Mathematics

The value of the integral $\underset{0}{\overset{\pi / 4}{\int}} \dfrac{\sin \, x + \cos \, x}{3 + \sin \, 2x} dx$, is

$\dfrac{1}{4} \log \, 3$

SOLUTION
$\int_{0}^{\frac{\pi}{4}}\dfrac{\sin x+\cos x}{3+\sin 2x}dx$

$=\int_{0}^{\frac{\pi}{4}}\dfrac{\sin x+\cos x}{4-(\sin x-\cos x)^2}dx$

substitute $u=\sin x-\cos x\Rightarrow (\sin x+\cos x)dx=dt$

$x(0,\dfrac{\pi}{4})\rightarrow t(-1,0)$

$=\int_{-1}^{0}\dfrac{dt}{4-t^2}$

$=\int_{-1}^{0}\dfrac{dt}{2^2-t^2}$

$=\dfrac{1}{2\times 2}\left [ \ln \dfrac{2+t}{2-t} \right ]_{-1}^0$

$=\dfrac{1}{4}\left \{ \ln 1 -\ln \dfrac{1}{3} \right \}$

$=\dfrac{1}{4}\ln 3$

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Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

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