Mathematics

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


ANSWER

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


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
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