Let \(I=\int_0^{\pi / 4} \frac{\sin x+\cos x}{16+9 \sin 2 x} d x=\int_0^{\pi / 4} \frac{\sin x+\cos x}{9(2 \sin x \cos x)+16} d x\)
\(=\int_0^{\pi / 4} \frac{\sin x+\cos x}{-9(-2 \sin x \cos x)+16} d x\) 
\(=\int_0^{\pi / 4} \frac{\sin x+\cos x}{-9\left(\sin ^2 x+\cos ^2 x-2 \sin x \cos x-1\right)+16} d x\)
\( {\left[\because \sin ^2 \theta+\cos ^2 \theta=1\right]}\)
\(=\int_0^{\pi / 4} \frac{\sin x+\cos x}{25-9(\sin x-\cos x)^2} d x \)
\({\left[\because a^2+b^2-2 a b=(a-b)^2\right]}\)
Put \(\sin x-\cos x=t \Rightarrow(\cos x+\sin x) d x=d t\)
Lower limit When x=0, then t=-1
Upper limit When \(x=\frac{\pi}{4}\), then t=0
\(\therefore \quad I=\int_{-1}^0 \frac{d t}{25-9 t^2}=\frac{1}{9} \int_{-1}^0 \frac{d t}{\left(\frac{5}{3}\right)^2-t^2}\)
\(=\frac{1}{9} \times \frac{1}{2 \times \frac{5}{3}}\left[\log \left|\frac{\frac{5}{3}+t}{\frac{5}{3}-t}\right|\right]_{-1}^0\)
\( {\left[\because \int \frac{d x}{a^2-x^2}=\frac{1}{2 a} \log \left|\frac{a+x}{a-x}\right|+C\right]} \)
\(=\frac{1}{30}\left[\log \left|\frac{5+3 t}{5-3 t}\right|\right]_{-1}^0=\frac{1}{30}\left[\log 1-\log \left(\frac{2}{8}\right)\right]\)
\(=\frac{1}{30}\left[0-\log \frac{1}{4}\right] \ {[\because \log 1=0]} \)
\(=\frac{1}{30}\left[-\log 4^{-1}\right]=\frac{1}{30} \log 4 \quad\left[\because \log m^n=n \log m\right]\)

For, \(2 \leq x<5,|x-2|=x-2\)
\(2 \leq x<3,|x-3|=-x-3\)
\(3 \leq x<5,|x-3|=x-3\)
and \(2 \leq x<5,|x-5|=5-x\)
\(\therefore I  =\int_2^5[|x-2|+|x-3|+|x-5|] d x\)
\(=\int_2^5(x-2) d x+\int_2^3(3-x) d x +\int_3^5(x-3) d x+\int_2^5(5-x) d x\)
\(=\left[\frac{x^2}{2}-2 x\right]_2^5+\left[3 x-\frac{x^2}{2}\right]_2^3+\left[\frac{x^2}{2}-3 x\right]_3^5+\left[5 x-\frac{x^2}{2}\right]_2^5\)
\(=\left[\left(\frac{25}{2}-10\right)-(2-4)\right]+\left[\left(9-\frac{9}{2}\right)-(6-2)\right] +\left[\left(\frac{25}{2}-15\right)-\left(\frac{9}{2}-9\right)\right]+\left[\left(25-\frac{25}{2}\right)-(10-2)\right]\)
\(=\left[\frac{25}{2}-8\right]+\left[\frac{9}{2}-4\right]+[8-6]+\left[\frac{25}{2}-8\right]\)
\(=\frac{9}{2}+\frac{1}{2}+2+\frac{9}{2}=\frac{19}{2}+2=\frac{23}{2}\)

Let \(I=\int_0^{\pi / 2} \frac{\sin ^2 x}{\sin x+\cos x} d x\)
\(\Rightarrow I=\int_0^{\pi / 2} \frac{\sin ^2\left(\frac{\pi}{2}-x\right)}{\sin \left(\frac{\pi}{2}-x\right)+\cos \left(\frac{\pi}{2}-x\right)} d x\)
\( {\left[\because \int_0^a f(x) d x=\int_0^a f(a-x) d x\right] }\)
\(\Rightarrow I=\int_0^{\pi / 2} \frac{\cos ^2 x}{\cos x+\sin x} d x\)
On adding Eqs. (i) and (ii), we get
\(2 I  =\int_0^{\pi / 2} \frac{\sin ^2 x+\cos ^2 x}{\sin x+\cos x} d x \)
\(\Rightarrow 2 I =\int_0^{\pi / 2} \frac{1}{\sin x+\cos x} d x\)
\(\Rightarrow 2 I  =\int_0^{\pi / 2} \frac{1}{\frac{2 \tan (x / 2)}{1+\tan ^2(x / 2)}+\frac{1-\tan ^2(x / 2)}{1+\tan ^2(x / 2)}} d x\)
\( {\left[\because \sin \theta=\frac{2 \tan \frac{\theta}{2}}{1+\tan ^2 \frac{\theta}{2}} \text { and } \cos \theta=\frac{1-\tan ^2 \frac{\theta}{2}}{1+\tan ^2 \frac{\theta}{2}}\right]} =\int_0^{\pi / 2} \frac{\sec ^2(x / 2)}{2 \tan (x / 2)+1-\tan ^2(x / 2)} d x\)
Put \(\tan \frac{x}{2}=t \Rightarrow \sec ^2 \frac{x}{2} \cdot \frac{1}{2} d x=d t \Rightarrow \sec ^2 \frac{x}{2} d x=2 d t\)
Lower limit When x=0, then t=tan 0=0
Upper limit When \(x=\frac{\pi}{2}\), then \(t=\tan \frac{\pi}{4}=1\).
\(\therefore 2 I  =\int_0^1 \frac{2 d t}{2 t+1-t^2}=2 \int_0^1 \frac{d t}{-\left[t^2-2 t-1\right]}\)
\(=2 \int_0^1 \frac{d t}{-\left[(t-1)^2-1-1\right]}=2 \int_0^1 \frac{d t}{(\sqrt{2})^2-(t-1)^2}\)
\(=\left[\frac{2}{2 \sqrt{2}} \log \left|\frac{\sqrt{2}+t-1}{\sqrt{2}-t+1}\right|\right]_0^1 \)
\(=\frac{1}{\sqrt{2}}\left[\log \frac{\sqrt{2}+1-1}{\sqrt{2}-1+1}-\log \frac{\sqrt{2}+0-1}{\sqrt{2}-0+1}\right] \)
\(=\frac{1}{\sqrt{2}}\left[\log 1-\log \frac{\sqrt{2}-1}{\sqrt{2}+1}\right]\)
\(\left.=-\frac{1}{\sqrt{2}} \log \left[\frac{\sqrt{2}-1}{\sqrt{2}+1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}\right]\left|\frac{a+x}{a-x}\right|+C\right] \ [(\because \log 1=0] \)
\(=\frac{-1}{\sqrt{2}} \log \frac{2-1}{(\sqrt{2}+1)^2}\ {\left[(a-b)(a+b)=a^2-b^2\right] }\)
\(=\frac{-1}{\sqrt{2}} \log \frac{1}{(\sqrt{2}+1)^2} \Rightarrow 2 I=\frac{2}{\sqrt{2}} \log (\sqrt{2}+1) \)
\(\Rightarrow I =\frac{1}{\sqrt{2}} \log (\sqrt{2}+1)\)
Hence proved. 

Let \(I=\int_0^{\pi / 2} x^2 \sin x d x\)
Now, \(\int_1 x_I^2 \sin x d x=-x^2 \cos x+2 \int_I^x \underset{\|}{\cos x} d x\)
[using integration by parts]
\(=-x^2 \cos x+2\left[x(\sin x)-\int 1 \cdot(\sin x) d x\right]\)
\(=-x^2 \cos x+2(x \sin x+\cos x)\)
\(\therefore I =\int_0^{\pi / 2} x^2 \sin x d x\)
\(=\left[-x^2 \cos x+2(x \sin x+\cos x)\right]_0^{\pi / 2} \)
\(=\left[-\left(\frac{\pi}{2}\right)^2 \cos \left(\frac{\pi}{2}\right)+2\left(\frac{\pi}{2} \sin \frac{\pi}{2}+\cos \frac{\pi}{2}\right)\right.\)
\(=-\frac{\pi^2}{4} \times 0+2\left(\frac{\pi}{2}+0\right)-2(0+1)=\pi-2\)

Let \(I=\int_{\pi / 6}^{\pi / 3} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x\)
Put \(\sin x-\cos x=t\)
\(\Rightarrow \quad(\cos x+\sin x) d x=d t\)
Lower limit When \(x=\frac{\pi}{6}\), then
\(t=\sin \frac{\pi}{6}-\cos \frac{\pi}{6}=\frac{1}{2}-\frac{\sqrt{3}}{2}=\frac{1-\sqrt{3}}{2}\)
Upper limit When \(x=\frac{\pi}{3}\), then
\(t=\sin \frac{\pi}{3}-\cos \frac{\pi}{3}=\frac{\sqrt{3}}{2}-\frac{1}{2}=\frac{\sqrt{3}-1}{2}\)
Also, \(\text { Also, } \quad(\sin x-\cos x)^2=t^2 \)
\( \Rightarrow \quad \sin ^2 x+\cos ^2 x-2 \sin x \cos x=t^2\)
\(\Rightarrow \quad 1-\sin 2 x=t^2\)
\(\Rightarrow \quad \sin 2 x=1-t^2\)
\(\therefore  =\int_{\frac{1-\sqrt{3}}{2}}^{\frac{\sqrt{3}-1}{2}} \frac{d t}{\sqrt{1-t^2}}=\left[\sin ^{-1} t\right]_{\frac{\sqrt{3}-\sqrt{3}}{2}}^2\)
\(=\sin ^{-1}\left(\frac{\sqrt{3}-1}{2}\right)-\sin ^{-1}\left(\frac{1-\sqrt{3}}{2}\right)\)
\(=\sin ^{-1}\left(\frac{\sqrt{3}-1}{2}\right)+\sin ^{-1}\left(\frac{\sqrt{3}-1}{2}\right) =2 \sin ^{-1}\left(\frac{\sqrt{3}-1}{2}\right)\)

Let \(I=\int_{-\pi / 2}^{\pi / 2} \frac{\cos x}{1+e^x} d x=\int_{-\pi / 2}^0 \frac{\cos x}{1+e^x} d x+\int_0^{\pi / 2} \frac{\cos x}{1+e^x} d x\)
In first integral, put \(x=-t \Rightarrow d x=-d t\)
Lower limit When \(x=-\frac{\pi}{2}\), then \(t=\frac{\pi}{2}\).
Upper limit When x=0, then t=0
\(\therefore I=\int_{\pi / 2}^0 \frac{\cos t}{1+e^{-t}}(-d t)+\int_0^{\pi / 2} \frac{\cos x}{1+e^x} d x\)
\(=\int_0^{\pi / 2} \frac{\cos t}{1+\frac{1}{e^t}} d t+\int_0^{\pi / 2} \frac{\cos x}{1+e^x} d x\)
\({\left[\because-\int_a^0 f(x) d x=\int_0^a f(x) d x\right]}\)
\(=\int_0^{\pi / 2} \frac{e^t \cos t}{e^t+1} d t+\int_0^{\pi / 2} \frac{\cos x}{1+e^x} d x \)
\(=\int_0^{\pi / 2} \frac{e^x \cos x}{e^x+1} d x+\int_0^{\pi / 2} \frac{\cos x}{1+e^x} d x\)
\( {\left[\because \int_a^b f(x) d x=\int_a^b f(t) d t\right]}\)
\(=\int_0^{\pi / 2} \frac{\left(1+e^x\right) \cos x}{\left(e^x+1\right)} d x=\int_0^{\pi / 2} \cos x d x\)
\(=[\sin x]_0^{\pi / 2}=\sin \frac{\pi}{2}-0=1\)

Let \(I =\int_0^{\pi / 4} \frac{d x}{\cos ^3 x \sqrt{2 \sin 2 x}} \)
\( =\int_0^{\pi / 4} \frac{d x}{\cos ^3 x \sqrt{2(2 \sin x \cos x)}}\)
\( =\frac{1}{2} \int_0^{\pi / 4} \frac{d x}{\cos ^3 x \cos ^{1 / 2} x \sin ^{1 / 2} x} \)
\( =\frac{1}{2} \int_0^{\pi / 4} \frac{d x}{\cos ^{7 / 2} x \sin ^{1 / 2} x} \)
\( =\frac{1}{2} \int_0^{\pi / 4} \frac{\sec ^4 x}{\cos ^{\frac{7}{2}-4} x \sin ^{1 / 2} x} d x\)
[dividing numerator and denominator by \(\cos ^4 x\) ]
\(=\frac{1}{2} \int_0^{\pi / 4} \frac{\sec ^4 x}{\cos ^{-1 / 2} x \sin ^{1 / 2} x} d x\)
\(=\frac{1}{2} \int_0^{\pi / 4} \frac{\sec ^2 x\left(1+\tan ^2 x\right)}{\tan ^{1 / 2} x} d x\)
\(\left[\because \sec ^2 \theta-\tan ^2 \theta=1\right]\)
Put \(\tan x=t \Rightarrow \sec ^2 x d x=d t\)
Lower limit When x=0, then t=tan 0=0
Upper limit When \(x=\frac{\pi}{4}\), then \(t=\tan \frac{\pi}{4}=1\)
\(\therefore I  =\frac{1}{2} \int_0^1\left(\frac{1+t^2}{t^{1 / 2}}\right) d t\)
\( =\frac{1}{2} \int_0^1\left(t^{-1 / 2}+t^{3 / 2}\right) d t \)
\(=\frac{1}{2}\left[2 t^{1 / 2}+\frac{2}{5} t^{5 / 2}\right]_0^1\)
\(=\left[t^{1 / 2}+\frac{1}{5} t^{5 / 2}\right]_0^1\)
\(=(1)^{1 / 2}+\frac{1}{5}(1)^{5 / 2}-0 =1+\frac{1}{5}=\frac{6}{5}\)

Let \(I=\int_{-\pi}^\pi(\cos a x-\sin b x)^2 d x\)
\(=\int_{-\pi}^\pi\left(\cos ^2 a x+\sin ^2 b x-2 \cos a x \sin b x\right) d x \)
\(=\int_{-\pi}^\pi\left(\cos ^2 a x+\sin ^2 b x\right) d x-2 \int_{-\pi}^\pi \cos a x \sin b x d x \)
\(=I_1-I_2\)
Now consider, 
\(I_1=  \int_{-\pi}^\pi\left(\cos ^2 a x+\sin ^2 b x\right) d x \) [be an even function] 
\(=  2 \int_0^\pi\left(\cos ^2 a x+\sin ^2 b x\right) d x\)
\(\left[\because \int_{-a}^a f(x) d x=2 \int_0^a f(x) d x, \text { if } f(x) \text { is even }\right]\)
\(=2 \int_0^\pi\left(\frac{1+\cos 2 a x}{2}+\frac{1-\cos 2 b x}{2}\right) d x\)
\(=\int_0^\pi(1+\cos 2 a x+1-\cos 2 b x) d x =\int_0^\pi(2+\cos 2 a x-\cos 2 b x) d x\)
\( =\left(2 x+\frac{\sin 2 a x}{2 a}-\frac{\sin 2 b x}{2 b}\right)_0^\pi\)
\(=\left(2 \pi+\frac{\sin 2 a \pi}{2 a}-\frac{\sin 2 b \pi}{2 b}\right)-0\)
\(=2 \pi+\frac{\sin 2 a \pi}{2 a}-\frac{\sin 2 b \pi}{2 b}\)
and \(I_2=2 \int_{-\pi}^\pi(\cos a x \sin b x) d x\) [be an odd function]
\(=0\left[\begin{array}{cc} \because \int_{-a}^a f(x) d x=2 \int_0^a f(x) d x \text {, if } f(x) \text { is even } \\ 0, \text { if } f(x) \text { is odd } \end{array}\right]\)
\(\therefore I=I_1-I_2=2 \pi+\frac{\sin 2 a \pi}{2 a}-\frac{\sin 2 b \pi}{2 b}\)

Let \(I=\int_0^\pi \frac{x}{1+\sin \alpha \sin x} d x\)
\(=\int_0^\pi \frac{(\pi-x)}{1+\sin \alpha \sin (\pi-x)} d x \) \({\left[\because \int_0^a f(x) d x=\int_0^a f(a-x) d x\right]}\)
\(=\int_0^\pi \frac{(\pi-x)}{1+\sin \alpha \sin x} d x \  {[\because \sin (\pi-x)=\sin x]}\)
On adding Eqs. (i) and (ii), we get
\(2 I=\int_0^\pi \frac{\pi}{1+\sin \alpha \sin x} d x=\pi \int_0^\pi \frac{d x}{1+\sin \alpha \sin x}\)
\(=\pi \int_0^\pi \frac{d x}{1+\sin \alpha\left(\frac{2 \tan (x / 2)}{1+\tan ^2(x / 2)}\right)}\)
\(=\pi \int_0^\pi \frac{1+\tan ^2(x / 2) d x}{1+\tan ^2(x / 2)+2 \sin \alpha \tan (x / 2)} \)
\(=\pi \int_0^\pi \frac{\sec ^2(x / 2) d x}{\tan ^2(x / 2)+2 \sin \alpha \cdot \tan (x / 2)+1}\)
\( \therefore \quad \text { Put } \tan \frac{x}{2}=t \Rightarrow \sec ^2 \frac{x}{2} \cdot \frac{1}{2} d x=d t \)
\(\Rightarrow \quad \sec ^2 \frac{x}{2} d x=2 d t\)
Lower limit When x=0, then t=0
Upper limit When \(x=\pi\), then \(t \rightarrow \infty\)
\( \therefore \quad 2 I=\pi \int_0^{\infty} \frac{2 d t}{t^2+2 \sin \alpha \cdot t+1}\)
\(\Rightarrow I  =\pi \int_0^m \frac{d t}{t^2+2 \sin \alpha \cdot t+\sin ^2 \alpha+\cos ^2 \alpha} \)
\(=\pi \int_0^{\infty} \frac{d}{(t+\sin \alpha)^2+(\cos \alpha)^2}\) \([ \because \sin ^2 \theta+\cos ^2 \theta=1]\)
\(=\frac{\pi}{\cos \alpha}\left[\tan ^{-1}\left(\frac{t+\sin \alpha}{\cos \alpha}\right)\right]_0^{\infty}\)
\(=\frac{\pi}{\cos \alpha}\left[\tan ^{-1}(\infty)-\tan ^{-1}(\tan \alpha)\right]\)
\(=\frac{\pi}{\cos \alpha}\left[\frac{\pi}{2}-\alpha\right]\)

Let \(I=\int_0^\pi e^{2 x} \sin \left(\frac{\pi}{4}+x\right) d x\)
Again, let \(I_1=\int e^{2 x} \sin \left(\frac{\pi}{4}+x\right) d x\)
\(=\sin \left(\frac{\pi}{4}+x\right) \int e^{2 x} d x-\int\left\{\frac{d}{d x} \sin \left(\frac{\pi}{4}+x\right) \int e^{2 x} d x\right\} d x\)
[using integration by parts]
\(=\sin \left(\frac{\pi}{4}+x\right) \frac{e^{2 x}}{2}-\int \cos \left(\frac{\pi}{4}+x\right) \frac{e^{2 x}}{2} d x\)
\(=\frac{e^{2 x}}{2} \sin \left(\frac{\pi}{4}+x\right)-\frac{1}{2} \int e^{2 x} \cos \left(\frac{\pi}{4}+x\right) d x \)
\(=\frac{e^{2 x}}{2} \sin \left(\frac{\pi}{4}+x\right) -\frac{1}{2}\left[\cos \left(\frac{\pi}{4}+x\right) \frac{e^{2 x}}{2}-\int-\sin \left(\frac{\pi}{4}+x\right) \frac{e^{2 x}}{2} d x\right]\)
\(=\frac{e^{2 x}}{2} \sin \left(\frac{\pi}{4}+x\right)-\frac{e^{2 x}}{4} \cos \left(\frac{\pi}{4}+x\right)-\frac{1}{4} \int e^{2 x} \sin \left(\frac{\pi}{4}+x\right) d x\)
\(\Rightarrow I_1=\frac{e^{2 x}}{4}\left\{2 \sin \left(\frac{\pi}{4}+x\right)-\cos \left(\frac{\pi}{4}+x\right)\right\}-\frac{1}{4} I_1\) [from Eq. (i)]
\(\Rightarrow I_1+\frac{1}{4} I_1=\frac{e^{2 x}}{4}\left\{2 \sin \left(\frac{\pi}{4}+x\right)-\cos \left(\frac{\pi}{4}+x\right)\right\}\)
\(\Rightarrow \quad \frac{5}{4} I_1=\frac{e^{2 x}}{4}\left\{2 \sin \left(\frac{\pi}{4}+x\right)-\cos \left(\frac{\pi}{4}+x\right)\right\}\)
\(\Rightarrow \quad I_1=\frac{e^{2 x}}{5}\left\{2 \sin \left(\frac{\pi}{4}+x\right)-\cos \left(\frac{\pi}{4}+x\right)\right\}\)
\(\therefore \quad I=\left[I_1\right]_0^\pi\)
\(=\left[\frac{e^{2 x}}{5}\left\{2 \sin \left(\frac{\pi}{4}+x\right)-\cos \left(\frac{\pi}{4}+x\right)\right\}\right]_0^\pi\)
\( =\frac{1}{5}\left[e^{2 \pi}\left\{2 \sin \left(\frac{\pi}{4}+\pi\right)-\cos \left(\frac{\pi}{4}+\pi\right)\right\}\right.\left.\left.-\quad e^0\left\{2 \sin \left(\frac{\pi}{4}+0\right)-\cos \left(\frac{\pi}{4}+0\right)\right\}\right]\right]\)
\(=\frac{1}{5}\left[c^{2 \pi}\left\{-2 \sin \frac{\pi}{4}+\cos \frac{\pi}{4}\right\}-e^0\left\{2 \sin \frac{\pi}{4}-\cos \frac{\pi}{4}\right\}\right] \)
\(=\frac{1}{5}\left[e^{2 \pi}\left\{-2 \times \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right\}-1\left\{2 \times \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\right\}\right]\)
\( =\frac{1}{5}\left[e^{2 \pi}\left\{-\frac{1}{\sqrt{2}}\right\}-\frac{1}{\sqrt{2}}\right] \)
\( =-\frac{1}{5 \sqrt{2}}\left[e^{2 \pi}+1\right]\)

Here, \(|x-1|,|x-2|\) and |x-4| occurs.
Now, define the absolute function as
\(|x-1|=\left\{\begin{array}{c} x-1, x \geq 1 \\ -(x-1), x<1 \end{array} ;|x-2|=\left\{\begin{array}{c} x-2, x \geq 2 \\ -(x-2), x<2 \end{array}\right.\right.\)
and \(|x-4|=\left\{\begin{array}{c}x-4, x \geq 4 \\ -(x-4), x<4\end{array}\right.\)
Let \(I=  \int_1^4(|x-1|+|x-2|+|x-4|) d x\)
\(=  \int_1^2(|x-1|+|x-2|+|x-4|) d x +\int_2^3(|x-1|+|x-2|+|x-4|) d x +\int_3^4(|x-1|+|x-2|+|x-4|) d x\)
\(=\int_1^2\{(x-1)-(x-2)-(x-4)\} d x+\int_2^3\{(x-1)+(x-2)-(x-4)\} d x+\int_3^4\{(x-1)+(x-2)-(x-4)\} d x\)
\(=\int_1^2(-x+5) d x+\int_2^3(x+1) d x+\int_3^4(x+1) d x\)
\(=\left(\frac{-x^2}{2}+5 x\right)_1^2+\left(\frac{x^2}{2}+x\right)_2^3+\left(\frac{x^2}{2}+x\right)_3^4\)
\(=\left(\frac{-4}{2}+10\right)-\left(\frac{-1}{2}+5\right)+\left(\frac{9}{2}+3\right)-\left(\frac{4}{2}+2\right)+\left(\frac{16}{2}+4\right)-\left(\frac{9}{2}+3\right)\)
\(=8-\frac{9}{2}+\frac{15}{2}-4+12-\frac{15}{2}=16-\frac{9}{2}=\frac{23}{2}\)

Let \(I=\int_{-\pi / 4}^{\pi / 4} \frac{x+\frac{\pi}{4}}{2-\cos 2 x} d x\)
\(=\int_{-\pi / 4}^{\pi / 4} \frac{x}{2-\cos 2 x} d x+\frac{\pi}{4} \int_{-\pi / 4}^{\pi / 4} \frac{d x}{2-\cos 2 x}\)
Let \(f(x)=\frac{x}{2-\cos 2 x}\)
\(\Rightarrow f(-x)=\frac{-x}{2-\cos 2(-x)}=\frac{-x}{2-\cos 2 x}=-f(x)\)
Hence, f is odd.
\(\therefore \quad \int_{-\pi / 4}^{\pi / 4} \frac{x}{2-\cos 2 x} d x=0\)
Let \(g(x)=\frac{1}{2-\cos 2 x}\) and \(g(-x)=\frac{1}{2-\cos 2 x}=g(x)\) 
Hence, g is even.
\(\therefore \frac{\pi}{4} \int_{-\pi / 4}^{\pi / 4} \frac{1}{2-\cos 2 x} d x=\frac{\pi}{4} \times 2 \int_0^{\pi / 4} \frac{1}{2-\cos 2 x} d x \)
\(=\frac{\pi}{2} \int_0^{\pi / 4} \frac{1}{2-\cos 2 x} d x\)
Hence, \( I=\frac{\pi}{2} \int_0^{\pi / 4} \frac{1}{2-\cos 2 x} d x\)
\( =\frac{\pi}{2} \int_0^{\pi / 4} \frac{1}{1+2 \sin ^2 x} d x \quad\left[\because \cos 2 \theta=1-2 \sin ^2 \theta\right]\)
\(=\frac{\pi}{2} \int_0^{\pi / 4} \frac{\sec ^2 x}{1+\tan ^2 x+2 \tan ^2 x} d x\)
[dividing numerator and denominator by \(\cos ^2 x\) ]
\(=\frac{\pi}{2} \int_0^{\pi / 4} \frac{\sec ^2 x}{1+3 \tan ^2 x} d x\)
Put \(\tan x=t \Rightarrow \sec ^2 x d x=d t\)
Lower limit When x=0, then t=0
Upper limit When \(x=\frac{\pi}{4}\), then t=1
\(\frac{\pi}{2} \int_0^1 \frac{1}{1+3 t^2} d t=\frac{\pi}{2} \times \frac{1}{3} \int_0^1 \frac{1}{\left(\frac{1}{\sqrt{3}}\right)^2+t^2} d t\)
\(=\frac{\pi}{6}(\sqrt{3})\left[\tan ^{-1} \sqrt{3} t\right]_0^1\)
\(\left[\because \int \frac{d x}{x^2+a^2}=\frac{1}{a} \tan ^{-1}\left(\frac{x}{a}\right)+C\right]\)
\(=\frac{\pi}{6}(\sqrt{3})\left(\tan ^{-1} \sqrt{3}-\tan ^{-1} 0\right)=\frac{\pi}{6}(\sqrt{3})\left(\frac{\pi}{3}-0\right)\)
\(=\frac{\pi}{6}(\sqrt{3})\left(\frac{\pi}{3}\right)=\frac{\sqrt{3} \pi^2}{18}\)

To prove \(\int_0^x f(x) d x=\int_0^x f(a-x) d x\)
Consider, RHS \(=\int_0^2 f(a-x) d x\)
Putting t=a-x, then d t=-d x
Lower limit When x=0, then t=a
Upper limit When x=a, then t=0
Now, RHS \(=-\int_0^0 f(t) d t=\int_0^2 f(t) d t\)
\(=\int_0^2 f(x) d x=\) LHS Hence proved.
Now, let \(l=\int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x\)
\(= \int_0^\pi \frac{(\pi-x) \sin (\pi-x) d x}{1+\cos ^2(\pi-x)}\)
\(\left[\because \int_0^a f(x) d x=\int_0^a f(a-x) d x\right]\)
\(=\int_0^\pi \frac{(\pi-x) \sin x d x}{1+\cos ^2 x} \)
\(=\pi \int_0^\pi \frac{\sin x d x}{1+\cos ^2 x}-\int_0^\pi \frac{x \sin x d x}{1+\cos ^2 x} \)
\( I=\pi \int_0^\pi \frac{\sin x d x}{1+\cos ^2 x}-1\)
\(\Rightarrow  2 I=\pi \int_0^\pi \frac{\sin x}{1+\cos ^2 x} d x\)
\(\Rightarrow  I=\frac{\pi}{2} \int_0^\pi \frac{\sin x}{1+\cos ^2 x} d x\)
Putting \(\cos x=t \Rightarrow-\sin x d x=d t\)
Lower limit When x=0, then t=1
Upper limit When \(x=\pi\), then t=-1
Now, \(I =-\frac{\pi}{2} \int_1^{-1} \frac{d t}{1+t^2}=\frac{\pi}{2} \int_{-1}^1 \frac{d t}{1+t^2} \)
\(=\frac{\pi}{2}\left[\tan ^{-1} t\right]_{-1}^1=\frac{\pi}{2}\left[\tan ^{-1}(1)-\tan ^{-1}(-1)\right] \)
\(=\frac{\pi}{2}\left[\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)\right]=\frac{\pi}{2}\left[\frac{\pi}{2}\right]=\frac{\pi^2}{4}\)

Let \( l=\int_0^\pi \frac{x \sin x}{1+\cos ^2 x} d x\)
\( \Rightarrow l=\int_0^\pi \frac{(\pi-x) \sin (\pi-x)}{1+\cos ^2(\pi-x)} d x\)
\(\left[\because \int_0^a f(x) d x=\int_0^2 f(a-x) d x\right]\)
\(=\int_0^\pi \frac{(\pi-x) \sin x d x}{1+\cos ^2 x}\)
On adding Eqs. (i) and (ii), we get
\(=\pi \int_0^\pi \frac{\sin x d x}{2} \Rightarrow l=\frac{\pi}{2} \int_0^\pi \frac{\sin x}{1+c} d x\)
Using \(\int_0^\pi f(x) d x=2 \int_0^\pi f(x) d x\), if f(2 a-x)=f(x)
\(\therefore \quad I=\frac{\pi}{2} \times 2 \int_0^\pi \frac{\sin x}{1+\cos ^2 x} d x\)
\(\Rightarrow \quad l=\pi \int_0^\pi \frac{\sin x}{1+\cos ^2 x} d x\)
Put \(\cos x=t\), then -sin x d x=d t
Lower limit When x=0 then t=1
Upper limit When \(x=\frac{\pi}{2}\), then t=0
\(\therefore I=\pi \int_1^0 \frac{-d t}{1+t^2} \Rightarrow I=-\pi\left[\tan ^{-1} t\right]_1^0\)
\(\Rightarrow  I=-\pi\left[\tan ^{-1} 0-\tan ^{-1} 1\right]\)
\(\Rightarrow  I=-\pi\left[0-\frac{\pi}{4}\right]=\frac{\pi^2}{4}\)

\(l =\int_{-2}^1 \sqrt{5-4 x-x^2} d x\)
\(=\int_{-2}^1 \sqrt{5-\left(x^2+4 x+4\right)+4} d x\)
\(=\int_{-2}^1 \sqrt{9-(x+2)^2} d x\)
Let \(x+2=t \Rightarrow d x=d t\)
Lower limit When x=-2, then t=0
Upper limit When x=1, then t=3
\(\therefore l  =\int_0^3 \sqrt{3^2-t^2} d t\)
\(=\left[\frac{1}{2} t \sqrt{3^2-t^2}+\frac{1}{2} 3^2 \sin ^{-1}\left(\frac{t}{3}\right)\right]_0^3\)
\(=0+\frac{9}{2} \sin ^{-1} 1-0=\frac{9}{2} \cdot \frac{\pi}{2}=\frac{9 \pi}{4}\)