Question

Evaluate : ${\int }_0^{\pi }\frac{xtanx}{secx+tanx}dx.$ OR Evaluate : ${\int }_1^4\text{{|x-1|+|x-2|+|x-4|}dx.}$

Answer 1

Let I= ${\int }_0^{\pi }\frac{xtanx}{secx+tanx}dx....\left(i\right)$ $\Rightarrow {\int }_0^{\pi }\frac{(\pi -x)tan(\pi -x)}{sec(\pi -x)tan(\pi -x)}dx={\int }_0^{\pi }\frac{(\pi -x)tanx}{secx+tanx}dx...\left(ii\right)$

Adding (i) and (ii),2I=$\pi {\int }_0^{\pi }\frac{tanx}{secx+tanx}dx$ $\Rightarrow 2I=\pi {\int }_0^{\pi }\frac{tanx}{secx+tanx}\times \frac{secx-tanx}{secx-tanx}dx$

$\Rightarrow 2I=\pi {\int }_0^{\pi }(secxtanx-ta{n}^{2}x)dx\Rightarrow 2I=\pi {\int }_0^{\pi }(secxtanx-se{c}^{2}x+1)dx$

$\Rightarrow 2I=\pi {[secx-tanx+x]}_0^{\pi }\Rightarrow 2I=\pi \left\{\right[sec\pi -tan\pi +\pi ]-[sec0-tan0+0\left]\right\}$

$\Rightarrow 2I=\pi \left\{\right[-1-0+\pi ]-[1-0+0\left]\right\}\therefore I=\frac{\pi (\pi -2)}{2}.$

OR Let I=${\int }_1^4\{|x-1|+|x-2|+|x-4|dx\}$

$\Rightarrow I={\int }_1^4|x-1|dx+={\int }_1^4|x-2|dx+{\int }_1^4|x-4|dx$

$\Rightarrow ={\int }_1^4(x-1)dx+{\int }_1^2|x-2|dx+{\int }_2^4|x-2|dx-{\int }_1^4(x-4)dx$

$\Rightarrow I=\frac{1}{2}{\left[\right(x-1{)}^2]}_1^4-{\int }_1^2(x-2)dx+{\int }_2^4(x-2)dx-\frac{1}{2}{\left[\right(x-4{)}^2]}_1^4$

$\Rightarrow I=\frac{1}{2}[9-0]-\frac{1}{2}{\left[\right(x-2{)}^2]}_1^2+\frac{1}{2}{\left[\right(x-2{)}^2]}_2^4-\frac{1}{2}[0-9]$

$\Rightarrow I=\frac{9}{2}-\frac{1}{2}[0-1]+\frac{1}{2}[4-0]+\frac{9}{2}=11+\frac{1}{2}=\frac{23}{2}$