Question

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

Answer 1

Step-by-step explanation:

We have,

$\displaystyle \: I = \int^{\pi}_{0} \left( \dfrac{x}{ \sin(x) + 1 } \right)dx \: \: \: \: \: \: \: ...(1)$

We know,

$\bigstar \: \boxed{ \displaystyle \bf{ \int^{b}_{a}f(x) \: dx = \int^{b}_{a}f(a + b - x) \: dx }}$

So,

$\displaystyle \: I = \int^{\pi}_{0} \left( \dfrac{\pi - x}{ \sin(\pi - x) + 1 } \right)dx$

$\displaystyle \implies I = \int^{\pi}_{0} \left( \dfrac{\pi - x}{ \sin( x) + 1 } \right)dx \: \: \: \: \: \: \: ...(2)$

Adding (1) and (2),

$\displaystyle \implies 2I = \int^{\pi}_{0} \left( \dfrac{ x}{ \sin( x) + 1 } \right)dx + \int^{\pi}_{0} \left( \dfrac{\pi - x}{ \sin( x) + 1 } \right)dx$

$\displaystyle \implies 2I = \int^{\pi}_{0} \left \{ \dfrac{ x}{ \sin( x) + 1 } + \dfrac{\pi - x}{ \sin( x) + 1 } \right \}dx$

$\displaystyle \implies 2I = \int^{\pi}_{0} \left \{ \dfrac{\pi }{ \sin( x) + 1 } \right \}dx$

$\displaystyle \implies 2I = \pi \int^{\pi}_{0} \left \{ \dfrac{1 }{ \sin( x) + 1 } \right \}dx$

$\displaystyle \implies 2I = \pi \int^{ \pi}_{0} \dfrac{1 }{ \sin( x) + 1 } dx$

$\displaystyle \implies2 I = \pi \int^{ \pi}_{0} \dfrac{1 - \sin(x) }{ \{\sin( x) + 1 \} \{ 1 - \sin(x) \}} dx$

$\displaystyle \implies 2I = \pi \int^{ \pi}_{0} \dfrac{1 - \sin(x) }{ 1 - \sin^{2} (x)} dx$

$\displaystyle \implies 2I = \pi \int^{ \pi}_{0} \dfrac{1 - \sin(x) }{ \cos^{2} (x)} dx$

$\displaystyle \implies 2I = \pi \int^{ \pi}_{0} \dfrac{1 }{ \cos^{2} (x)} dx -\pi \int^{ \frac{\pi}{2}}_{0} \dfrac{ \sin(x) }{ \cos^{2} (x)} dx$

$\displaystyle \implies 2I = \pi \int^{ \pi}_{0} \sec^{2} (x) \: dx -\pi \int^{ \frac{\pi}{2}}_{0} \sec(x) \tan(x) \: dx$

$\displaystyle \implies 2I = \pi [ \tan(x) ] ^{ \pi}_{0} -\pi [ \sec(x) ] ^{ \pi}_{0}$

$\displaystyle \implies 2I = \pi [ \tan(\pi) - \tan(0) ] -\pi [ \sec(\pi) - \sec(0) ]$

$\displaystyle \implies 2I = \pi [ 0- 0 ] -\pi [ - 1 - 1 ]$

$\displaystyle \implies 2I = -\pi ( - 2 )$

$\displaystyle \implies 2I = 2\pi$

$\displaystyle \implies I = \pi$

Answer 2

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

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.

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

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

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

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

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

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

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