If the two lines ${\mathcal{l}}_1:\frac{x-2}{3}=\frac{y+1}{-2},z=2and{\mathcal{l}}_2:\frac{x-1}{1}=\frac{2y+3}{\alpha }=\frac{z+5}{2}$ are perpendicular, then an angle between the lines ${\mathcal{l}}_{2}and{\mathcal{l}}_3:\frac{1-x}{3}=\frac{2y-1}{-4}=\frac{z}{4}is:$

$\underset{h\to 0}{lim}\frac{{\displaystyle {\int }_{\left(\frac{\pi }{2}-h\right)}^{{\left(\frac{\pi }{2}\right)}^3}cos\left(t^{1/3}\right)dt}}{h^2}$

$=\underset{h\to 0}{lim}\frac{0+3{\left(\frac{\pi }{2}-h\right)}^{2}cos\left(\frac{\pi }{2}-h\right)}{2h}$

$=\underset{h\to 0}{lim}\frac{3{\left(\frac{\pi }{2}-h\right)}^{2}sinh}{2h}$

$=\frac{3{\pi }^2}{8}$

${\displaystyle \underset{-\frac{\pi }{2}}{\overset{\frac{\pi }{2}}{\int }}\frac{8\sqrt[]{2}cosx}{\left(1+e^{sinx}\right)\left(1+si{n}^{4}x\right)}}$ dx

$={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\left\{\left(\frac{8\sqrt[]{2}cosx}{\left(1+e^{sinx}\right)\left(1+si{n}^{4}x\right)}+\frac{8\sqrt[]{2}cosx}{\left(1+e^{-sinx}\right)\left(1+si{n}^{4}x\right)}\right)\right\}dx}$            

$=8\sqrt[]{2}{\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{cosx}{1+si{n}^{4}x}dx}$            

Let sin x = t

$I=8\sqrt[]{2}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{1+t^4}}$            

$=4\sqrt[]{2}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{\left(1+\frac{1}{t^2}\right)-\left(1-\frac{1}{t^2}\right)}{t^2+\frac{1}{t^2}}dt}$       

$=4\sqrt[]{2}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{\left(1+\frac{1}{t^2}\right)dt}{{\left(t-\frac{1}{t}\right)}^2+2}}-4\sqrt[]{2}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{\left(1-\frac{1}{t^2}\right)dt}{{\left(t+\frac{1}{t}\right)}^2-2}}$            

$=4\sqrt[]{2}\cdot \frac{1}{\sqrt[]{2}}{\left(ta{n}^{-1}\frac{t-\frac{1}{t}}{\sqrt[]{2}}\right)}_0^1-4\sqrt[]{2}\cdot \frac{1}{2\sqrt[]{2}}{\left[log\left|\frac{t+\frac{1}{t}-\sqrt[]{2}}{t+\frac{1}{t}+\sqrt[]{2}}\right|\right]}_0^1$         

$=2\pi -2log\left|\frac{2-\sqrt[]{2}}{2+\sqrt[]{2}}\right|$       &nbs

${\displaystyle \underset{-a}{\overset{a}{\int }}\left(\left|x\right|+\left|x-2\right|\right)dx=22,a>2}$

${\displaystyle \underset{-a}{\overset{0}{\int }}\left(-2x+2\right)dx+{\displaystyle \underset{0}{\overset{2}{\int }}\left(x-x+2\right)dx+{\displaystyle \underset{2}{\overset{a}{\int }}\left(2x-2\right)dx=22}}}$

$\Rightarrow 2a^2+2=20\Rightarrow a^2=9\Rightarrow a=3$

$\therefore {\displaystyle \underset{3}{\overset{-3}{\int }}\left(x+\left[x\right]\right)dx=-{\displaystyle \underset{-3}{\overset{3}{\int }}\left(2x-\left\{x\right\}\right)dx=}}-{\displaystyle \underset{-3}{\overset{3}{\int }}2xdx+6}{{\displaystyle \underset{0}{\overset{1}{\int }}xdx=6.\frac{x^2}{2}}|}_0^1=3$

Let sin = t

${\displaystyle \frac{}{}}$ ${\displaystyle \int \frac{sin\theta \left(2sin\theta .cos\theta \right)\left(si{n}^6\theta +si{n}^4\theta +si{n}^2\theta \right)\sqrt[]{2si{n}^4\theta +3si{n}^2\theta +6}}{2si{n}^2\theta }}$  d

sin = t

cos . d = dt

${\displaystyle \int \frac{u^{1/2}}{12}}du=\frac{u^{3/2}}{18}+C=\frac{{\left(2t^6+3t^4+6t^2\right)}^{3/2}}{18}+C=\frac{{\left(2si{n}^6\theta +3si{n}^4\theta +6si{n}^2\theta \right)}^{3/2}}{18}+C$

$x^4-2x^3+2x-1={\left(x-1\right)}^2\left(x^2-1\right)$

$sin\pi x=sin(\pi \left(1-x\right)$

$=-sin\left(sin\pi \left(x-1\right)\right)$

$\underset{x\to 1}{lim}\frac{\left(x^2-1\right)\cdot si{n}^2\pi x}{\left(x^2-1\right){\left(x-1\right)}^2}=\underset{x\to 1}{lim}\frac{si{n}^2\left(\pi \left(x-1\right)\right)}{{\left(x-1\right)}^2}$

$=\underset{x\to 1}{lim}\frac{si{n}^2\left(\pi \left(x-1\right)\right)}{{\left(\pi \left(x-1\right)\right)}^2}\cdot {\pi }^2$

=2