JEE Main 2024 — Mathematics Calculus

Let, $\displaystyle I={\int }_{0}^{\pi }\frac{{x}^{2}\mathrm{sin}x\mathrm{cos}x}{{\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x}dx$I=∫0π​sin4x+cos4xx2sinxcosx​dx

Now using the property $\displaystyle {\int }_{0}^{2a}f(x)dx={\int }_{0}^{a}f(x)dx+{\int }_{0}^{a}f(2a-x)dx$∫02a​f(x)dx=∫0a​f(x)dx+∫0a​f(2a−x)dx we get,

$$\Rightarrow I={\int }_{0}^{\frac{\pi }{2}}\frac{\mathrm{sin}x\mathrm{cos}x}{{\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x}({x}^{2}-{(\pi -x)}^{2})dx$$⇒I=∫02π​​sin4x+cos4xsinxcosx​(x2−(π−x)2)dx

$$\Rightarrow I={\int }_{0}^{\frac{\pi }{2}}\frac{\mathrm{sin}x\mathrm{cos}x}{{\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x}(2\pi x-{\pi }^{2})dx$$⇒I=∫02π​​sin4x+cos4xsinxcosx​(2πx−π2)dx

$$\Rightarrow I=2\pi \underset{{I}_{1}}{\underset{⏟}{{\int }_{0}^{\frac{\pi }{2}}\frac{x\cdot \mathrm{sin}x\mathrm{cos}x}{{\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x}dx}}-{\pi }^{2}{\int }_{0}^{\frac{\pi }{2}}\frac{\mathrm{sin}x\mathrm{cos}x}{{\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x}dx$$⇒I=2πI1​⏟∫02π​​sin4x+cos4xx⋅sinxcosx​dx​​−π2∫02π​​sin4x+cos4xsinxcosx​dx

Now, solving $\displaystyle {I}_{1}={\int }_{0}^{\frac{\pi }{2}}\frac{x\cdot \mathrm{sin}x\mathrm{cos}x}{{\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x}dx....(1)$I1​=∫02π​​sin4x+cos4xx⋅sinxcosx​dx....(1)

$$\Rightarrow {I}_{1}={\int }_{0}^{\frac{\pi }{2}}\frac{(\frac{\pi }{2}-x)\cdot \mathrm{sin}x\mathrm{cos}x}{{\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x}dx....(2)$$⇒I1​=∫02π​​sin4x+cos4x(2π​−x)⋅sinxcosx​dx....(2)

Adding both equations we get,

$$\Rightarrow 2{I}_{1}=\frac{\pi }{2}{\int }_{0}^{\frac{\pi }{2}}\frac{\mathrm{sin}x\mathrm{cos}x}{{\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x}dx$$⇒2I1​=2π​∫02π​​sin4x+cos4xsinxcosx​dx

$$\Rightarrow {I}_{1}=\frac{\pi }{4}{\int }_{0}^{\frac{\pi }{2}}\frac{\mathrm{sin}x\mathrm{cos}x}{{\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x}dx$$⇒I1​=4π​∫02π​​sin4x+cos4xsinxcosx​dx

Now, putting the value in $I$I we get,

$$\Rightarrow I=2\pi \cdot \frac{\pi }{4}{\int }_{0}^{\frac{\pi }{2}}\frac{x\cdot \mathrm{sin}x\mathrm{cos}x}{{\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x}dx-{\pi }^{2}{\int }_{0}^{\frac{\pi }{2}}\frac{\mathrm{sin}x\mathrm{cos}x}{{\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x}dx$$⇒I=2π⋅4π​∫02π​​sin4x+cos4xx⋅sinxcosx​dx−π2∫02π​​sin4x+cos4xsinxcosx​dx

$$\Rightarrow I=-\frac{{\pi }^{2}}{2}{\int }_{0}^{\frac{\pi }{2}}\frac{\mathrm{sin}x\mathrm{cos}x}{{\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x}dx$$⇒I=−2π2​∫02π​​sin4x+cos4xsinxcosx​dx

$$\Rightarrow I=-\frac{{\pi }^{2}}{2}{\int }_{0}^{\frac{\pi }{2}}\frac{\mathrm{sin}x\mathrm{cos}x}{1-2{\mathrm{sin}}^{2}x{\mathrm{cos}}^{2}x}dx$$⇒I=−2π2​∫02π​​1−2sin2xcos2xsinxcosx​dx

$$\Rightarrow I=-\frac{{\pi }^{2}}{2}{\int }_{0}^{\frac{\pi }{2}}\frac{\frac{1}{2}\mathrm{sin}2x}{1-\frac{1}{2}{\mathrm{sin}}^{2}2x}dx$$⇒I=−2π2​∫02π​​1−21​sin22x21​sin2x​dx

$$\Rightarrow I=-\frac{{\pi }^{2}}{2}{\int }_{0}^{\frac{\pi }{2}}\frac{\mathrm{sin}2x}{2-{\mathrm{sin}}^{2}2x}dx$$⇒I=−2π2​∫02π​​2−sin22xsin2x​dx

$$\Rightarrow I=-\frac{{\pi }^{2}}{2}{\int }_{0}^{\frac{\pi }{2}}\frac{\mathrm{sin}2x}{1+{\mathrm{cos}}^{2}2x}dx$$⇒I=−2π2​∫02π​​1+cos22xsin2x​dx

Now, let $\mathrm{cos}2x=t\Rightarrow -2\mathrm{sin}2xdx=dt$cos2x=t⇒−2sin2xdx=dt

$$\Rightarrow I=-\frac{{\pi }^{2}}{2}{\int }_{1}^{-1}\frac{-\frac{1}{2}}{1+{t}^{2}}dt$$⇒I=−2π2​∫1−1​1+t2−21​​dt

$$\Rightarrow I=-\frac{{\pi }^{2}}{4}{\int }_{-1}^{1}\frac{1}{1+{t}^{2}}dt$$⇒I=−4π2​∫−11​1+t21​dt

$$\Rightarrow I=-\frac{{\pi }^{2}}{4}\cdot \frac{\pi }{2}=-\frac{{\pi }^{3}}{8}$$⇒I=−4π2​⋅2π​=−8π3​

Hence, $\displaystyle \frac{120}{{\pi }^{3}}|{\int }_{0}^{\pi }\frac{{x}^{2}\mathrm{sin}x\mathrm{cos}x}{{\mathrm{sin}}^{4}x+{\mathrm{cos}}^{4}x}dx|=\frac{120}{{\pi }^{3}}\times \frac{{\pi }^{3}}{8}=15$π3120​∣∫0π​sin4x+cos4xx2sinxcosx​dx∣=π3120​×8π3​=15