Question

please solve it fast☹️​

Answer 1

Step-by-step explanation:

this question's answer is (1+c)

Answer 2

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$\displaystyle \rm \int_{0}^{\frac{\pi}{4}} \dfrac{1+sin\ 2x}{cos\ x+sin\ x}\ dx$

$\bullet\ \; \rm \orange{sin^2 \theta +cos^2 \theta =1}$

$\bullet\ \; \rm \orange{sin\ 2 \theta = 2\ sin\ \theta\ .\ cos\ \theta}$

$\to \displaystyle \rm \int_{0}^{\frac{\pi}{4}} \dfrac{sin^2x+cos^2x+2\ sin\ x\ .\ cos\ x}{sin\ x+cos\ x}\ dx$

$\bullet\ \; \rm \orange{(A+B)^2=A^2+B^2+2AB}$

$\to \displaystyle \rm \int_{0}^{\frac{\pi}{4}} \dfrac{(sin\ x+cos\ x)^2}{sin\ x+cos\ x}\ dx$

$\to \displaystyle \rm \int_{0}^{\frac{\pi}{4}} sin\ x+cos\ x\ dx$

$\to \displaystyle \rm \int_{0}^{\frac{\pi}{4}} sin\ x\ dx + \int_{0}^{\frac{\pi}{4}} cos\ x\ dx$

$\bullet\ \; \orange{ \rm \int sin\ x\ dx= - cos\ x}\\\\\bullet\ \; \orange{ \rm \int cos\ x\ dx = sin\ x}$

$\to \displaystyle \rm - cos\ x \big]_{0}^{\frac{\pi}{4}} + sin\ x \big]_{0}^{\frac{\pi}{4}} + c$

$\to \displaystyle \rm - \big[ cos\ \frac{\pi}{4} -cos\ 0 \big] + \big[ sin\ \frac{\pi}{4} - sin\ 0 \big] + c$

$\to \rm - \bigg[ \dfrac{1}{\sqrt{2}} - 1 \bigg] + \bigg[ \dfrac{1}{\sqrt{2}} - 0 \bigg] + c$

$\leadsto \rm 1+c\ \; \pink{\bigstar}$

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