The value of integral $\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{x}{1+\sin x} d x$ is

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Aicharya · Accepted Answer

Let $I=\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{x}{1+\sin x} d x$ also let $K=\frac{x}{1+\sin x}$ Multiplying numerator and denominator by $(1-\sin x)$, 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 	an \ x & \sec x \end{aligned} $ Now, $I=\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} x \sec ^2 x d x-\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} x \sec x 	an x d x$ $ =\left[x 	an x-\int \frac{d x}{d x} 	an x d xight]_{\frac{\pi}{4}}^{\frac{3 \pi}{4}}-\left[x \sec x-\int \frac{d x}{d x} \sec x d xight]_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} $ $ \begin{aligned} &=[x 	an x-\ln |\sec x|]_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \ &-[x \sec x-\ln |\sec x+	an x|]_{\frac{\pi}{4}}^{\frac{3 \pi}{4}}+c \ &\Rightarrow I=\left\{\left[\frac{3 \pi}{4} 	an \frac{3 \pi}{4}-\ln \left|\frac{3 \pi}{4}ight|ight.ight. \ &\left.-\left[\frac{3 \pi}{4} \sec \frac{3 \pi}{4}-\ln \left|\sec \frac{3 \pi}{4}+	an \frac{3 \pi}{4}ight|ight]ight\} \ &-\left\{\left[\frac{\pi}{4} 	an \frac{\pi}{4}-\ln \left|\frac{\pi}{4}ight|ight.ight. \ &\left.-\left[\frac{\pi}{4} \sec \frac{\pi}{4}-\ln \left|\sec \frac{\pi}{4}+	an \frac{\pi}{4}ight|ight]ight\} \ &=\frac{\pi}{2}(\sqrt{2}+1) \end{aligned} $

The value of integral $\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{x}{1+\sin x} d x$ is

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