Evaluate the definite integral in Exercise: ^ 4 _ 0 + 9+16 2x dx - Vidyadip

Question

Evaluate the definite integral in Exercise:

$\int^{\frac{\pi}{4}}_{0}\frac{\sin\text{x}+\cos\text{x}}{9+16\sin2\text{x}}\text{dx}$

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Answer

$\text{Let I}=\int^{\frac{\pi}{4}}_{0}\frac{\sin\text{x}+\cos\text{x}}{9+16\sin2\text{x}}\text{dx}$
Also, let $\sin\text{x}-\cos\text{x}=\text{t}\Rightarrow(\cos\text{x}+\sin\text{x})\text{dx}=\text{dt}$
when $\text{x}=0,\text{t}=-1$ and when $\text{x}=\frac{\pi}{4},\text{t}=0$
$\Rightarrow(\sin\text{x}-\cos\text{x})^{2}=\text{t}^{2}$
$\Rightarrow\sin^{2}\text{x}+\cos^{2}\text{x}-2\sin\text{x}\cos\text{x}=\text{t}^{2}$
$\Rightarrow1-\sin2\text{x}=\text{t}^{2}$
$\Rightarrow\sin2\text{x}=1-\text{t}^{2}$
$\therefore\text{I}=\int^{0}\limits_{-1}\frac{\text{dt}}{9+16(1-\text{t}^{2})}$
$=\int^{0}\limits_{-1}\frac{\text{dt}}{9+16-16\text{t}^{2}}$
$=\int^{0}\limits_{-1}\frac{\text{dt}}{25-16\text{t}^{2}}=\int^{0}\limits_{-1}\frac{\text{dt}}{(5)^{2}-(4\text{t})^{2}}$
$=\frac{1}{4}\Bigg[\frac{1}{2(5)}\log\Bigg|\frac{5+4\text{t}}{5-4\text{t}}\Bigg|\Bigg]^{0}_{-1}$
$=\frac{1}{40}\bigg[\log(1)-\log\bigg|\frac{1}{9}\bigg|\bigg]^{0}_{-1}$
$=\frac{1}{40}\log9$

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