\displaystyle \text{Evaluate each of the following integrals (1-15):}

\displaystyle \textbf{Question 1: }~\int_{0}^{2\pi}\frac{e^{\sin x}}{e^{\sin x}+e^{-\sin x}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int_{0}^{2\pi}\frac{e^{\sin x}}{e^{\sin x}+e^{-\sin x}}\,dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)
\displaystyle \text{Then, using the property } \int_{0}^{a} f(x)\,dx=\int_{0}^{a} f(a-x)\,dx
\displaystyle I=\int_{0}^{2\pi}\frac{e^{\sin(2\pi-x)}}{e^{\sin(2\pi-x)}+e^{-\sin(2\pi-x)}}\,dx
\displaystyle \Rightarrow I=\int_{0}^{2\pi}\frac{e^{-\sin x}}{e^{-\sin x}+e^{\sin x}}\,dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)
\displaystyle \text{Adding (1) and (2), we get}
\displaystyle 2I=\int_{0}^{2\pi}\left(\frac{e^{\sin x}}{e^{\sin x}+e^{-\sin x}}+\frac{e^{-\sin x}}{e^{-\sin x}+e^{\sin x}}\right)\,dx
\displaystyle \Rightarrow 2I=\int_{0}^{2\pi}\frac{e^{\sin x}+e^{-\sin x}}e^{\sin x}+e^{-\sin x}\,dx
\displaystyle \Rightarrow 2I=\int_{0}^{2\pi} 1\,dx
\displaystyle \Rightarrow 2I=\left[x\right]_{0}^{2\pi}
\displaystyle \Rightarrow 2I=2\pi-0
\displaystyle \Rightarrow I=\pi

\displaystyle \textbf{Question 2: }~\int_{0}^{2\pi}\log(\sec x+\tan x)\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int_{0}^{2\pi} \log(\sec x+\tan x)\,dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)
\displaystyle \text{Then, using the property } \int_{0}^{a} f(x)\,dx=\int_{0}^{a} f(a-x)\,dx
\displaystyle I=\int_{0}^{2\pi} \log\big(\sec(2\pi-x)+\tan(2\pi-x)\big)\,dx
\displaystyle \Rightarrow I=\int_{0}^{2\pi} \log(\sec x-\tan x)\,dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)
\displaystyle \text{Adding (1) and (2), we get}
\displaystyle 2I=\int_{0}^{2\pi}\left[\log(\sec x+\tan x)+\log(\sec x-\tan x)\right]\,dx
\displaystyle \Rightarrow 2I=\int_{0}^{2\pi} \log\big((\sec x+\tan x)(\sec x-\tan x)\big)\,dx
\displaystyle \Rightarrow 2I=\int_{0}^{2\pi} \log(\sec^{2}x-\tan^{2}x)\,dx
\displaystyle \Rightarrow 2I=\int_{0}^{2\pi} \log 1\,dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1+\tan^{2}x=\sec^{2}x)
\displaystyle \Rightarrow 2I=0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\log 1=0)
\displaystyle \Rightarrow I=0

\displaystyle \textbf{Question 3: }~\int_{\pi/6}^{\pi/3}\frac{\sqrt{\tan x}}{\sqrt{\tan x}+\sqrt{\cot x}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\tan x}}{\sqrt{\tan x}+\sqrt{\cot x}}\,dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)
\displaystyle \text{Then, using the property } \int_{a}^{b} f(x)\,dx=\int_{a}^{b} f(a+b-x)\,dx
\displaystyle I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\tan\left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}}{\sqrt{\tan\left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}+\sqrt{\cot\left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}}\,dx
\displaystyle \Rightarrow I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\tan\left(\frac{\pi}{2}-x\right)}}{\sqrt{\tan\left(\frac{\pi}{2}-x\right)}+\sqrt{\cot\left(\frac{\pi}{2}-x\right)}}\,dx
\displaystyle \Rightarrow I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cot x}}{\sqrt{\cot x}+\sqrt{\tan x}}\,dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)
\displaystyle \text{Adding (1) and (2), we get}
\displaystyle 2I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\tan x}+\sqrt{\cot x}}{\sqrt{\tan x}+\sqrt{\cot x}}\,dx
\displaystyle \Rightarrow 2I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1\,dx
\displaystyle \Rightarrow 2I=\left[x\right]_{\frac{\pi}{6}}^{\frac{\pi}{3}}
\displaystyle \Rightarrow 2I=\frac{\pi}{3}-\frac{\pi}{6}
\displaystyle \Rightarrow 2I=\frac{\pi}{6}
\displaystyle \Rightarrow I=\frac{\pi}{12}

\displaystyle \textbf{Question 4: }~\int_{\pi/6}^{\pi/3}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)
\displaystyle \text{Then, using the property } \int_{a}^{b} f(x)\,dx=\int_{a}^{b} f(a+b-x)\,dx
\displaystyle I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin\left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}}{\sqrt{\sin\left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}+\sqrt{\cos\left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}}\,dx
\displaystyle \Rightarrow I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin\left(\frac{\pi}{2}-x\right)}}{\sqrt{\sin\left(\frac{\pi}{2}-x\right)}+\sqrt{\cos\left(\frac{\pi}{2}-x\right)}}\,dx
\displaystyle \Rightarrow I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)
\displaystyle \text{Adding (1) and (2), we get}
\displaystyle 2I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx
\displaystyle \Rightarrow 2I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1\,dx
\displaystyle \Rightarrow 2I=\left[x\right]_{\frac{\pi}{6}}^{\frac{\pi}{3}}
\displaystyle \Rightarrow 2I=\frac{\pi}{3}-\frac{\pi}{6}
\displaystyle \Rightarrow 2I=\frac{\pi}{6}
\displaystyle \Rightarrow I=\frac{\pi}{12}

\displaystyle \textbf{Question 5: }~\int_{-\pi/4}^{\pi/4}\frac{\tan^2 x}{1+e^x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\tan^{2}x}{1+e^{x}}\,dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)
\displaystyle \text{Then, using the property } \int_{a}^{b} f(x)\,dx=\int_{a}^{b} f(a+b-x)\,dx
\displaystyle I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\tan^{2}\!\left(\frac{\pi}{4}+(-\frac{\pi}{4})-x\right)}{1+e^{\frac{\pi}{4}+(-\frac{\pi}{4})-x}}\,dx
\displaystyle \Rightarrow I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\tan^{2}(-x)}{1+e^{-x}}\,dx
\displaystyle \Rightarrow I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\tan^{2}x}{1+e^{-x}}\,dx
\displaystyle \Rightarrow I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{e^{x}\tan^{2}x}{1+e^{x}}\,dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)
\displaystyle \text{Adding (1) and (2), we get}
\displaystyle 2I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\frac{\tan^{2}x}{1+e^{x}}+\frac{e^{x}\tan^{2}x}{1+e^{x}}\right)\,dx
\displaystyle \Rightarrow 2I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{(1+e^{x})\tan^{2}x}{1+e^{x}}\,dx
\displaystyle \Rightarrow 2I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \tan^{2}x\,dx
\displaystyle \Rightarrow 2I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} (\sec^{2}x-1)\,dx
\displaystyle \Rightarrow 2I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec^{2}x\,dx-\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} dx
\displaystyle \Rightarrow 2I=\left[\tan x\right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}-\left[x\right]_{-\frac{\pi}{4}}^{\frac{\pi}{4}}
\displaystyle \Rightarrow 2I=\left(\tan\frac{\pi}{4}-\tan\!\left(-\frac{\pi}{4}\right)\right)-\left(\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)\right)
\displaystyle \Rightarrow 2I=(1+1)-\frac{\pi}{2}
\displaystyle \Rightarrow 2I=2-\frac{\pi}{2}
\displaystyle \Rightarrow I=1-\frac{\pi}{4}

\displaystyle \textbf{Question 6: }~\int_{-a}^{a}\frac{1}{1+a^x}\,dx,\ a>0.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int_{-a}^{a}\frac{1}{1+a^{x}}\,dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)
\displaystyle \text{Using the property } \int_{-a}^{a} f(x)\,dx=\int_{-a}^{a} f(-x)\,dx
\displaystyle I=\int_{-a}^{a}\frac{1}{1+a^{-x}}\,dx
\displaystyle \Rightarrow I=\int_{-a}^{a}\frac{a^{x}}{a^{x}+1}\,dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)
\displaystyle \text{Adding (1) and (2), we get}
\displaystyle 2I=\int_{-a}^{a}\left(\frac{1}{1+a^{x}}+\frac{a^{x}}{1+a^{x}}\right)\,dx
\displaystyle \Rightarrow 2I=\int_{-a}^{a}\frac{1+a^{x}}{1+a^{x}}\,dx
\displaystyle \Rightarrow 2I=\int_{-a}^{a} 1\,dx
\displaystyle \Rightarrow 2I=\left[x\right]_{-a}^{a}
\displaystyle \Rightarrow 2I=a-(-a)
\displaystyle \Rightarrow 2I=2a
\displaystyle \Rightarrow I=a

\displaystyle \textbf{Question 7: }~\int_{-\pi/3}^{\pi/3}\frac{1}{1+e^{\tan x}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \frac{1}{1+e^{\tan x}}\,dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)
\displaystyle \text{Then, using the property } \int_{-a}^{a} f(x)\,dx=\int_{-a}^{a} f(-x)\,dx
\displaystyle I=\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \frac{1}{1+e^{\tan(-x)}}\,dx
\displaystyle \Rightarrow I=\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \frac{1}{1+e^{-\tan x}}\,dx
\displaystyle \Rightarrow I=\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \frac{e^{\tan x}}{1+e^{\tan x}}\,dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)
\displaystyle \text{Adding (1) and (2), we get}
\displaystyle 2I=\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\left(\frac{1}{1+e^{\tan x}}+\frac{e^{\tan x}}{1+e^{\tan x}}\right)\,dx
\displaystyle \Rightarrow 2I=\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} \frac{1+e^{\tan x}}{1+e^{\tan x}}\,dx
\displaystyle \Rightarrow 2I=\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}} 1\,dx
\displaystyle \Rightarrow 2I=\left[x\right]_{-\frac{\pi}{3}}^{\frac{\pi}{3}}
\displaystyle \Rightarrow 2I=\frac{\pi}{3}-\left(-\frac{\pi}{3}\right)
\displaystyle \Rightarrow 2I=\frac{2\pi}{3}
\displaystyle \Rightarrow I=\frac{\pi}{3}

\displaystyle \textbf{Question 8: }~\int_{-\pi/2}^{\pi/2}\frac{\cos^2 x}{1+e^x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos^{2}x}{1+e^{x}}\,dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)
\displaystyle \text{Then, using the property } \int_{a}^{b} f(x)\,dx=\int_{a}^{b} f(a+b-x)\,dx
\displaystyle I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos^{2}\!\left(\frac{\pi}{2}+(-\frac{\pi}{2})-x\right)}{1+e^{\frac{\pi}{2}+(-\frac{\pi}{2})-x}}\,dx
\displaystyle \Rightarrow I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos^{2}(-x)}{1+e^{-x}}\,dx
\displaystyle \Rightarrow I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\cos^{2}x}{1+e^{-x}}\,dx
\displaystyle \Rightarrow I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{e^{x}\cos^{2}x}{1+e^{x}}\,dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)
\displaystyle \text{Adding (1) and (2), we get}
\displaystyle 2I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{\cos^{2}x}{1+e^{x}}+\frac{e^{x}\cos^{2}x}{1+e^{x}}\right)\,dx
\displaystyle \Rightarrow 2I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{(1+e^{x})\cos^{2}x}{1+e^{x}}\,dx
\displaystyle \Rightarrow 2I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos^{2}x\,dx
\displaystyle \Rightarrow 2I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1+\cos 2x}{2}\,dx
\displaystyle \Rightarrow 2I=\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} dx+\frac{1}{2}\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cos 2x\,dx
\displaystyle \Rightarrow 2I=\frac{1}{2}\left[x\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}+\frac{1}{2}\left[\frac{\sin 2x}{2}\right]_{-\frac{\pi}{2}}^{\frac{\pi}{2}}
\displaystyle \Rightarrow 2I=\frac{1}{2}\left(\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right)+\frac{1}{4}\left(\sin\pi-\sin(-\pi)\right)
\displaystyle \Rightarrow 2I=\frac{\pi}{2}+\frac{1}{4}(0-0)
\displaystyle \Rightarrow 2I=\frac{\pi}{2}
\displaystyle \Rightarrow I=\frac{\pi}{4}

\displaystyle \textbf{Question 9: }~\int_{-\pi/4}^{\pi/4}\frac{x^{11}-3x^9+5x^7-x^5+1}{\cos^2 x}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x^{11}-3x^{9}+5x^{7}-x^{5}+1}{\cos^{2}x}\,dx
\displaystyle \Rightarrow I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x^{11}-3x^{9}+5x^{7}-x^{5}}{\cos^{2}x}\,dx+\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{\cos^{2}x}\,dx
\displaystyle \Rightarrow I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x^{11}-3x^{9}+5x^{7}-x^{5}}{\cos^{2}x}\,dx+\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec^{2}x\,dx
\displaystyle \Rightarrow I=I_{1}+I_{2}
\displaystyle \text{Now, consider } f(x)=\frac{x^{11}-3x^{9}+5x^{7}-x^{5}}{\cos^{2}x}
\displaystyle \Rightarrow f(-x)=\frac{(-x)^{11}-3(-x)^{9}+5(-x)^{7}-(-x)^{5}}{\cos^{2}(-x)}
\displaystyle \Rightarrow f(-x)=\frac{-x^{11}+3x^{9}-5x^{7}+x^{5}}{\cos^{2}x}=-\frac{x^{11}-3x^{9}+5x^{7}-x^{5}}{\cos^{2}x}=-f(x)
\displaystyle \Rightarrow f(x)\text{ is an odd function}
\displaystyle \therefore I_{1}=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} f(x)\,dx=0
\displaystyle \text{Let } g(x)=\sec^{2}x
\displaystyle \Rightarrow g(-x)=\sec^{2}(-x)=\sec^{2}x=g(x)
\displaystyle \Rightarrow g(x)\text{ is an even function}
\displaystyle \therefore I_{2}=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec^{2}x\,dx=2\int_{0}^{\frac{\pi}{4}} \sec^{2}x\,dx
\displaystyle \Rightarrow I_{2}=2\left[\tan x\right]_{0}^{\frac{\pi}{4}}
\displaystyle \Rightarrow I_{2}=2\left(\tan\frac{\pi}{4}-\tan 0\right)
\displaystyle \Rightarrow I_{2}=2(1-0)
\displaystyle \Rightarrow I_{2}=2
\displaystyle \therefore I=I_{1}+I_{2}=0+2=2

\displaystyle \textbf{Question 10: }~\int_{a}^{b}\frac{x^{1/n}}{x^{1/n}+(a+b-x)^{1/n}}\,dx,\ n\in\mathbb{N},\ n\ge 2.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int_{a}^{b}\frac{x^{\frac{1}{n}}}{x^{\frac{1}{n}}+(a+b-x)^{\frac{1}{n}}}\,dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)
\displaystyle \text{Then, using the property } \int_{a}^{b} f(x)\,dx=\int_{a}^{b} f(a+b-x)\,dx
\displaystyle I=\int_{a}^{b}\frac{(a+b-x)^{\frac{1}{n}}}{(a+b-x)^{\frac{1}{n}}+x^{\frac{1}{n}}}\,dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)
\displaystyle \text{Adding (1) and (2), we get}
\displaystyle 2I=\int_{a}^{b}\frac{x^{\frac{1}{n}}+(a+b-x)^{\frac{1}{n}}}{x^{\frac{1}{n}}+(a+b-x)^{\frac{1}{n}}}\,dx
\displaystyle \Rightarrow 2I=\int_{a}^{b} 1\,dx
\displaystyle \Rightarrow 2I=\left[x\right]_{a}^{b}
\displaystyle \Rightarrow 2I=b-a
\displaystyle \Rightarrow I=\frac{b-a}{2}

\displaystyle \textbf{Question 11: }~\int_{0}^{\pi/2}\left(2\log(\cos x)-\log(\sin 2x)\right)\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int_{0}^{\frac{\pi}{2}}\left(2\log\cos x-\log\sin 2x\right)\,dx
\displaystyle \Rightarrow I=\int_{0}^{\frac{\pi}{2}}\left(2\log\cos x-\log(2\sin x\cos x)\right)\,dx
\displaystyle \Rightarrow I=\int_{0}^{\frac{\pi}{2}}\left(2\log\cos x-\log 2-\log\sin x-\log\cos x\right)\,dx
\displaystyle \Rightarrow I=\int_{0}^{\frac{\pi}{2}}\left(\log\cos x-\log 2-\log\sin x\right)\,dx
\displaystyle \Rightarrow I=\int_{0}^{\frac{\pi}{2}} \log\cos x\,dx-\int_{0}^{\frac{\pi}{2}} \log 2\,dx-\int_{0}^{\frac{\pi}{2}} \log\sin x\,dx
\displaystyle \Rightarrow I=\int_{0}^{\frac{\pi}{2}} \log\cos x\,dx-\int_{0}^{\frac{\pi}{2}} \log 2\,dx-\int_{0}^{\frac{\pi}{2}} \log\sin\left(\frac{\pi}{2}-x\right)\,dx
\displaystyle \Rightarrow I=\int_{0}^{\frac{\pi}{2}} \log\cos x\,dx-\int_{0}^{\frac{\pi}{2}} \log 2\,dx-\int_{0}^{\frac{\pi}{2}} \log\cos x\,dx
\displaystyle \Rightarrow I=-\int_{0}^{\frac{\pi}{2}} \log 2\,dx
\displaystyle \Rightarrow I=-\log 2\int_{0}^{\frac{\pi}{2}} 1\,dx
\displaystyle \Rightarrow I=-\log 2\left[x\right]_{0}^{\frac{\pi}{2}}
\displaystyle \Rightarrow I=-\log 2\left(\frac{\pi}{2}-0\right)
\displaystyle \Rightarrow I=-\frac{\pi}{2}\log 2

\displaystyle \textbf{Question 12: }~\int_{0}^{a}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int_{0}^{a}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}\,dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)
\displaystyle \Rightarrow I=\int_{0}^{a}\frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}\,dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [\text{Using } \int_{0}^{a} f(x)\,dx=\int_{0}^{a} f(a-x)\,dx]
\displaystyle \Rightarrow I=\int_{0}^{a}\frac{\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}}\,dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)
\displaystyle \text{Adding (1) and (2)}
\displaystyle 2I=\int_{0}^{a}\frac{\sqrt{x}+\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}}\,dx
\displaystyle \Rightarrow 2I=\int_{0}^{a} 1\,dx
\displaystyle \Rightarrow 2I=\left[x\right]_{0}^{a}
\displaystyle \Rightarrow 2I=a
\displaystyle \Rightarrow I=\frac{a}{2}

\displaystyle \textbf{Question 13: }~\int_{0}^{5}\frac{\sqrt[4]{x+4}}{\sqrt[4]{x+4}+\sqrt[4]{9-x}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int_{0}^{5}\frac{\sqrt[4]{x+4}}{\sqrt[4]{x+4}-\sqrt[4]{9-x}}\,dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)
\displaystyle \Rightarrow I=\int_{0}^{5}\frac{\sqrt[4]{9-x}}{\sqrt[4]{9-x}-\sqrt[4]{x+4}}\,dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [\text{Using } \int_{0}^{a} f(x)\,dx=\int_{0}^{a} f(a-x)\,dx]
\displaystyle \Rightarrow I=-\int_{0}^{5}\frac{\sqrt[4]{9-x}}{\sqrt[4]{x+4}-\sqrt[4]{9-x}}\,dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)
\displaystyle \text{Adding (1) and (2)}
\displaystyle 2I=\int_{0}^{5}\frac{\sqrt[4]{x+4}-\sqrt[4]{9-x}}{\sqrt[4]{x+4}-\sqrt[4]{9-x}}\,dx
\displaystyle \Rightarrow 2I=\int_{0}^{5} 1\,dx
\displaystyle \Rightarrow 2I=\left[x\right]_{0}^{5}
\displaystyle \Rightarrow 2I=5
\displaystyle \Rightarrow I=\frac{5}{2}

\displaystyle \textbf{Question 14: }~\int_{0}^{7}\frac{\sqrt[3]{x}}{\sqrt[3]{x}+\sqrt[3]{7-x}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int_{0}^{7}\frac{\sqrt[3]{x}}{\sqrt[3]{x}+\sqrt[3]{7-x}}\,dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)
\displaystyle \Rightarrow I=\int_{0}^{7}\frac{\sqrt[3]{7-x}}{\sqrt[3]{7-x}+\sqrt[3]{x}}\,dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ [\text{Using } \int_{0}^{a} f(x)\,dx=\int_{0}^{a} f(a-x)\,dx]
\displaystyle \Rightarrow I=\int_{0}^{7}\frac{\sqrt[3]{7-x}}{\sqrt[3]{x}+\sqrt[3]{7-x}}\,dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)
\displaystyle \text{Adding (1) and (2), we get}
\displaystyle 2I=\int_{0}^{7}\frac{\sqrt[3]{x}+\sqrt[3]{7-x}}{\sqrt[3]{x}+\sqrt[3]{7-x}}\,dx
\displaystyle \Rightarrow 2I=\int_{0}^{7} 1\,dx
\displaystyle \Rightarrow 2I=\left[x\right]_{0}^{7}
\displaystyle \Rightarrow 2I=7
\displaystyle \Rightarrow I=\frac{7}{2}

\displaystyle \textbf{Question 15: }~\int_{\pi/6}^{\pi/3}\frac{1}{1+\sqrt{\tan x}}\,dx.
\displaystyle \text{Answer:}
\displaystyle \text{Let } I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\sqrt{\tan x}}\,dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)
\displaystyle \text{Using the property } \int_{a}^{b} f(x)\,dx=\int_{a}^{b} f(a+b-x)\,dx
\displaystyle I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\sqrt{\tan\left(\frac{\pi}{3}+\frac{\pi}{6}-x\right)}}\,dx
\displaystyle \Rightarrow I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\sqrt{\tan\left(\frac{\pi}{2}-x\right)}}\,dx
\displaystyle \Rightarrow I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1}{1+\sqrt{\cot x}}\,dx\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (2)
\displaystyle \text{Adding (1) and (2)}
\displaystyle 2I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\left(\frac{1}{1+\sqrt{\tan x}}+\frac{1}{1+\sqrt{\cot x}}\right)\,dx
\displaystyle \Rightarrow 2I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\left(\frac{1}{1+\sqrt{\tan x}}+\frac{1}{1+\frac{1}{\sqrt{\tan x}}}\right)\,dx
\displaystyle \Rightarrow 2I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\left(\frac{1}{1+\sqrt{\tan x}}+\frac{\sqrt{\tan x}}{1+\sqrt{\tan x}}\right)\,dx
\displaystyle \Rightarrow 2I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{1+\sqrt{\tan x}}{1+\sqrt{\tan x}}\,dx
\displaystyle \Rightarrow 2I=\int_{\frac{\pi}{6}}^{\frac{\pi}{3}} 1\,dx
\displaystyle \Rightarrow 2I=\left[x\right]_{\frac{\pi}{6}}^{\frac{\pi}{3}}
\displaystyle \Rightarrow 2I=\frac{\pi}{3}-\frac{\pi}{6}
\displaystyle \Rightarrow 2I=\frac{\pi}{6}
\displaystyle \Rightarrow I=\frac{\pi}{12}

\displaystyle \textbf{Question 16: }~\text{If }f(a+b-x)=f(x),\ \text{then prove that } \\ \int_{a}^{b}x f(x)\,dx=\frac{a+b}{2}\int_{a}^{b}f(x)\,dx.
\displaystyle \text{Answer:}
\displaystyle \int_{a}^{b} x f(x)\,dx
\displaystyle =\int_{a}^{b} (a+b-x)\,f(a+b-x)\,dx \ \ \ [\text{Using } \int_{a}^{b} f(x)\,dx=\int_{a}^{b} f(a+b-x)\,dx]
\displaystyle =\int_{a}^{b} (a+b-x)\,f(x)\,dx\ \ \ \ \ \ \ \ \ \ \ \ \ [\text{Given } f(a+b-x)=f(x)]
\displaystyle \Rightarrow \int_{a}^{b} x f(x)\,dx=\int_{a}^{b} (a+b)f(x)\,dx-\int_{a}^{b} x f(x)\,dx
\displaystyle \Rightarrow \int_{a}^{b} x f(x)\,dx+\int_{a}^{b} x f(x)\,dx=(a+b)\int_{a}^{b} f(x)\,dx
\displaystyle \Rightarrow 2\int_{a}^{b} x f(x)\,dx=(a+b)\int_{a}^{b} f(x)\,dx
\displaystyle \Rightarrow \int_{a}^{b} x f(x)\,dx=\frac{a+b}{2}\int_{a}^{b} f(x)\,dx

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