JEE Main 2021 — Mathematics Calculus

Let $\displaystyle I={\int }_{0}^{\pi }{\mathrm{sin}}^{3}x{e}^{-{\mathrm{sin}}^{2}x}dx$I=∫0π​sin3xe−sin2xdx

$$\Rightarrow I=2{\int }_{0}^{\frac{\pi }{2}}{\mathrm{sin}}^{3}x{e}^{-{\mathrm{sin}}^{2}x}dx{\text{applying}{\int }_{0}^{a}f(x)=2{\int }_{0}^{\frac{a}{2}}f(x)\text{when}f(x)=f(a-x)}$$⇒I=2∫02π​​sin3xe−sin2xdxapplying∫0a​f(x)=2∫02a​​f(x)whenf(x)=f(a−x)

$$=2{\int }_{0}^{\frac{\pi }{2}}\mathrm{sin}x(1-{\mathrm{cos}}^{2}x){e}^{-{\mathrm{sin}}^{2}x}dx$$=2∫02π​​sinx(1−cos2x)e−sin2xdx

$$=2{\int }_{0}^{\frac{\pi }{2}}\mathrm{sin}x{e}^{-{\mathrm{sin}}^{2}x}dx-{\int }_{0}^{\frac{\pi }{2}}2\mathrm{sin}x\mathrm{cos}x\cdot \mathrm{cos}x{e}^{-{\mathrm{sin}}^{2}x}dx$$=2∫02π​​sinxe−sin2xdx−∫02π​​2sinxcosx⋅cosxe−sin2xdx

$$=2{\int }_{0}^{\frac{\pi }{2}}\mathrm{sin}x{e}^{-{\mathrm{sin}}^{2}x}dx+{\int }_{0}^{\frac{\pi }{2}}\underset{I}{\underset{⏟}{\mathrm{cos}x}}\cdot \underset{\mathrm{II}}{\underset{⏟}{{e}^{-{\mathrm{sin}}^{2}x}(-\mathrm{sin}2x)}}dx$$=2∫02π​​sinxe−sin2xdx+∫02π​​I⏟cosx​​⋅II⏟e−sin2x(−sin2x)​​dx

$$=2{\int }_{0}^{\frac{\pi }{2}}\mathrm{sin}x{e}^{-{\mathrm{sin}}^{2}x}dx+{[\mathrm{cos}x{e}^{-{\mathrm{sin}}^{2}x}]}_{0}^{\frac{\pi }{2}}+{\int }_{0}^{\frac{\pi }{2}}\mathrm{sin}x{e}^{-{\mathrm{sin}}^{2}x}dx$$=2∫02π​​sinxe−sin2xdx+[cosxe−sin2x]02π​​+∫02π​​sinxe−sin2xdx

$$=3{\int }_{0}^{\frac{\pi }{2}}\mathrm{sin}x{e}^{-{\mathrm{sin}}^{2}x}dx-1$$=3∫02π​​sinxe−sin2xdx−1

$$=\frac{3}{2}{\int }_{-1}^{0}\frac{{e}^{\alpha }d\alpha }{\sqrt{1+\alpha }}-1(\mathrm{Put}-{\mathrm{sin}}^{2}x=\alpha )$$=23​∫−10​1+α​eαdα​−1(Put−sin2x=α)

$$=\frac{3}{2e}{\int }_{0}^{1}\frac{{e}^{x}}{\sqrt{x}}dx-1(\mathrm{Put}1+\alpha =x)$$=2e3​∫01​x​ex​dx−1(Put1+α=x)

$$=\frac{3}{2e}{\int }_{0}^{1}{e}^{x}\frac{1}{\sqrt{x}}dx-1=\frac{3}{2e}({(2\sqrt{x}{e}^{x})}_{0}^{1}-{\int }_{0}^{1}2\sqrt{x}{e}^{x}dx)-1$$=2e3​∫01​exx​1​dx−1=2e3​((2x​ex)01​−∫01​2x​exdx)−1

$$=\frac{3}{2e}((2e)-{\int }_{0}^{1}2\sqrt{x}{e}^{x}dx)-1$$=2e3​((2e)−∫01​2x​exdx)−1

$$=2-\frac{3}{e}{\int }_{0}^{1}{e}^{x}\sqrt{x}dx$$=2−e3​∫01​exx​dx

Hence, $\alpha +\beta =5.$α+β=5.