In this exercise, we are given two integrals $I$ and $J$:
$$I = \int_{0}^{\pi/2}\frac{\sin^nx}{\sin^nx+\cos^nx} dx, \quad n\in \mathbb{N^*},$$ and $$J = \int_{1}^{2} \frac{\arctan(x)}{\arctan \left (\frac{1}{x^2-3x+3} \right )} \, dx.$$
I know that $I = \pi/4$ but I do not understand the association between $I$ and $J$.
This exercise is found in a lesson on substitution so maybe there is no association between the two integrals, but $J$ should be $1.108205454429995$.
As you suspected, there may be one more typo in the textbook.
As I commented, I was wondering if approximation would be require. So, for the fun of it, I tried to. $$I=\int_1^2\frac{\tan ^{-1}(x)}{\tan ^{-1}(2-x)-\tan ^{-1}(1-x)}\,dx=\int_0^1 \frac{\tan ^{-1}(y+1)}{\tan ^{-1}(y)-\tan ^{-1}(y-1)}\,dy$$
Using Taylor expansions built around $y=0$ $$\tan(y+a)=\tan ^{-1}(a)+\frac{y}{a^2+1}-\frac{a y^2}{\left(a^2+1\right)^2}+\frac{\left(3 a^2-1\right) y^3}{3 \left(a^2+1\right)^3}+\frac{\left(a-a^3\right) y^4}{\left(a^2+1\right)^4}+O\left(y^5\right)$$ leads to $$\frac{\tan ^{-1}(y+1)}{\tan ^{-1}(y)-\tan ^{-1}(y-1)}=\frac{\frac{\pi }{4}+\frac{y}{2}-\frac{y^2}{4}+\frac{y^3}{12}+O\left(y^5\right) }{\frac{\pi }{4}+\frac{y}{2}-\frac{y^2}{4}-\frac{5 y^3}{12}+O\left(y^5\right) }$$ Using long division, the integrand is $$1+\frac{2 y^3}{\pi }-\frac{4 y^4}{\pi ^2}+O\left(y^5\right)$$ leading to $$I_{(5)}=1+\frac{1}{2 \pi }-\frac{4}{5 \pi ^2}\approx 1.07810$$ which is not fantastic.
For sure, we could continue the expansions and get $$I_{(6)}=1+\frac{1}{3 \pi }+\frac{31}{105 \pi ^2}+\frac{4}{21 \pi ^3}-\frac{16}{7 \pi ^4}\approx 1.11870$$ $$I_{(7)}=1+\frac{19}{48 \pi }+\frac{143}{840 \pi ^2}-\frac{145}{84 \pi ^3}+\frac{5}{7 \pi ^4}+\frac{4}{\pi ^5}\approx 1.10798$$ $$I_{(8)}=1+\frac{19}{48 \pi }-\frac{493}{1512 \pi ^2}-\frac{409}{756 \pi ^3}+\frac{33}{7 \pi ^4}-\frac{28}{9 \pi ^5}-\frac{64}{9 \pi ^6}\approx 1.10635$$
Now, I give up !
$$=\arctan(x-1)-\arctan(x-2)$$
– lab bhattacharjee Jan 24 '18 at 12:23