JEE Main 2018 — Mathematics Calculus

Let $\displaystyle I=\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{x}{1+\sin x} d x$I=∫4π​43π​​1+sinxx​dx also let $K=\frac{x}{1+\sin x}$K=1+sinxx​ Multiplying numerator and denominator by $(1-\sin x)$(1−sinx), we get; $$\begin{aligned} K &=\frac{x(1-\sin x)}{1-(\sin x)^2}=\frac{x(1-\sin x)}{(\cos x)^2} \\ &=x(1-\sin x) \sec ^2 x \\ &=x \sec ^2 x-x \sin x \sec ^2 x=x \sec ^2 x-x \tan \\ x & \sec x \end{aligned}$$Kx​=1−(sinx)2x(1−sinx)​=(cosx)2x(1−sinx)​=x(1−sinx)sec2x=xsec2x−xsinxsec2x=xsec2x−xtansecx​ Now, $\displaystyle I=\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} x \sec ^2 x d x-\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} x \sec x \tan x d x$I=∫4π​43π​​xsec2xdx−∫4π​43π​​xsecxtanxdx $$=\left[x \tan x-\int \frac{d x}{d x} \tan x d x\right]_{\frac{\pi}{4}}^{\frac{3 \pi}{4}}-\left[x \sec x-\int \frac{d x}{d x} \sec x d x\right]_{\frac{\pi}{4}}^{\frac{3 \pi}{4}}$$=[xtanx−∫dxdx​tanxdx]4π​43π​​−[xsecx−∫dxdx​secxdx]4π​43π​​ $$\begin{aligned} &=[x \tan x-\ln |\sec x|]_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \\ &-[x \sec x-\ln |\sec x+\tan x|]_{\frac{\pi}{4}}^{\frac{3 \pi}{4}}+c \\ &\Rightarrow I=\left\{\left[\frac{3 \pi}{4} \tan \frac{3 \pi}{4}-\ln \left|\frac{3 \pi}{4}\right|\right.\right. \\ &\left.-\left[\frac{3 \pi}{4} \sec \frac{3 \pi}{4}-\ln \left|\sec \frac{3 \pi}{4}+\tan \frac{3 \pi}{4}\right|\right]\right\} \\ &-\left\{\left[\frac{\pi}{4} \tan \frac{\pi}{4}-\ln \left|\frac{\pi}{4}\right|\right.\right. \\ &\left.-\left[\frac{\pi}{4} \sec \frac{\pi}{4}-\ln \left|\sec \frac{\pi}{4}+\tan \frac{\pi}{4}\right|\right]\right\} \\ &=\frac{\pi}{2}(\sqrt{2}+1) \end{aligned}$$​=[xtanx−ln∣secx∣]4π​43π​​−[xsecx−ln∣secx+tanx∣]4π​43π​​+c⇒I={[43π​tan43π​−ln​43π​​−[43π​sec43π​−ln​sec43π​+tan43π​​]}−{[4π​tan4π​−ln​4π​​−[4π​sec4π​−ln​sec4π​+tan4π​​]}=2π​(2​+1)​