Let I= ^ 4 _ - 4 ( + )dx (i) I= _ - 4 ^ 4 ( 4 - 4 -x )+ ( 4 - 4 -x ) dx = ^ 4 _ - 4 (-x)+ (-x) dx and I= ^ 4 _ - 4 ( - )dx (ii) From Eqs. (i) and (ii) 2I= ^ 4 _ - 4 2x dx 2I= ^ 4 _ 0 2x dx (iii) [ ^ a _ -a f(x)dx=2 ^ a _ 0 f(x), if f(-x)=f(x) ] Put 2x=t dx= dt 2 As x 0, then t 0 and x 4 then t 2 2I=1 2 ^ 2 _ 0 dt (iv) 2I=1 2 ^ 2 _ 0 ( 2 -t ) dt [ ^ a _ 0 f(x)dx= ^ a _ 0 f(a-x)dx ] 2I=1 2 ^ 2 _ 0 dt (v) On adding Eqs. (iv) and (v), we get 4I=1 2 ^ 2 _ 0 dt 4I=1 2 ^ 2 _ 0 2t 2 dt 4I=1 2 ^ 2 _ 0 2x dx-1 2 ^ 2 _ 0 2dx 4I=1 2 ^ 2 _ 0 ( 2 -2x ) dx- 2 4 4I=1 2 ^ 2 _ 0 2x dx- 4 2 4I= ^ 4 _ 0 2x dx- 4 2 [ ^ 2a _ 0 f(x)dx=2 ^ a _ 0 f(x)dx ] 4I=2I- 4 2 [from Eq.(iii)] I= 8 2 8 (1 2 )

^ 4 _ - 4 ( + )dx

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$\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\log(\sin\text{x}+\cos\text{x})\text{dx}$

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Answer

Let $\text{I}=\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\log(\sin\text{x}+\cos\text{x})\text{dx}\ \ \dots(\text{i})$
$ \text{I}=\int\limits_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\log\bigg\{\sin\Big(\frac{\pi}{4}-\frac{\pi}{4}-\text{x}\Big)+\cos\Big(\frac{\pi}{4}-\frac{\pi}{4}-\text{x}\Big)\bigg\}\text{dx}$
$ =\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\log\{\sin(-\text{x})+\cos(-\text{x})\}\text{dx} $
and $\text{I}=\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\log(\cos\text{x}-\sin\text{x})\text{dx}\ \ \dots(\text{ii})$
From Eqs. (i) and (ii)
$2\text{I}=\int\limits^{\frac{\pi}{4}}_{-\frac{\pi}{4}}\log\cos2\text{x dx}$
$2\text{I}=\int\limits^{\frac{\pi}{4}}_{0}\log\cos2\text{x dx}\ \ \dots(\text{iii})$
$ \bigg[\because\ \int\limits^{\text{a}}_{-\text{a}}\text{f(x)dx}=2\int\limits^{\text{a}}_{0}\text{f(x), if f}(-\text{x})=\text{f(x)}\bigg]$
Put $2\text{x}=\text{t}\Rightarrow\ \text{dx}=\frac{\text{dt}}{2} $
As $\text{x}\rightarrow0,$ then $\text{t}\rightarrow0$
and $\text{x}\rightarrow\frac{\pi}{4}$ then $\text{t}\rightarrow\frac{\pi}{2}$
$2\text{I}=\frac{1}{2}\int\limits^{\frac{\pi}{2}}_{0}\log\cos\text{t dt}\ \ \dots(\text{iv})$
$2\text{I}=\frac{1}{2}\int\limits^{\frac{\pi}{2}}_{0}\log\cos\bigg(\frac{\pi}{2}-\text{t}\bigg)\text{ dt}$ $ \bigg[\because\ \int\limits^{\text{a}}_{0}\text{f(x)dx}=\int\limits^{\text{a}}_{0}\text{f}(\text{a}-\text{x})\text{dx}\bigg] $
$ \Rightarrow\ 2\text{I}=\frac{1}{2}\int\limits^{\frac{\pi}{2}}_{0}\log\sin\text{t dt}\ \ \dots(\text{v}) $
On adding Eqs. (iv) and (v), we get
$ 4\text{I}=\frac{1}{2}\int\limits^{\frac{\pi}{2}}_{0}\log\sin\text{t}\cos\text{t dt} $
$\Rightarrow\ 4\text{I}=\frac{1}{2}\int\limits^{\frac{\pi}{2}}_{0}\log\frac{\sin2\text{t}}{2}\text{dt} $
$ \Rightarrow\ 4\text{I}=\frac{1}{2}\int\limits^{\frac{\pi}{2}}_{0}\log\sin2\text{x dx}-\frac{1}{2}\int\limits^{\frac{\pi}{2}}_{0}\log2\text{dx}$
$ \Rightarrow\ 4\text{I}=\frac{1}{2}\int\limits^{\frac{\pi}{2}}_{0}\log\sin\Big(\frac{\pi}{2}-2\text{x}\Big)\text{ dx}-\log2\cdot\frac{\pi}{4} $
$ \Rightarrow\ 4\text{I}=\frac{1}{2}\int\limits^{\frac{\pi}{2}}_{0}\log\cos2\text{x}\text{ dx}-\frac{\pi}{4}\log2$
$\Rightarrow\ 4\text{I}=\int\limits^{\frac{\pi}{4}}_{0}\log\cos2\text{x}\text{ dx}-\frac{\pi}{4}\log2$ $\bigg[\because\ \int\limits^{2\text{a}}_{0}\text{f(x)dx}=2\int\limits^{\text{a}}_{0}\text{f(x)dx}\bigg]$
$\Rightarrow\ 4\text{I}=2\text{I}-\frac{\pi}{4}\log2$ [from Eq.(iii)]
$\text{I}=\frac{\pi}{8}\log2\frac{\pi}{8}\log\bigg(\frac{1}{2}\bigg)$

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