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  • The value of     <img alt="\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{(2+3 \sin x)}{\sin x(1+\cos x)} d x" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cint_%7B%5Cfrac

The value of     \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{(2+3 \sin x)}{\sin x(1+\cos x)} d x    is equal to

 

Option: 1

\frac{10}{3}-\sqrt{3}-\log _e \sqrt{3}


Option: 2

\frac{7}{2}-\sqrt{3}-\log _e \sqrt{3}


Option: 3

-2+3\sqrt{3}+\log _e \sqrt{3}


Option: 4

\frac{10}{3}-\sqrt{3}+\log _e \sqrt{3}


Answers (1)

best_answer

\begin{aligned} & \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{(2+3 \sin x)}{\sin x(1+\cos x)} d x \\ & =\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{2}{\sin x+\sin x \cos x} d x+\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{3}{1+\cos x} d x \\ & =I_1+I_2 \end{aligned}

\begin{aligned} & I_1=\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{2 d x}{\sin x(1+\cos x)}=2 \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\left(1+\tan ^2 \frac{x}{2}\right) d x}{2 \tan \frac{x}{2} \times\left(1+\frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}\right)} \\ & =2 \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\left(1+\tan ^2 \frac{x}{2}\right)\left(1+\tan ^2 \frac{x}{2}\right) d x}{2 \tan \frac{x}{2} \times 2} \\ & =2 \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\sec ^2 \frac{x}{2}\left(1+\tan ^2 \frac{x}{2}\right)}{4 \tan \frac{x}{2}} d x \\ & \end{aligned}

Let,  \tan \frac{x}{2}=t    then   \sec ^2 \frac{x}{2} \times \frac{1}{2} d x=d t  

 

\begin{aligned} & =2 \int_{\frac{1}{\sqrt{3}}}^1 \frac{1+\mathrm{t}^2}{2 \mathrm{t}} \mathrm{dt} \\ & =\left[\ell \mathrm{nt}+\frac{\mathrm{t}^2}{2}\right]_{\frac{1}{\sqrt{3}}}^1 \\ & =\left[\frac{1}{2}-\ell \mathrm{n} \frac{1}{\sqrt{3}}-\frac{1}{6}\right] \\ \end{aligned}

\begin{aligned} & \mathrm{I}_2=3 \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{\mathrm{dx}}{1+\cos \mathrm{x}}=3 \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{1-\cos \mathrm{x}}{\sin ^2 \mathrm{x}} \mathrm{dx} \end{aligned}

\begin{gathered} \mathrm{I}_2=3[\operatorname{cosec} x-\cot x]_{\frac{\pi}{3}}^{\frac{\pi}{2}}=3-\sqrt{3} \\ \mathrm{I}_1+\mathrm{I}_2=\ln \sqrt{3}+\frac{1}{3}+3-\sqrt{3} \\ =\frac{10}{3}+\ln \sqrt{3}-\sqrt{3} \end{gathered}

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Riya

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