Question

❖ᴏɴʟʏ ᴘʀᴏᴘᴇʀ ꜱᴏʟᴠᴇᴅ ᴀɴꜱᴡᴇʀ ᴡɪᴛʜ ɢᴏᴏᴅ ᴇxᴘʟᴀɴᴀɪᴏɴ ɴᴇᴇᴅᴇᴅ
❖ ɴᴏ ꜱᴘᴀᴍᴍɪɴɢ
❖ᴏɴʟʏ ꜰᴏʀ ᴍᴏᴅᴇʀᴀᴛᴏʀꜱ, ʙʀᴀɪɴʟʏ ꜱᴛᴀʀꜱ ᴀɴᴅ ᴏᴛʜᴇʀ ʙᴇꜱᴛ ᴜꜱᴇʀꜱ​​

Answer 1

$\displaystyle \bold \red{\int_{0}^{\pi}\left( \frac{x}{\sin{\left(x \right)} + 1} \right)dx}$

First, calculate the corresponding indefinite integral:

$\displaystyle\bold\red{\int{\frac{x}{\sin{\left(x \right)} + 1} d x}}$

$\bold\red{\frac{- 2 x + \left(\tan{\left(\frac{x}{2} \right)} + 1\right) \left(x + \ln{\left(\cos{\left(x \right)} + 1 \right)} + 2 \ln{\left(\left|{\tan{\left(\frac{x}{2} \right)} + 1}\right| \right)} - \ln{\left(2 \right)}\right)}{\tan{\left(\frac{x}{2} \right)} + 1}}$

$\displaystyle \bold \red{ \int{\frac{x}{\sin{\left(x \right)} + 1} d x}} \\ \bold \red{=\frac{- 2 x + \left(\tan{\left(\frac{x}{2} \right)} + 1\right) \left(x + \ln{\left(\cos{\left(x \right)} + 1 \right)} + 2 \ln{\left(\left|{\tan{\left(\frac{x}{2} \right)} + 1}\right| \right)} - \ln{\left(2 \right)}\right)}{\tan{\left(\frac{x}{2} \right)} + 1}}$

According to the Fundamental Theorem of Calculus,

$\displaystyle \bold \red{\int_a^b F(x) dx=f(b)-f(a), }$

so just evaluate the integral at the endpoints, and that's the answer.

$\bold \red{\left(\frac{- 2 x + \left(\tan{\left(\frac{x}{2} \right)} + 1\right) \left(x + \ln{\left(\cos{\left(x \right)} + 1 \right)} + 2 \ln{\left(\left|{\tan{\left(\frac{x}{2} \right)} + 1}\right| \right)} - \ln{\left(2 \right)}\right)}{\tan{\left(\frac{x}{2} \right)} + 1}\right)}$

$\bold \red{|_{\left(x=\pi\right)}=\pi}$

$\bold \red{\left(\frac{- 2 x + \left(\tan{\left(\frac{x}{2} \right)} + 1\right) \left(x + \ln{\left(\cos{\left(x \right)} + 1 \right)} + 2 \ln{\left(\left|{\tan{\left(\frac{x}{2} \right)} + 1}\right| \right)} - \ln{\left(2 \right)}\right)}{\tan{\left(\frac{x}{2} \right)} + 1}\right)}$

$\bold \red{|_{\left(x=0\right)}=0}$

$\bold \red{ \int_{0}^{\pi}\left( \frac{x}{\sin{\left(x \right)} + 1} \right)dx }\\ \bold \red{ =\left(\frac{- 2 x + \left(\tan{\left(\frac{x}{2} \right)} + 1\right) \left(x + \ln{\left(\cos{\left(x \right)} + 1 \right)} + 2 \ln{\left(\left|{\tan{\left(\frac{x}{2} \right)} + 1}\right| \right)} - \ln{\left(2 \right)}\right)}{\tan{\left(\frac{x}{2} \right)} + 1}\right)} \\ \bold \red{|_{\left(x=\pi\right)}} \\ \bold \red{-\left(\frac{- 2 x + \left(\tan{\left(\frac{x}{2} \right)} + 1\right) \left(x + \ln{\left(\cos{\left(x \right)} + 1 \right)} + 2 \ln{\left(\left|{\tan{\left(\frac{x}{2} \right)} + 1}\right| \right)} - \ln{\left(2 \right)}\right)}{\tan{\left(\frac{x}{2} \right)} + 1}\right)}$

$\bold \red{|_{\left(x=0\right)}=\pi}$

$\bold \red{\int_{0}^{\pi}\left( \frac{x}{\sin{\left(x \right)} + 1} \right)dx=\pi}$

Answer 2

Answer:

$\displaystyle \bold \red{\int_{0}^{\pi}\left( \frac{x}{\sin{\left(x \right)} + 1} \right)dx}$

First, calculate the corresponding indefinite integral:

$\displaystyle\bold\red{\int{\frac{x}{\sin{\left(x \right)} + 1} d x}}$

$\bold\red{\frac{- 2 x + \left(\tan{\left(\frac{x}{2} \right)} + 1\right) \left(x + \ln{\left(\cos{\left(x \right)} + 1 \right)} + 2 \ln{\left(\left|{\tan{\left(\frac{x}{2} \right)} + 1}\right| \right)} - \ln{\left(2 \right)}\right)}{\tan{\left(\frac{x}{2} \right)} + 1}}$

$\displaystyle \bold \red{ \int{\frac{x}{\sin{\left(x \right)} + 1} d x}} \\ \bold \red{=\frac{- 2 x + \left(\tan{\left(\frac{x}{2} \right)} + 1\right) \left(x + \ln{\left(\cos{\left(x \right)} + 1 \right)} + 2 \ln{\left(\left|{\tan{\left(\frac{x}{2} \right)} + 1}\right| \right)} - \ln{\left(2 \right)}\right)}{\tan{\left(\frac{x}{2} \right)} + 1}}$

According to the Fundamental Theorem of Calculus,

$\displaystyle \bold \red{\int_a^b F(x) dx=f(b)-f(a), }$

so just evaluate the integral at the endpoints, and that's the answer.

$\bold \red{\left(\frac{- 2 x + \left(\tan{\left(\frac{x}{2} \right)} + 1\right) \left(x + \ln{\left(\cos{\left(x \right)} + 1 \right)} + 2 \ln{\left(\left|{\tan{\left(\frac{x}{2} \right)} + 1}\right| \right)} - \ln{\left(2 \right)}\right)}{\tan{\left(\frac{x}{2} \right)} + 1}\right)}$

$\bold \red{|_{\left(x=\pi\right)}=\pi}$

$\bold \red{\left(\frac{- 2 x + \left(\tan{\left(\frac{x}{2} \right)} + 1\right) \left(x + \ln{\left(\cos{\left(x \right)} + 1 \right)} + 2 \ln{\left(\left|{\tan{\left(\frac{x}{2} \right)} + 1}\right| \right)} - \ln{\left(2 \right)}\right)}{\tan{\left(\frac{x}{2} \right)} + 1}\right)}$

$\bold \red{|_{\left(x=0\right)}=0}$

$\bold \red{ \int_{0}^{\pi}\left( \frac{x}{\sin{\left(x \right)} + 1} \right)dx }\\ \bold \red{ =\left(\frac{- 2 x + \left(\tan{\left(\frac{x}{2} \right)} + 1\right) \left(x + \ln{\left(\cos{\left(x \right)} + 1 \right)} + 2 \ln{\left(\left|{\tan{\left(\frac{x}{2} \right)} + 1}\right| \right)} - \ln{\left(2 \right)}\right)}{\tan{\left(\frac{x}{2} \right)} + 1}\right)} \\ \bold \red{|_{\left(x=\pi\right)}} \\ \bold \red{-\left(\frac{- 2 x + \left(\tan{\left(\frac{x}{2} \right)} + 1\right) \left(x + \ln{\left(\cos{\left(x \right)} + 1 \right)} + 2 \ln{\left(\left|{\tan{\left(\frac{x}{2} \right)} + 1}\right| \right)} - \ln{\left(2 \right)}\right)}{\tan{\left(\frac{x}{2} \right)} + 1}\right)}$

$\bold \red{|_{\left(x=0\right)}=\pi}$

$\bold \red{\int_{0}^{\pi}\left( \frac{x}{\sin{\left(x \right)} + 1} \right)dx=\pi}$