JEE Main 2023 — Mathematics Calculus

Given that $\displaystyle I(x)=\int {x}^{2}(\frac{x{\mathrm{sec}}^{2}x+\mathrm{tan}x}{(x\mathrm{tan}x+1{)}^{2}})dx$I(x)=∫x2((xtanx+1)2xsec2x+tanx​)dx

Apply Integration by parts.

$$\int f(x)g(x)dx=f(x)\int g(x)dx-\int (f'(x)\int g(x)dx)dx$$∫f(x)g(x)dx=f(x)∫g(x)dx−∫(f′(x)∫g(x)dx)dx

$$={x}^{2}\int (\frac{x{\mathrm{sec}}^{2}x+\mathrm{tan}x}{(x\mathrm{tan}x+1{)}^{2}})dx-\int (\frac{d{x}^{2}}{dx}\int (\frac{x{\mathrm{sec}}^{2}x+\mathrm{tan}x}{(x\mathrm{tan}x+1{)}^{2}})dx)dx$$=x2∫((xtanx+1)2xsec2x+tanx​)dx−∫(dxdx2​∫((xtanx+1)2xsec2x+tanx​)dx)dx

Let $x\mathrm{tan}x+1=p$xtanx+1=p

$$\Rightarrow (x{\mathrm{sec}}^{2}x+\mathrm{tan}x)dp=dx$$⇒(xsec2x+tanx)dp=dx

$$={x}^{2}\int \frac{dp}{{p}^{2}}-\int (2x\int \frac{dp}{{p}^{2}})dx$$=x2∫p2dp​−∫(2x∫p2dp​)dx

$$=\frac{-{x}^{2}}{(x\mathrm{tan}x+1)}+\int \frac{2x}{x\mathrm{tan}x+1}dx$$=(xtanx+1)−x2​+∫xtanx+12x​dx

Let $\displaystyle {I}_{1}=2\int \frac{x}{x\mathrm{tan}x+1}dx$I1​=2∫xtanx+1x​dx

$$=2\int \frac{x\mathrm{cos}x}{x\mathrm{sin}x+\mathrm{cos}x}dx$$=2∫xsinx+cosxxcosx​dx

Let $x\mathrm{sin}x+\mathrm{cos}x=t$xsinx+cosx=t

$$\Rightarrow (x\mathrm{cos}x+\mathrm{sin}x-\mathrm{sin}x)dx=dt$$⇒(xcosx+sinx−sinx)dx=dt

$$\Rightarrow (x\mathrm{cos}x)dx=dt$$⇒(xcosx)dx=dt

$$=2\int \frac{dt}{t}=2\mathrm{log}t+c$$=2∫tdt​=2logt+c

$$=2\mathrm{log}|x\mathrm{sin}x+\mathrm{cos}x|+c$$=2log∣xsinx+cosx∣+c

$$\therefore \int \frac{{x}^{2}(x{\mathrm{sec}}^{2}x+\mathrm{tan}x)}{(x\mathrm{tan}x+1{)}^{2}}dx$$∴∫(xtanx+1)2x2(xsec2x+tanx)​dx

$$=\frac{-{x}^{2}}{x\mathrm{tan}x+1}+2\mathrm{log}|x\mathrm{sin}x+\mathrm{cos}x|+c$$=xtanx+1−x2​+2log∣xsinx+cosx∣+c

But $I(0)=0$I(0)=0

$$\Rightarrow c=0$$⇒c=0

Also,$I(\frac{\pi }{4})=-\frac{{(\frac{\pi }{4})}^{2}}{\frac{\pi }{4}\times 1+1}+2\mathrm{log}|\frac{1}{\sqrt{2}}(\frac{\pi }{4})+\frac{1}{\sqrt{2}}|$I(4π​)=−4π​×1+1(4π​)2​+2log∣2​1​(4π​)+2​1​∣

$$={\mathrm{log}}_{e}\frac{(\pi +4{)}^{2}}{32}-\frac{{\pi }^{2}}{4(\pi +4)}$$=loge​32(π+4)2​−4(π+4)π2​

Hence, this is the correct option.