Evaluate the following integrals: _ 0 ^ 4 + 3+ 2x dx - Vidyadip

Question

Evaluate the following integrals:
$\int\limits_{0}^{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{3+\sin2\text{x}}\text{ dx}$

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Answer

Let $\text{I}=\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{3+\sin2\text{x}}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{4-(1-\sin2\text{x})}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{4-(\sin^2\text{x}+\cos^2\text{x}-2\sin\text{x}\cos\text{x})}\text{ dx}$
$=\int_{0}^\limits{\frac{\pi}{4}}\frac{\sin\text{x}+\cos\text{x}}{4-(\sin\text{x}-\cos\text{x})^2}\text{ dx}$
Put $\sin\text{x}-\cos\text{x}=\text{z}$
$\therefore\ (\cos\text{x}+\sin\text{x})\text{dx}=\text{dz}$
When $\text{x}\rightarrow0,\text{z}\rightarrow-1$ $(\text{z}=\sin0-\cos0=0-1=-1)$
When $\text{x}\rightarrow\frac{\pi}{4},\text{z}\rightarrow0$ $\Big(\text{z}=\sin\frac{\pi}{4}-\cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}=0\Big)$
$\therefore\ \text{I}=\int_{-1}^\limits{0}\frac{\text{dz}}{2^2-\text{z}^2}$
$=\frac{1}{2\times2}\log\Big[\log\Big(\frac{2+\text{z}}{2-\text{z}}\Big)\Big]^0_{-1}$
$=\frac{1}{4}\Big(\log1-\log\frac{1}{3}\Big)$
$=\frac{1}{4}\big[0-\big(\log1-\log3\big)\big]$
$=-\frac{1}{4}\big(0-\log3\big)$
$=\frac{1}{4}\log3$

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