\sin (\pi -t))(-\mathrm{d}t)={\int }_0^{\pi }(\pi -t)\mathrm{f}(\sin t)\mathrm{d}t,$$using the identity $\sin (\pi -t)=\sin t$. Since $t$ is a dummy variable, ${\int }_0^{\pi }t\mathrm{f}(\sin t)\mathrm{d}t=I$, so expanding the right-hand side gives$$I=\pi {\int }_0^{\pi }\mathrm{f}(\sin t)\mathrm{d}t-I.$$Rearranging: $2I=\pi {\int }_0^{\pi }\mathrm{f}(\sin t)\mathrm{d}t$, which yields$${\int }_0^{\pi }x\mathrm{f}(\sin x)\mathrm{d}x=\frac{\pi }{2}{\int }_0^{\pi }\mathrm{f}(\sin x)\mathrm{d}x.$$Part (i).

Here $\mathrm{f}(\sin x)=\frac{\sin x}{3+{\sin }^{2}x}$, so the identity gives$${\int }_0^{\pi }\frac{x\sin x}{3+{\sin }^{2}x}\mathrm{d}x=\frac{\pi }{2}{\int }_0^{\pi }\frac{\sin x}{3+{\sin }^{2}x}\mathrm{d}x.$$Substitute $c=\cos x$, $\mathrm{d}c=-\sin x\mathrm{d}x$. The limits $x=0,\pi$ become $c=1,-1$, and ${\sin }^{2}x=1-c^2$, so the right-hand integral becomes$$\frac{\pi }{2}{\int }_{-1}^1\frac{\mathrm{d}c}{3+(1-c^2)}=\frac{\pi }{2}{\int }_{-1}^1\frac{\mathrm{d}c}{4-c^2}.$$Using partial fractions, $\frac{1}{4-c^2}=\frac{1}{4}\left(\frac{1}{2-c}+\frac{1}{2+c}\right)$, so$$\frac{\pi }{2}\cdot \frac{1}{4}[-\ln |2-c|+\ln |2+c|{]}_{-1}^1=\frac{\pi }{8}[\ln \left(\frac{2+c}{2-c}\right){]}_{-1}^1=\frac{\pi }{8}\left(\ln 3-\ln \frac{1}{3}\right)=\frac{\pi }{8}\cdot 2\ln 3=\frac{\pi \ln 3}{4}.$$Part (ii).

Split the integral at $\pi$:$${\int }_0^{2\pi }\frac{x\sin x}{3+{\sin }^{2}x}\mathrm{d}x={\int }_0^{\pi }\frac{x\sin x}{3+{\sin }^{2}x}\mathrm{d}x+{\int }_{\pi }^{2\pi }\frac{x\sin x}{3+{\sin }^{2}x}\mathrm{d}x.$$For the second piece, substitute $y=x-\pi$, so $x=y+\pi$ and $\sin x=\sin (y+\pi )=-\sin y$:$${\int }_0^{\pi }\frac{(y+\pi )(-\sin y)}{3+{\sin }^{2}y}\mathrm{d}y=-{\int }_0^{\pi }\frac{y\sin y}{3+{\sin }^{2}y}\mathrm{d}y-\pi {\int }_0^{\pi }\frac{\sin y}{3+{\sin }^{2}y}\mathrm{d}y.$$The first integral is the result from part (i), $\frac{\pi \ln 3}{4}$. For the second, the same $c=\cos y$ substitution gives $\pi \cdot \frac{\ln 3}{4}$ (using the same working as part (i) without the $\frac{\pi }{2}$ factor). Therefore the total is$$\frac{\pi \ln 3}{4}-\frac{\pi \ln 3}{4}-\frac{\pi \ln 3}{2}=-\frac{\pi \ln 3}{2}.$$Part (iii).

Write $|\sin 2x|=2|\sin x\cos x|$. On $[0,\pi ]$ one has $\sin x\ge 0$, so$$\frac{x|\sin 2x|}{3+{\sin }^{2}x}=\frac{2x|\cos x|\sin x}{3+{\sin }^{2}x}.$$Now $|\cos x|=\sqrt{1-{\sin }^{2}x}$, which is a function of $\sin x$, so the identity applies with $\mathrm{f}(\sin x)=\frac{2|\cos x|\sin x}{3+{\sin }^{2}x}$:$${\int }_0^{\pi }\frac{x|\sin 2x|}{3+{\sin }^{2}x}\mathrm{d}x=\pi {\int }_0^{\pi }\frac{|\cos x|\sin x}{3+{\sin }^{2}x}\mathrm{d}x.$$Split at $x=\frac{\pi }{2}$ where $\cos x$ changes sign, and use $c=\cos x$ on each half:$$\pi \left({\int }_0^{\pi /2}\frac{\cos x\sin x}{3+{\sin }^{2}x}\mathrm{d}x+{\int }_{\pi /2}^{\pi }\frac{-\cos x\sin x}{3+{\sin }^{2}x}\mathrm{d}x\right)=2\pi {\int }_0^{\pi /2}\frac{\cos x\sin x}{3+{\sin }^{2}x}\mathrm{d}x.$$With $u={\sin }^{2}x$, $\mathrm{d}u=2\sin x\cos x\mathrm{d}x$, the integral becomes$$2\pi {\int }_0^1\frac{\mathrm{d}u}{2(3+u)}=\pi [\ln (3+u){]}_0^1=\pi (\ln 4-\ln 3)=\pi \ln \frac{4}{3}.$$

Part (i).

To show the first result, substitute $\theta \mapsto -\theta$ in the original integral $I$. Since ${\cos }^2(-\theta )={\cos }^2\theta$ and $\sin (-\theta )=-\sin \theta$, the integrand becomes $\frac{{\cos }^2\theta }{1+\sin \theta \sin 2\alpha }$, and the limits $-\frac{\pi }{2}$ to $\frac{\pi }{2}$ are unchanged. Hence$$\begin{gathered} I={\int }_{-\pi /2}^{\pi /2}\frac{{\cos }^2\theta }{1+\sin \theta \sin 2\alpha } \\ ,\mathrm{d}\theta . \end{gathered}$$Adding this to the original expression for $I$ gives$$\begin{gathered} 2I={\int }_{-\pi /2}^{\pi /2}{\cos }^2\theta \left(\frac{1}{1-\sin \theta \sin 2\alpha }+\frac{1}{1+\sin \theta \sin 2\alpha }\right)\mathrm{d}\theta ={\int }_{-\pi /2}^{\pi /2}\frac{2{\cos }^2\theta }{1-{\sin }^2\theta {\sin }^{2}2\alpha } \\ ,\mathrm{d}\theta . \end{gathered}$$Divide numerator and denominator by ${\cos }^2\theta$:$$\begin{gathered} 2I={\int }_{-\pi /2}^{\pi /2}\frac{2}{{\sec }^2\theta -{\tan }^2\theta {\sin }^{2}2\alpha } \\ ,\mathrm{d}\theta . \end{gathered}$$Since ${\sec }^2\theta =1+{\tan }^2\theta$, the denominator becomes $1+{\tan }^2\theta (1-{\sin }^{2}2\alpha )=1+{\tan }^2\theta {\cos }^{2}2\alpha$, giving$$\begin{gathered} 2I={\int }_{-\pi /2}^{\pi /2}\frac{2}{1+{\tan }^2\theta {\cos }^{2}2\alpha } \\ ,\mathrm{d}\theta . \end{gathered}$$Part (ii).

To evaluate $$\begin{gathered} J={\int }_{-\pi /2}^{\pi /2}\frac{{\sec }^2\theta }{1+{\tan }^2\theta {\cos }^{2}2\alpha } \\ ,\mathrm{d}\theta \end{gathered}$$, set $u=\cos 2\alpha \tan \theta$. Then $$\begin{gathered} \mathrm{d}u=\cos 2\alpha {\sec }^2\theta \\ ,\mathrm{d}\theta \end{gathered}$$, so $$\begin{gathered} {\sec }^2\theta \\ ,\mathrm{d}\theta =\frac{\mathrm{d}u}{\cos 2\alpha } \end{gathered}$$. Since $0<\alpha <\frac{\pi }{4}$, we have $\cos 2\alpha >0$, so the substitution is orientation-preserving and as $\theta$ ranges from $-\frac{\pi }{2}$ to $\frac{\pi }{2}$, $u$ ranges from $-\infty$ to $+\infty$. Therefore$$J=\frac{1}{\cos 2\alpha }{\int }_{-\infty }^{\infty }\frac{\mathrm{d}u}{1+u^2}=\frac{1}{\cos 2\alpha }\cdot \pi =\pi \sec 2\alpha .$$Part (iii).

From Part (i), $$\begin{gathered} 2I={\int }_{-\pi /2}^{\pi /2}\frac{2}{1+{\tan }^2\theta {\cos }^{2}2\alpha } \\ ,\mathrm{d}\theta \end{gathered}$$, so $I=J\cdot \frac{{\cos }^{2}2\alpha +1}{\cdots }$... A cleaner route is to form the combination $I{\sin }^{2}2\alpha +J{\cos }^{2}2\alpha$.

Note that $$\begin{gathered} J={\int }_{-\pi /2}^{\pi /2}\frac{{\sec }^2\theta }{1+{\tan }^2\theta {\cos }^{2}2\alpha } \\ ,\mathrm{d}\theta \end{gathered}$$ and, from Part (i), $$\begin{gathered} 2I={\int }_{-\pi /2}^{\pi /2}\frac{2}{1+{\tan }^2\theta {\cos }^{2}2\alpha } \\ ,\mathrm{d}\theta \end{gathered}$$. Multiply the second by $\frac{1}{2}{\sin }^{2}2\alpha$ and add ${\cos }^{2}2\alpha$ times the $J$-integral:$$\begin{gathered} I{\sin }^{2}2\alpha +J{\cos }^{2}2\alpha ={\int }_{-\pi /2}^{\pi /2}\frac{{\sin }^{2}2\alpha +{\cos }^{2}2\alpha {\sec }^2\theta }{1+{\tan }^2\theta {\cos }^{2}2\alpha } \\ ,\mathrm{d}\theta . \end{gathered}$$The numerator equals ${\sin }^{2}2\alpha +{\cos }^{2}2\alpha (1+{\tan }^2\theta )=1+{\tan }^2\theta {\cos }^{2}2\alpha$, which cancels the denominator exactly. Hence$$I{\sin }^{2}2\alpha +J{\cos }^{2}2\alpha ={\int }_{-\pi /2}^{\pi /2}\mathrm{d}\theta =\pi .$$Substituting $J=\pi \sec 2\alpha$:$$I{\sin }^{2}2\alpha =\pi -\pi \sec 2\alpha {\cos }^{2}2\alpha =\pi (1-\cos 2\alpha ).$$Using the identity $1-\cos 2\alpha =2{\sin }^2\alpha$ and ${\sin }^{2}2\alpha =4{\sin }^2\alpha {\cos }^2\alpha$:$$I=\frac{2\pi {\sin }^2\alpha }{4{\sin }^2\alpha {\cos }^2\alpha }=\frac{\pi }{2{\cos }^2\alpha }=\frac{1}{2}\pi {\sec }^2\alpha .$$Part (iv).

For $\frac{\pi }{4}<\alpha <\frac{\pi }{2}$, the constant $\cos 2\alpha <0$. In the substitution for $J$, the factor $\cos 2\alpha$ is now negative, so $u=\cos 2\alpha \tan \theta$ runs from $+\infty$ to $-\infty$ as $\theta$ increases, reversing orientation. Thus $J=-\pi \sec 2\alpha$ in this range.

Returning to $I{\sin }^{2}2\alpha +J{\cos }^{2}2\alpha =\pi$ (this combinatorial identity holds regardless of the sign of $\cos 2\alpha$, since it was established by purely algebraic cancellation in the integrand):$$I{\sin }^{2}2\alpha =\pi -(-\pi \sec 2\alpha ){\cos }^{2}2\alpha =\pi (1+\cos 2\alpha ).$$Using $1+\cos 2\alpha =2{\cos }^2\alpha$:$$\begin{gathered} I=\frac{2\pi {\cos }^2\alpha }{4{\sin }^2\alpha {\cos }^2\alpha }=\frac{\pi }{2{\sin }^2\alpha }=\frac{1}{2}\pi \\ ,{\mathrm{cosec}}^2\alpha . \end{gathered}$$

Part (i).

The key observation is that the integrand has a hidden symmetry when you swap $x$ with $a-x$. To exploit it, substitute $x=a-y$ (so $\mathrm{d}x=-\mathrm{d}y$), and note that the limits flip: when $x=0$, $y=a$, and when $x=a$, $y=0$. Therefore$$I={\int }_0^a\frac{\mathrm{f}(x)}{\mathrm{f}(x)+\mathrm{f}(a-x)}\mathrm{d}x={\int }_a^0\frac{\mathrm{f}(a-y)}{\mathrm{f}(a-y)+\mathrm{f}(y)}\cdot (-\mathrm{d}y)={\int }_0^a\frac{\mathrm{f}(a-y)}{\mathrm{f}(y)+\mathrm{f}(a-y)}\mathrm{d}y.$$Since $y$ is just a dummy variable, this is exactly$$I={\int }_0^a\frac{\mathrm{f}(a-x)}{\mathrm{f}(x)+\mathrm{f}(a-x)}\mathrm{d}x.$$Now add the two expressions for $I$:$$2I={\int }_0^a\frac{\mathrm{f}(x)+\mathrm{f}(a-x)}{\mathrm{f}(x)+\mathrm{f}(a-x)}\mathrm{d}x={\int }_0^{a}1\mathrm{d}x=a.$$Hence $I=\frac{a}{2}$, regardless of what $\mathrm{f}$ is (as long as the denominator never vanishes).

First integral. We need to match $\frac{\ln (x+1)}{\ln (2+x-x^2)}$ to the form $\frac{\mathrm{f}(x)}{\mathrm{f}(x)+\mathrm{f}(a-x)}$. Factorise the denominator's argument:$$2+x-x^2=(1+x)(2-x).$$So $\ln (2+x-x^2)=\ln (1+x)+\ln (2-x)$. Set $\mathrm{f}(x)=\ln (1+x)$ and $a=1$. Then $\mathrm{f}(a-x)=\ln (2-x)$, and the integral becomes ${\int }_0^1\frac{\mathrm{f}(x)}{\mathrm{f}(x)+\mathrm{f}(1-x)}\mathrm{d}x=\frac{1}{2}$.

Second integral. For ${\int }_0^{\pi /2}\frac{\sin x}{\sin (x+\frac{\pi }{4})}\mathrm{d}x$, expand the denominator:$$\sin \left(x+\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}\sin x+\frac{1}{\sqrt{2}}\cos x.$$So the denominator is $\frac{1}{\sqrt{2}}(\sin x+\cos x)$. Set $\mathrm{f}(x)=\sin x$ and $a=\frac{\pi }{2}$; then $\mathrm{f}(a-x)=\sin (\frac{\pi }{2}-x)=\cos x$. The integral is$${\int }_0^{\pi /2}\frac{\sin x}{\sin x+\cos x}\mathrm{d}x=\frac{a}{2}=\frac{\pi }{4}.$$Part (ii).

The structure is different: the limits are $\frac{1}{2}$ and $2$, and the integrand involves $\sin x$ and $\sin \frac{1}{x}$. The hint is to mimic the same averaging trick with the substitution $u=\frac{1}{x}$.

Let $u=\frac{1}{x}$, so $\mathrm{d}x=-\frac{1}{u^2}\mathrm{d}u$. When $x=\frac{1}{2}$, $u=2$; when $x=2$, $u=\frac{1}{2}$. Write $J$ for the integral:$$J={\int }_{1/2}^2\frac{\sin x}{x(\sin x+\sin \frac{1}{x})}\mathrm{d}x={\int }_2^{1/2}\frac{\sin \frac{1}{u}}{\frac{1}{u}(\sin \frac{1}{u}+\sin u)}\cdot \left(-\frac{1}{u^2}\right)\mathrm{d}u.$$Simplifying (noting $\frac{1}{u}\cdot u^2=u$):$$J={\int }_{1/2}^2\frac{\sin \frac{1}{u}}{u(\sin u+\sin \frac{1}{u})}\mathrm{d}u.$$Adding this to the original:$$2J={\int }_{1/2}^2\frac{\sin x+\sin \frac{1}{x}}{x(\sin x+\sin \frac{1}{x})}\mathrm{d}x={\int }_{1/2}^2\frac{1}{x}\mathrm{d}x=[\ln x{]}_{1/2}^2=\ln 2-\ln \frac{1}{2}=2\ln 2.$$Hence $J=\ln 2$.

Part (i). We use the compound-angle identity for cosine. Write $\frac{1}{4}\pi -\frac{1}{2}x=\theta$; then$$\cos 2\theta =1-2{\sin }^2\theta ,$$so $\cos \left(\frac{\pi }{2}-x\right)=1-2{\sin }^2\left(\frac{\pi }{4}-\frac{x}{2}\right)$. Since $\cos \left(\frac{\pi }{2}-x\right)=\sin x$, this gives$$\sin x=1-2{\sin }^2\left(\frac{\pi }{4}-\frac{x}{2}\right).$$Divide through by ${\cos }^2\left(\frac{\pi }{4}-\frac{x}{2}\right)$ and use ${\sec }^2=1+{\tan }^2$:$${\sec }^2\left(\frac{\pi }{4}-\frac{x}{2}\right)=\frac{1}{{\cos }^2(\frac{\pi }{4}-\frac{x}{2})}.$$Equivalently, $1+\sin x=2{\cos }^2\left(\frac{\pi }{4}-\frac{x}{2}\right)$, so$${\sec }^2\left(\frac{\pi }{4}-\frac{x}{2}\right)=\frac{2}{1+\sin x}.✓$$Hence $\frac{1}{1+\sin x}=\frac{1}{2}{\sec }^2\left(\frac{\pi }{4}-\frac{x}{2}\right)$. Since $\frac{\mathrm{d}}{\mathrm{d}x}\tan \left(\frac{\pi }{4}-\frac{x}{2}\right)=-\frac{1}{2}{\sec }^2\left(\frac{\pi }{4}-\frac{x}{2}\right)$,$$\int \frac{\mathrm{d}x}{1+\sin x}=-\tan \left(\frac{\pi }{4}-\frac{x}{2}\right)+C.$$Part (ii). Set $y=\pi -x$, so $x=\pi -y$ and $\mathrm{d}x=-\mathrm{d}y$. When $x=0$, $y=\pi$; when $x=\pi$, $y=0$. Also $\sin (\pi -y)=\sin y$. Therefore$${\int }_0^{\pi }x\mathrm{f}(\sin x)\mathrm{d}x={\int }_{\pi }^0(\pi -y)\mathrm{f}(\sin y)(-\mathrm{d}y)={\int }_0^{\pi }(\pi -y)\mathrm{f}(\sin y)\mathrm{d}y.$$Call this common value $I$. Adding the two expressions:$$2I={\int }_0^{\pi }x\mathrm{f}(\sin x)\mathrm{d}x+{\int }_0^{\pi }(\pi -x)\mathrm{f}(\sin x)\mathrm{d}x=\pi {\int }_0^{\pi }\mathrm{f}(\sin x)\mathrm{d}x,$$yielding $I=\frac{\pi }{2}{\int }_0^{\pi }\mathrm{f}(\sin x)\mathrm{d}x$. $✓$

Apply this with $\mathrm{f}(\sin x)=\frac{1}{1+\sin x}$:$${\int }_0^{\pi }\frac{x}{1+\sin x}\mathrm{d}x=\frac{\pi }{2}{\int }_0^{\pi }\frac{\mathrm{d}x}{1+\sin x}=\frac{\pi }{2}[-\tan \left(\frac{\pi }{4}-\frac{x}{2}\right){]}_0^{\pi }.$$At $x=\pi$: $\tan (-\frac{\pi }{4})=-1$. At $x=0$: $\tan (\frac{\pi }{4})=1$. So$${\int }_0^{\pi }\frac{x}{1+\sin x}\mathrm{d}x=\frac{\pi }{2}(1-(-1)\cdot (-1))=\frac{\pi }{2}\cdot 2=\pi .$$Wait — more carefully: ${\left[-\tan (\frac{\pi }{4}-\frac{x}{2})\right]}_0^{\pi }=\left(-\tan (-\frac{\pi }{4})\right)-\left(-\tan (\frac{\pi }{4})\right)=1-(-1)=2$. Hence the integral equals $\frac{\pi }{2}\cdot 2=\pi$.$$\boxed{{\int }_0^{\pi }\frac{x}{1+\sin x}\mathrm{d}x=\pi .}$$Part (iii). Write $2x^3-3\pi x^2=x^2(2x-3\pi )$. Try the substitution $y=\pi -x$ again. Under $x\mapsto \pi -y$, we have $(1+\sin x{)}^2\to (1+\sin y{)}^2$ and $\mathrm{d}x=-\mathrm{d}y$, so$$J={\int }_0^{\pi }\frac{2x^3-3\pi x^2}{(1+\sin x{)}^2}\mathrm{d}x={\int }_0^{\pi }\frac{2(\pi -y{)}^3-3\pi (\pi -y{)}^2}{(1+\sin y{)}^2}\mathrm{d}y.$$Adding the two expressions for $J$:$$2J={\int }_0^{\pi }\frac{\left[2x^3-3\pi x^2\right]+\left[2(\pi -x{)}^3-3\pi (\pi -x{)}^2\right]}{(1+\sin x{)}^2}\mathrm{d}x.$$Expand the numerator: $2x^3-3\pi x^2+2(\pi -x{)}^3-3\pi (\pi -x{)}^2$. Direct expansion gives $-{\pi }^3$, a constant. Hence$$2J=-{\pi }^3{\int }_0^{\pi }\frac{\mathrm{d}x}{(1+\sin x{)}^2}=-{\pi }^3{\int }_0^{\pi }\frac{1}{4}{\sec }^4\left(\frac{\pi }{4}-\frac{x}{2}\right)\mathrm{d}x.$$Substitute $u=\frac{\pi }{4}-\frac{x}{2}$, so $\mathrm{d}x=-2\mathrm{d}u$; limits $u:\frac{\pi }{4}\to -\frac{\pi }{4}$:$${\int }_0^{\pi }\frac{\mathrm{d}x}{(1+\sin x{)}^2}=\frac{1}{4}\cdot 2{\int }_{-\pi /4}^{\pi /4}{\sec }^{4}u\mathrm{d}u=\frac{1}{2}{\int }_{-\pi /4}^{\pi /4}(1+{\tan }^{2}u){\sec }^{2}u\mathrm{d}u.$$This equals $\frac{1}{2}{\left[\tan u+\frac{{\tan }^{3}u}{3}\right]}_{-\pi /4}^{\pi /4}=\frac{1}{2}\cdot 2\left(1+\frac{1}{3}\right)=\frac{4}{3}$.

Therefore $2J=-{\pi }^3\cdot \frac{4}{3}$, giving$$\boxed{{\int }_0^{\pi }\frac{2x^3-3\pi x^2}{(1+\sin x{)}^2}\mathrm{d}x=-\frac{2{\pi }^3}{3}.}$$

Preliminary result. Let $u=a-x$, so $du=-dx$. When $x=0$, $u=a$; when $x=a$, $u=0$. Thus$${\int }_0^a\mathrm{f}(x)dx={\int }_a^0\mathrm{f}(a-u)(-du)={\int }_0^a\mathrm{f}(a-u)du={\int }_0^a\mathrm{f}(a-x)dx,$$where the last step renames the dummy variable. $\square$

Part (i). Let $I={\int }_0^{\pi /2}\frac{\sin x}{\cos x+\sin x}dx$. Applying $(\ast )$ with $a=\frac{\pi }{2}$, and using $\sin (\frac{\pi }{2}-x)=\cos x$, $\cos (\frac{\pi }{2}-x)=\sin x$:$$I={\int }_0^{\pi /2}\frac{\cos x}{\sin x+\cos x}dx.$$Adding the two expressions for $I$:$$2I={\int }_0^{\pi /2}\frac{\sin x+\cos x}{\cos x+\sin x}dx={\int }_0^{\pi /2}1dx=\frac{\pi }{2}.$$Hence $\boxed{I=\frac{\pi }{4}}$.

Part (ii). Let $J={\int }_0^{\pi /4}\frac{\sin x}{\cos x+\sin x}dx$. Write $\cos x+\sin x=\sqrt{2}\sin (x+\frac{\pi }{4})$, and expand:$$\frac{\sin x}{\cos x+\sin x}=\frac{1}{2}-\frac{1}{2}\cot \left(x+\frac{\pi }{4}\right).$$Integrating term by term:$$J={\left[\frac{x}{2}-\frac{1}{2}\ln \left|\sin \left(x+\frac{\pi }{4}\right)\right|\right]}_0^{\pi /4}.$$At $x=\frac{\pi }{4}$: $\frac{\pi }{8}-\frac{1}{2}\ln 1=\frac{\pi }{8}$. At $x=0$: $0-\frac{1}{2}\ln \frac{1}{\sqrt{2}}=\frac{\ln 2}{4}$. Hence$$\boxed{J=\frac{\pi }{8}-\frac{\ln 2}{4}}.$$Part (iii). Let $K={\int }_0^{\pi /4}\ln (1+\tan x)dx$. Apply $(\ast )$ with $a=\frac{\pi }{4}$. Since $\tan (\frac{\pi }{4}-x)=\frac{1-\tan x}{1+\tan x}$:$$K={\int }_0^{\pi /4}\ln \left(1+\frac{1-\tan x}{1+\tan x}\right)dx={\int }_0^{\pi /4}\ln \left(\frac{2}{1+\tan x}\right)dx.$$Adding the two expressions for $K$:$$2K={\int }_0^{\pi /4}\left[\ln (1+\tan x)+\ln 2-\ln (1+\tan x)\right]dx=\frac{\pi }{4}\ln 2.$$Hence $\boxed{K=\frac{\pi \ln 2}{8}}$.

Part (iv). Let $L={\int }_0^{\pi /4}\frac{x}{\cos x(\cos x+\sin x)}dx$. Apply $(\ast )$ with $a=\frac{\pi }{4}$. Note that$$\cos \left(\frac{\pi }{4}-x\right)+\sin \left(\frac{\pi }{4}-x\right)=\sqrt{2}\cos x,\cos \left(\frac{\pi }{4}-x\right)=\frac{\cos x+\sin x}{\sqrt{2}},$$so the symmetrised integrand simplifies to$$\frac{\frac{\pi }{4}-x}{\frac{\cos x+\sin x}{\sqrt{2}}\cdot \sqrt{2}\cos x}=\frac{\frac{\pi }{4}-x}{\cos x(\cos x+\sin x)}.$$Adding the two expressions for $L$:$$2L=\frac{\pi }{4}{\int }_0^{\pi /4}\frac{dx}{\cos x(\cos x+\sin x)}.$$To evaluate this remaining integral, substitute $u=\tan x$ (so $dx=\frac{du}{1+u^2}$ and the limits become $0$ to $1$):$${\int }_0^{\pi /4}\frac{dx}{\cos x(\cos x+\sin x)}={\int }_0^{\pi /4}\frac{\sec x}{1+\tan x}dx={\int }_0^1\frac{du}{(1+u)(1+u^2)}.$$Partial fractions: $\frac{1}{(1+u)(1+u^2)}=\frac{1/2}{1+u}+\frac{1/2-u/2}{1+u^2}$. Integrating:$${\int }_0^1\frac{du}{(1+u)(1+u^2)}={\left[\frac{\ln (1+u)}{2}+\frac{\arctan u}{2}-\frac{\ln (1+u^2)}{4}\right]}_0^1=\frac{\ln 2}{2}+\frac{\pi }{8}-\frac{\ln 2}{4}=\frac{\ln 2}{4}+\frac{\pi }{8}.$$Therefore$$2L=\frac{\pi }{4}\left(\frac{\ln 2}{4}+\frac{\pi }{8}\right)=\frac{\pi \ln 2}{16}+\frac{{\pi }^2}{32},$$giving$$\boxed{L=\frac{\pi \ln 2}{32}+\frac{{\pi }^2}{64}}.$$

Part (i). We differentiate $\mathrm{f}(x)=\frac{1}{1+\tan x}$ using the quotient rule. The derivative of $\tan x$ is ${\sec }^{2}x$, so$${\mathrm{f}}^{\prime }(x)=-\frac{{\sec }^{2}x}{(1+\tan x{)}^2}.$$To reach the target form, multiply top and bottom by ${\cos }^{2}x$:$${\mathrm{f}}^{\prime }(x)=-\frac{1}{(\cos x+\sin x{)}^2}.$$Expanding the denominator and applying the double-angle identity $\sin 2x=2\sin x\cos x$ gives $(\cos x+\sin x{)}^2=1+\sin 2x$, so$${\mathrm{f}}^{\prime }(x)=-\frac{1}{1+\sin 2x}.$$For $0⩽x<\frac{1}{2}\pi$, the argument $2x$ ranges over $[0,\pi )$, so $\sin 2x\in [0,1]$ and hence $1+\sin 2x\in [1,2]$. Therefore ${\mathrm{f}}^{\prime }(x)\in [-1,-\frac{1}{2}]$.

For the sketch: $\mathrm{f}$ is strictly decreasing from $\mathrm{f}(0)=1$ down to $0$ as $x\to {\frac{\pi }{2}}^{-}$. The derivative is most negative (steepest) where $\sin 2x=1$, i.e. at $x=\frac{\pi }{4}$, giving the point $(\frac{\pi }{4},\frac{1}{2})$. Since ${\mathrm{f}}^{\prime \prime }$ changes sign there, this is a point of inflexion. The curve is concave up near $x=0$ and concave down near $x=\frac{\pi }{2}$.

Part (ii). A $180^{\circ }$ rotation about $(a,b)$ sends $(x,y)$ to $$\begin{gathered} (2a-x, \\ ,2b-y) \end{gathered}$$. The curve $y=\mathrm{g}(x)$ is invariant under this rotation if and only if, whenever $(x,\mathrm{g}(x))$ lies on the curve, so does $$\begin{gathered} (2a-x, \\ ,2b-\mathrm{g}(x)) \end{gathered}$$. That second point lies on the curve precisely when $\mathrm{g}(2a-x)=2b-\mathrm{g}(x)$, i.e.$$\mathrm{g}(x)+\mathrm{g}(2a-x)=2b.$$This must hold for all $x$ in the domain, establishing the equivalence.

Now suppose $\mathrm{g}$ passes through the origin and has rotational symmetry of order 2 about the origin, so $(a,b)=(0,0)$. The condition becomes $\mathrm{g}(x)+\mathrm{g}(-x)=0$, meaning $\mathrm{g}$ is an odd function. For any odd continuous function on a symmetric interval,$$\begin{gathered} {\int }_{-1}^1\mathrm{g}(x) \\ ,\mathrm{d}x=0. \end{gathered}$$Part (iii). Set $\mathrm{h}(x)=\frac{1}{1+{\tan }^{k}x}$ on $(0,\frac{\pi }{2})$. We test the symmetry condition about the candidate point $(\frac{\pi }{4},\frac{1}{2})$. We need $\mathrm{h}(x)+\mathrm{h}(\frac{\pi }{2}-x)=1$.

Using $\tan (\frac{\pi }{2}-x)=\cot x={\tan }^{-1}x$:$$\begin{gathered} \mathrm{h} \\ !\left(\frac{\pi }{2}-x\right)=\frac{1}{1+{\cot }^{k}x}=\frac{{\tan }^{k}x}{{\tan }^{k}x+1}. \end{gathered}$$Therefore$$\begin{gathered} \mathrm{h}(x)+\mathrm{h} \\ !\left(\frac{\pi }{2}-x\right)=\frac{1}{1+{\tan }^{k}x}+\frac{{\tan }^{k}x}{1+{\tan }^{k}x}=1. \end{gathered}$$By part (ii) with $a=\frac{\pi }{4}$, $b=\frac{1}{2}$, the curve has rotational symmetry of order 2 about $(\frac{\pi }{4},\frac{1}{2})$.

For the integral, the interval $[\frac{\pi }{6},\frac{\pi }{3}]$ is centred on $\frac{\pi }{4}$ and has width $\frac{\pi }{6}$. Since the curve has rotational symmetry about the midpoint of this interval at height $\frac{1}{2}$, the area under the curve equals the area of the rectangle of width $\frac{\pi }{6}$ and height $\frac{1}{2}$:$$\begin{gathered} {\int }_{\pi /6}^{\pi /3}\frac{1}{1+{\tan }^{k}x} \\ ,\mathrm{d}x=\frac{1}{2}\cdot \frac{\pi }{6}=\frac{\pi }{12}. \end{gathered}$$