JEE Main 2023 — Mathematics Calculus

Let, $\displaystyle I={\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{2+3\mathrm{sin}x}{\mathrm{sin}x(1+\mathrm{cos}x)}dx$I=∫3π​2π​​sinx(1+cosx)2+3sinx​dx

Now rearranging the above integral we get,

$$I={\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{2}{\mathrm{sin}x(1+\mathrm{cos}x)}dx+{\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{3}{(1+\mathrm{cos}x)}dx$$I=∫3π​2π​​sinx(1+cosx)2​dx+∫3π​2π​​(1+cosx)3​dx

$$\Rightarrow I={\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{2\mathrm{sin}x}{{\mathrm{sin}}^{2}x(1+\mathrm{cos}x)}dx+{\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{3}{(2{\mathrm{cos}}^{2}\frac{x}{2})}dx$$⇒I=∫3π​2π​​sin2x(1+cosx)2sinx​dx+∫3π​2π​​(2cos22x​)3​dx

$$\Rightarrow I=\underset{{I}_{1}}{\underset{⏟}{{\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{2\mathrm{sin}x}{{\mathrm{sin}}^{2}x(1+\mathrm{cos}x)}dx}}+\frac{3}{2}\underset{{I}_{2}}{\underset{⏟}{{\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}{\mathrm{sec}}^{2}(\frac{x}{2})dx}}$$⇒I=I1​⏟∫3π​2π​​sin2x(1+cosx)2sinx​dx​​+23​I2​⏟∫3π​2π​​sec2(2x​)dx​​

$$\Rightarrow I=\underset{{I}_{1}}{\underset{⏟}{{\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{2\mathrm{sin}x}{{\mathrm{sin}}^{2}x(1+\mathrm{cos}x)}dx}}+\frac{3}{2}\times 2\underset{{I}_{2}}{\underset{⏟}{{[\mathrm{tan}\frac{x}{2}]}_{\frac{\pi }{3}}^{\frac{\pi }{2}}}}$$⇒I=I1​⏟∫3π​2π​​sin2x(1+cosx)2sinx​dx​​+23​×2I2​⏟[tan2x​]3π​2π​​​​

$$\Rightarrow I=\underset{{I}_{1}}{\underset{⏟}{{\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{2\mathrm{sin}x}{{\mathrm{sin}}^{2}x(1+\mathrm{cos}x)}dx}}+3(1-\frac{1}{\sqrt{3}})$$⇒I=I1​⏟∫3π​2π​​sin2x(1+cosx)2sinx​dx​​+3(1−3​1​)

Now solving ${I}_{1}$I1​ we get,

$${I}_{1}={\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{2\mathrm{sin}x}{{\mathrm{sin}}^{2}x(1+\mathrm{cos}x)}dx$$I1​=∫3π​2π​​sin2x(1+cosx)2sinx​dx

$$\Rightarrow {I}_{1}=2{\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{1}{\mathrm{sin}x(1+\mathrm{cos}x)}dx$$⇒I1​=2∫3π​2π​​sinx(1+cosx)1​dx

$$\Rightarrow {I}_{1}=2{\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{1}{(\frac{2\mathrm{tan}(\frac{x}{2})}{1+{\mathrm{tan}}^{2}(\frac{x}{2})})(1+\frac{1-\mathrm{tan}(\frac{x}{2})}{1+{\mathrm{tan}}^{2}(\frac{x}{2})})}dx$$⇒I1​=2∫3π​2π​​(1+tan2(2x​)2tan(2x​)​)(1+1+tan2(2x​)1−tan(2x​)​)1​dx

$$\Rightarrow {I}_{1}=\frac{1}{2}{\int }_{\frac{\pi }{3}}^{\frac{\pi }{2}}\frac{[1+{\mathrm{tan}}^{2}(\frac{x}{2})]{\mathrm{sec}}^{2}(\frac{x}{2})}{\mathrm{tan}(\frac{x}{2})}dx$$⇒I1​=21​∫3π​2π​​tan(2x​)[1+tan2(2x​)]sec2(2x​)​dx

Put $\mathrm{tan}(\frac{x}{2})=v\Rightarrow \frac{1}{2}{\mathrm{sec}}^{2}(\frac{x}{2})dx=dv$tan(2x​)=v⇒21​sec2(2x​)dx=dv

$$\Rightarrow {I}_{1}={\int }_{\frac{1}{\sqrt{3}}}^{1}\frac{(1+{v}^{2})dv}{v}$$⇒I1​=∫3​1​1​v(1+v2)dv​

$$\Rightarrow {I}_{1}={[{\mathrm{log}}_{e}(v)+\frac{{v}^{2}}{2}]}_{\frac{1}{\sqrt{3}}}^{1}$$⇒I1​=[loge​(v)+2v2​]3​1​1​

$$\Rightarrow {I}_{1}={\mathrm{log}}_{e}(1)-{\mathrm{log}}_{e}(\frac{1}{\sqrt{3}})+\frac{1}{2}(1-\frac{1}{3})$$⇒I1​=loge​(1)−loge​(3​1​)+21​(1−31​)

$$\Rightarrow {I}_{1}={\mathrm{log}}_{e}(\frac{1}{\sqrt{3}})+\frac{1}{3}$$⇒I1​=loge​(3​1​)+31​

So,

$$I=\frac{1}{3}+{\mathrm{log}}_{e}\sqrt{3}+3(1-\frac{1}{\sqrt{3}})$$I=31​+loge​3​+3(1−3​1​)

$$\Rightarrow I=\frac{10}{3}-\sqrt{3}+{\mathrm{log}}_{e}\sqrt{3}$$⇒I=310​−3​+loge​3​