Mathematics

Evaluate $$\displaystyle\int_{0}^{\pi/2}\dfrac{\sin x+\cos x}{3+\sin 2x}dx$$


SOLUTION
Let $$I=\displaystyle\int_{0}^{\pi/4}\dfrac{\sin x+\cos x}{3+\sin 2x}\ dx$$

$$I=\displaystyle\int_{0}^{\pi/4}\dfrac{\sin x+\cos x}{3+1-1+\sin 2x}\ dx$$

$$I=\displaystyle\int_{0}^{\pi/4}\dfrac{\sin x+\cos x}{4-(1-\sin 2x)} \ dx$$

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

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

Let $$(\sin x-\cos x)=t \Rightarrow (\cos x+\sin x)=\dfrac {dt}{dx} $$

$$\therefore I_t=\displaystyle\int\dfrac{1}{4-(t)^{2}}\ dt$$

$$\therefore I=\dfrac{1}{4}\left[\log\left|\dfrac{2+\sin x-\cos x}{2-\sin x+\cos x}\right|\right]_{0}^{\pi/4}=-\dfrac{1}{4}\log\dfrac{1}{3}=\dfrac{1}{4}\log_{e}3$$
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Subjective Medium Published on 17th 09, 2020
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