JEE Main 2022 — Mathematics Calculus

Let $\displaystyle I=\frac{48}{{\pi }^{4}}{\int }_{0}^{\pi }(\frac{3\pi }{2}{x}^{2}-{x}^{3})\frac{\mathrm{sin}x}{1+{\mathrm{cos}}^{2}x}dx\ldots (1)$I=π448​∫0π​(23π​x2−x3)1+cos2xsinx​dx…(1)

Using $\displaystyle {\int }_{a}^{b}f(x)dx={\int }_{a}^{b}f(a+b-x)dx$∫ab​f(x)dx=∫ab​f(a+b−x)dx

We get $\displaystyle I=\frac{48}{{\pi }^{4}}{\int }_{0}^{\pi }(\frac{3\pi }{2}{(\pi -x)}^{2}-{(\pi -x)}^{3})\frac{\mathrm{sin}(0+\pi -x)}{1+{\mathrm{cos}}^{2}(0+\pi -x)}dx$I=π448​∫0π​(23π​(π−x)2−(π−x)3)1+cos2(0+π−x)sin(0+π−x)​dx

$\displaystyle I=\frac{48}{{\pi }^{4}}{\int }_{0}^{\pi }(\frac{3{\pi }^{3}}{2}-3{\pi }^{2}x+\frac{3\pi {x}^{2}}{2}-{\pi }^{3}+{x}^{3}+3{\pi }^{2}x-3\pi {x}^{2})\frac{\mathrm{sin}x}{1+{\mathrm{cos}}^{2}x}dx$I=π448​∫0π​(23π3​−3π2x+23πx2​−π3+x3+3π2x−3πx2)1+cos2xsinx​dx ..............equation $2$2

Now adding equation (1) + equation (2)

$$2I=\frac{48}{{\pi }^{4}}{\int }_{0}^{\pi }(\frac{3{\pi }^{3}}{2}-{\pi }^{3})\frac{\mathrm{sin}x}{1+{\mathrm{cos}}^{2}x}dx$$2I=π448​∫0π​(23π3​−π3)1+cos2xsinx​dx

$$2I=\frac{48}{{\pi }^{4}}\times \frac{{\pi }^{3}}{2}{\int }_{0}^{\pi }\frac{\mathrm{sin}x}{1+{\mathrm{cos}}^{2}x}dx$$2I=π448​×2π3​∫0π​1+cos2xsinx​dx

$$2I=\frac{24}{\pi }\times 2\times {\int }_{0}^{\frac{\pi }{2}}\frac{\mathrm{sin}x}{1+{\mathrm{cos}}^{2}x}dx$$2I=π24​×2×∫02π​​1+cos2xsinx​dx

Let $\mathrm{cos}x=t\Rightarrow -\mathrm{sin}xdx=dt$cosx=t⇒−sinxdx=dt

So $\displaystyle I=\frac{24}{\pi }\times {\int }_{1}^{0}\frac{-dt}{1+{t}^{2}}$I=π24​×∫10​1+t2−dt​

$$\Rightarrow I=\frac{-24}{\pi }{[{\mathrm{tan}}^{1}t]}_{1}^{0}$$⇒I=π−24​[tan1t]10​

$$\Rightarrow I=-\frac{24}{\pi }[0-\frac{\pi }{4}]=6$$⇒I=−π24​[0−4π​]=6