CUET PG 2024 — Mathematics PYQ

$I={\int }_0^{\pi /2}\frac{{\sin }^{4}x}{{\sin }^{4}x+{\cos }^{4}x}dx$

$I={\int }_0^{\pi /2}\frac{{\sin }^4\left(\frac{\pi }{2}-x\right)}{{\sin }^4\left(\frac{\pi }{2}-x\right)+{\cos }^4\left(\frac{\pi }{2}-x\right)}dx$

$I={\int }_0^{\pi /2}\frac{{\cos }^{4}x}{{\cos }^{4}x+{\sin }^{4}x}dx$

$I={\int }_0^{\pi /2}\frac{{\cos }^{4}x}{{\cos }^{4}x+{\sin }^{4}x}dx$

Adding (1) and (3)

$2I={\int }_0^{\pi /2}\frac{{\sin }^{4}x+{\cos }^{4}x}{{\sin }^{4}x+{\cos }^{4}x}dx$

$2I={\int }_0^{\pi /2}dx={\left[x\right]}_0^{\pi /2}$

$2I=\frac{\pi }{2}$

$I=\frac{\pi }{4}$

$I={\int }_{\pi /6}^{\pi /3}\frac{1}{1+\sqrt{\tan x}}dx$

$I={\int }_{\pi /6}^{\pi /3}\frac{1+\sqrt{\sin x}}{\cos x}dx$

$I={\int }_{\pi /6}^{\pi /3}\frac{1}{\sqrt{\cos x}+\sqrt{\sin x}}dx\text{........ (1)}$

Using property 

${\int }_b^{a}f(x)dx={\int }_b^{a}f(a+b-x)dx$

$\text{So }I={\int }_{\pi /6}^{\pi /3}\frac{1}{\sqrt{\cos x}+\sqrt{\sin x}}dx$

$={\int }_{\pi /6}^{\pi /3}\frac{1}{\sqrt{\cos \left(\frac{\pi }{6}+\frac{\pi }{3}-x\right)}+\sqrt{\sin \left(\frac{\pi }{6}+\frac{\pi }{3}-x\right)}}dx$

$={\int }_{\pi /6}^{\pi /3}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}dx\text{........ (2)}$

Adding (1) and (2)

$2I={\int }_{\pi /6}^{\pi /3}\left[\frac{\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}+\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\right]dx$

$2I={\int }_{\pi /6}^{\pi /3}dx$

$I=\frac{1}{2}{\left[x\right]}_{\pi /6}^{\pi /3}$

$I=\frac{1}{2}\left(\frac{\pi }{3}-\frac{\pi }{6}\right)$

$I=\frac{\pi }{12}$

$I={\int }_0^{1}x{e}^{x}dx$

$I={\left[x\int e^{x}dx-\int \frac{d}{dx}\left(x\int e^{x}dx\right)dx\right]}_0^1$

$I={\left[x{e}^x-\int 1\cdot e^{x}dx\right]}_0^1$

$I={\left[x{e}^x-e^x\right]}_0^1$

$I=\left[(e^1-e^1)-(0e^0-e^0)\right]$

$=(0)-(0-1)$

$=1$