If I_m,n = ${\displaystyle \underset{0}{\overset{1}{\int }}{x}^{m-1}}{\left(1-x\right)}^{n-1}dx,form,n\ge 1,and{\displaystyle \underset{0}{\overset{1}{\int }}\frac{x^{m-1}+x^{n-1}}{{\left(1+x\right)}^{m+n}}}dx=\alpha I_{m,n,\alpha \in R},$ then a equals…....

${\displaystyle \int \frac{\left(sinx-cosx\right)si{n}^{2}x}{tanx\left(si{n}^{3}x+co{s}^{3}x\right)}dx}$

${\displaystyle \int \frac{\left(sinx-cosx\right)sinxcosx}{si{n}^{3}x+co{s}^{3}x}dx}$ , put sin³x + cos³x = t(3 sin²x×cosx – 3cos²xsinx) dx = dt

-> $\frac{1}{3}{\displaystyle \int \frac{dt}{t}}$

$=\frac{lnt}{3}+c$

$=\frac{ln\left|si{n}^{3}x+co{s}^{3}x\right|}{3}+c$

$l={\displaystyle \underset{1}{\overset{3}{\int }}\left[x^2-2x+1-3\right]dx=}{\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2-3\right]dx={\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2\right]dx-3{\displaystyle \underset{1}{\overset{3}{\int }}dx}}}$ ...........(A)

$l_1={\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2\right]dxPut{\left(x-1\right)}^2=t}$

$l_1=\frac{1}{2}\left[0{\displaystyle \underset{0}{\overset{1}{\int }}\frac{dt}{\sqrt[]{t}}}{\displaystyle \underset{1}{\overset{2}{\int }}\frac{dt}{\sqrt[]{t}}}+2{\displaystyle \underset{2}{\overset{3}{\int }}\frac{dt}{\sqrt[]{t}}}+3{\displaystyle \underset{3}{\overset{4}{\int }}\frac{dt}{\sqrt[]{t}}}\right]=\frac{1}{2}\left\{{\left|\frac{t^{\frac{-1}{2}+1}}{-\frac{1}{2}+1}\right|}_1^2{\left|+2\frac{t^{\frac{-1}{2}+1}}{-\frac{1}{2}+1}\right|}_2^3{+3\frac{t^{\frac{-1}{2}+1}}{-\frac{1}{2}+1}]}_3^4\right\}$

Hence from (A)

= $5-\sqrt[]{2}-\sqrt[]{3}-\sqrt[]{3}-6=-1-\sqrt[]{2}-\sqrt[]{3}$

2^nd method           

${\displaystyle \underset{1}{\overset{3}{\int }}\left[{\left(x-1\right)}^2\right]dx-6..............\left(A\right)}$

From (A), $l=5-\sqrt[]{2}-\sqrt[]{3}-6=-1-\sqrt[]{2}-\sqrt[]{3}$

    $l={\displaystyle \underset{0}{\overset{2}{\int }}f\left(x\right)dx={\left[xf\left(x\right)\right]}_0^2-{\displaystyle \underset{0}{\overset{2}{\int }}xf\text{'}\left(x\right)dx=2e^2-{\displaystyle \underset{0}{\overset{2}{\int }}xf\text{'}\left(x\right)dx}}}$    ........(A)

Put $l_1={\displaystyle \underset{0}{\overset{2}{\int }}xf\text{'}\left(x\right)dx}$               .........(i)

Using properties ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx={\displaystyle \underset{a}{\overset{b}{\int }}f\left(a+b-x\right)dx}}$

$l_1={\displaystyle \underset{0}{\overset{2}{\int }}\left(2-x\right)f\text{'}\left(2-x\right)dx={\displaystyle \underset{0}{\overset{2}{\int }}\left(2-x\right)f\text{'}\left(x\right)dx}}$ ...........(ii)

Adding (i) and (ii) we get

$2l_1=2{\displaystyle \underset{0}{\overset{2}{\int }}f\text{'}\left(x\right)dx\Rightarrow l_1={\left[f\left(x\right)\right]}_0^2}$

f(2) – f(0) = e² – 1

From (A) l = 2e² – e² + 1 = e² + 1

$l=\frac{48}{{\pi }^4}{\displaystyle {\int }_0^{\pi }\left[{\left(\frac{\pi }{2}-x\right)}^3-\frac{3{\pi }^2}{4}\left(\frac{\pi }{2}-x\right)+\frac{{\pi }^3}{4}\right]\frac{sinxdx}{1+co{s}^{2}x}}$

Using ${\displaystyle {\int }_a^{b}f\left(x\right)dx={\displaystyle {\int }_a^{b}f\left(a+b-x\right)dx}}$

we get $l=\frac{48}{{\pi }^4}{\displaystyle {\int }_0^{\pi }\left[-{\left(\frac{\pi }{2}-x\right)}^3+\frac{3{\pi }^2}{4}\left(\frac{\pi }{2}-x\right)+\frac{{\pi }^3}{4}\right]\frac{sinxdx}{1+co{s}^{2}x}}$

Adding these two equations, we get

$\Rightarrow l=\frac{12}{\pi }{\left[-ta{n}^{-1}\left(cosx\right)\right]}_0^{\pi }=\frac{12}{\pi }.\frac{\pi }{2}=6$

$\therefore {\displaystyle \underset{-6}{\overset{0}{\int }}f\left(x\right)dx=2\times \frac{1}{2}\left(2+5\right)\times 3=21}$