Question:

<p>Let&nbsp;<span class="math-tex">\(I=\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{x}{1+\sin x} d x\)</span><br /> also let&nbsp;<span class="math-tex">\(K=\frac{x}{1+\sin x}\)</span><br /> Multiplying numerator and denominator by (1 - sinx), we get;<br /> <span class="math-tex">\(K=\frac{x(1-\sin x)}{1-(\sin x)^2}=\frac{x(1-\sin x)}{(\cos x)^2}\)</span><br /> = x(1 - sin x) sec²x<br /> =&nbsp;x sec² x - x sin x sec² x = x sec² x - x tan x sec x<br /> Now,&nbsp;<span class="math-tex">\(I=\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} x \sec ^2 x d x-\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}}\)</span>&nbsp;x se c x tan x dx<br /> =<span class="math-tex">\(\left[x \tan x-\int \frac{d x}{d x} \tan x d x\right]_{\frac{\pi}{4}}^{\frac{3 \pi}{4}}\)</span>&nbsp;-&nbsp;<span class="math-tex">\(\left[x \sec x-\int \frac{d x}{d x} \sec x d x\right]_{\frac{\pi}{4}}^{\frac{3 x}{4}}\)</span><br /> <span class="math-tex">\(=[x \tan x-\ln |\sec x|]_{\frac{\pi}{4}}^{\frac{3 \pi}{4}}\)</span>&nbsp;-&nbsp;<span class="math-tex">\([x \sec x-\ln |\sec x+\tan x|]_{\frac{\pi}{4}}^{\frac{3 \pi}{4}}+c\)</span><br /> <span class="math-tex">\(\Rightarrow I=\left\{\left[\frac{3 \pi}{4} \tan \frac{3 \pi}{4}-\ln \left|\frac{3 \pi}{4}\right|\right.\right.\)</span><span class="math-tex">\(\left.-\left[\frac{3 \pi}{4} \sec \frac{3 \pi}{4}-\ln \left|\sec \frac{3 \pi}{4}+\tan \frac{3 \pi}{4}\right|\right]\right\}\)</span><span class="math-tex">\(-\left\{\left[\frac{\pi}{4} \tan \frac{\pi}{4}-\ln \left|\frac{\pi}{4}\right|\right.\right.\)</span><br /> <span class="math-tex">\(\left.-\left[\frac{\pi}{4} \sec \frac{\pi}{4}-\ln \left|\sec \frac{\pi}{4}+\tan \frac{\pi}{4}\right|\right]\right\}\)</span><br /> <span class="math-tex">\(=\frac{\pi}{2}(\sqrt{2}+1)\)</span></p>