JEE Main 2023 — Definite Integration Question with Solution

Let, $$I=\frac{8}{\pi }\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{{(\cos x)}^{2023}}{{(\sin x)}^{2023}+{(\cos x)}^{2023}}dx$$ ..... (1)

Using formula,

$$\underset{a}{\overset{b}{\int }}f(x)dx=\underset{a}{\overset{b}{\int }}f(a+b-x)dx$$

$$I=\frac{8}{\pi }\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{{\left[\cos \left(\frac{\pi }{2}-x\right)\right]}^{2023}}{{\left[\sin \left(\frac{\pi }{2}-x\right)\right]}^{2023}+{\left[\cos \left(\frac{\pi }{2}-x\right)\right]}^{2023}}dx$$

$$=\frac{8}{\pi }\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{{(\sin x)}^{2023}}{{(\cos x)}^{2023}+{(\sin x)}^{2023}}dx$$ ..... (2)

Adding equation (1) and (2), we get

$$2I=\frac{8}{\pi }\underset{0}{\overset{\frac{\pi }{2}}{\int }}\frac{{(\sin x)}^{2023}+{(\cos x)}^{2023}}{{(\sin x)}^{2023}+{(\cos x)}^{203}}dx$$

$$\Rightarrow 2I=\frac{8}{\pi }\underset{0}{\overset{\frac{\pi }{2}}{\int }}dx$$

$$\Rightarrow 2I=\frac{8}{\pi }{\left[x\right]}_0^{\frac{\pi }{2}}$$

$$\Rightarrow 2I=\frac{8}{\pi }\times \frac{\pi }{2}$$

$$\Rightarrow 2I=4$$

$$\Rightarrow I=2$$