JEE Main 2024 — Mathematics Calculus

Let, $\displaystyle I=\int \frac{{\mathrm{sin}}^{\frac{3}{2}}x+{\mathrm{cos}}^{\frac{3}{2}}x}{\sqrt{{\mathrm{sin}}^{3}x{\mathrm{cos}}^{3}x\mathrm{sin}(x-\theta )}}dx$I=∫sin3xcos3xsin(x−θ)​sin23​x+cos23​x​dx

$$\Rightarrow I=\int \frac{{\mathrm{sin}}^{\frac{3}{2}}x+{\mathrm{cos}}^{\frac{3}{2}}x}{\sqrt{{\mathrm{sin}}^{3}x{\mathrm{cos}}^{3}x(\mathrm{sin}x\mathrm{cos}\theta -\mathrm{cos}x\mathrm{sin}\theta )}}dx$$⇒I=∫sin3xcos3x(sinxcosθ−cosxsinθ)​sin23​x+cos23​x​dx

$$\Rightarrow I=\int \frac{{\mathrm{sin}}^{\frac{3}{2}}x+{\mathrm{cos}}^{\frac{3}{2}}x}{\sqrt{{\mathrm{sin}}^{3}x{\mathrm{cos}}^{4}x(\mathrm{tan}x\mathrm{cos}\theta -\mathrm{sin}\theta )}}dx$$⇒I=∫sin3xcos4x(tanxcosθ−sinθ)​sin23​x+cos23​x​dx

$$\Rightarrow I=\int \frac{{\mathrm{sin}}^{\frac{3}{2}}x}{{\mathrm{sin}}^{\frac{3}{2}}x{\mathrm{cos}}^{2}x\sqrt{\mathrm{tan}x\mathrm{cos}\theta -\mathrm{sin}\theta }}dx+\int \frac{{\mathrm{cos}}^{\frac{3}{2}}x}{{\mathrm{sin}}^{2}x{\mathrm{cos}}^{\frac{3}{2}}x\sqrt{\mathrm{cos}\theta -\mathrm{cot}x\mathrm{sin}\theta }}dx$$⇒I=∫sin23​xcos2xtanxcosθ−sinθ​sin23​x​dx+∫sin2xcos23​xcosθ−cotxsinθ​cos23​x​dx

$$\Rightarrow I=\int \frac{1}{{\mathrm{cos}}^{2}x\sqrt{\mathrm{tan}x\mathrm{cos}\theta -\mathrm{sin}\theta }}dx+\int \frac{1}{{\mathrm{sin}}^{2}x\sqrt{\mathrm{cos}\theta -\mathrm{cot}x\mathrm{sin}\theta }}dx$$⇒I=∫cos2xtanxcosθ−sinθ​1​dx+∫sin2xcosθ−cotxsinθ​1​dx

$$\Rightarrow I=\int \frac{{\mathrm{sec}}^{2}x}{\sqrt{\mathrm{tan}x\mathrm{cos}\theta -\mathrm{sin}\theta }}dx+\int \frac{{cosec}^{2}x}{\sqrt{\mathrm{cos}\theta -\mathrm{cot}x\mathrm{sin}\theta }}dx$$⇒I=∫tanxcosθ−sinθ​sec2x​dx+∫cosθ−cotxsinθ​cosec2x​dx

So, $I={I}_{1}+{I}_{2}....{\mathrm{let}}$I=I1​+I2​....let

For ${I}_{1}$I1​, let $\mathrm{tan}x\mathrm{cos}\theta -\mathrm{sin}\theta ={t}^{2}$tanxcosθ−sinθ=t2

$$\Rightarrow \mathrm{cos}\theta \times {\mathrm{sec}}^{2}xdx=2tdt$$⇒cosθ×sec2xdx=2tdt

$$\Rightarrow {\mathrm{sec}}^{2}xdx=\frac{2tdt}{\mathrm{cos}\theta }$$⇒sec2xdx=cosθ2tdt​

For ${I}_{2}$I2​, let $\mathrm{cos}\theta -\mathrm{cot}x\mathrm{sin}\theta ={z}^{2}$cosθ−cotxsinθ=z2

$$\Rightarrow {\mathrm{cosec}}^{2}x\times \mathrm{sin}\theta dx=2zdz$$⇒cosec2x×sinθdx=2zdz

$$\Rightarrow {cosec}^{2}xdx=\frac{2zdz}{\mathrm{sin}\theta }$$⇒cosec2xdx=sinθ2zdz​

$$\Rightarrow I=\int \frac{2tdt}{\mathrm{cos}\theta \times t}+\int \frac{2zdz}{\mathrm{sin}\theta \times z}$$⇒I=∫cosθ×t2tdt​+∫sinθ×z2zdz​

$$\Rightarrow I=2[(\mathrm{sec}\theta )\int (1)dt+\mathrm{cosec}\theta \int (1)dz]$$⇒I=2[(secθ)∫(1)dt+cosecθ∫(1)dz]

$$\Rightarrow I=2\mathrm{sec}\theta \sqrt{\mathrm{tan}x\mathrm{cos}\theta -\mathrm{sin}\theta }+2cosec\theta \sqrt{\mathrm{cos}\theta -\mathrm{cot}x\mathrm{sin}\theta }$$⇒I=2secθtanxcosθ−sinθ​+2cosecθcosθ−cotxsinθ​

$$\Rightarrow A\times B=4\mathrm{sec}\theta \mathrm{cosec}\theta$$⇒A×B=4secθcosecθ

$$\Rightarrow A\times B=\frac{2}{2}\times \frac{4}{\mathrm{sin}\theta \mathrm{cos}\theta }$$⇒A×B=22​×sinθcosθ4​

$$\Rightarrow A\times B=8cosec2\theta$$⇒A×B=8cosec2θ