Evaluate: _0^ 4 + 16+9 2x - Vidyadip

Question

Evaluate: $\int\limits_0^{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{16+9\sin2\text{x}}$

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Answer

$\int\limits_0^\frac{\pi}{4}\frac{\sin\text{x}+\cos\text{x}}{16+9\sin2\text{x}}\text{dx}$
$\int\limits_0^{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{16+9\big[1-(\sin\text{x}-\cos\text{x})^2\big]}\text{dx}$
$\int\limits_0^{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{25-9(\sin\text{x}-\cos\text{x})^2}\text{dx}$
$\sin\text{x}=\cos\text{x}=\text{t}$
$(\cos\text{x}+\sin\text{x})\text{dx}=\text{dt}$
$\int\limits_{-1}^0\frac{\text{dx}}{25-9\text{t}^2}$
$\int\limits_{-1}^0\frac{1}{9\Big(\frac{25}{9}-\text{t}^2\Big)}\text{dt}$
$\frac{1}{9}\int\limits_{-1}^0\frac{1}{\Big(\frac{5}{3}\Big)^2-\text{t}^2}$
$\frac{1}{9}\Bigg[\frac{1}{2\times\frac{5}{3}}\log\begin{vmatrix}\frac{\frac{5}{3}+\text{t}}{\frac{5}{3}-\text{t}}\end{vmatrix}\Bigg]_{-1}^0$
$\frac{1}{9}\times\frac{1}{\frac{10}{3}}\Bigg(\log\begin{vmatrix}\frac{\frac{5}{3}}{\frac{5}{3}}\end{vmatrix}\Bigg)-\log\begin{vmatrix}\frac{\frac{2}{3}}{\frac{8}{3}} \end{vmatrix}$
$\frac{1}{30}\Big[\log1-\log\frac{1}{4}\Big]$
$\frac{1}{30}\log4=\frac{1}{15}\log2$

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