Question

Evaluate integratin ki power π\4 or 0. √1+sin2x.dx​

Answer 1

Step-by-step explanation:

$\huge\bf{\mid{\overline{\underline{ANSWER:-}}\mid}}$

$\int \limits_{0}^ { \frac{\pi}{4} } \sqrt{1 + \sin(2x) } \\ \\ we \: \: know \: \: \\ \boxed{\sin(2x) = 2 \sin(x) \cos(x) } \\ \\ \int \limits_{0}^ { \frac{\pi}{4} } \sqrt{1 + 2\sin(x) \cos(x) } \\ \\ \boxed{ { \sin}^{2} (x) + { \cos}^{2} (x) = 1} \\ \\ \int \limits_{0}^ { \frac{\pi}{4} } \sqrt{{ \sin}^{2} (x) + { \cos}^{2} (x) + 2\sin(x) \cos(x) } \\ \\ \boxed{ {a}^{2} + {b}^{2} + 2ab = {(a + b)}^{2} } \\ \\ \int \limits_{0}^ { \frac{\pi}{4} } \sqrt{ {(\sin x+ \cos x )}^{2} } \\ \\ \int \limits_{0}^ { \frac{\pi}{4} }( \sin(x) ) + \int \limits_{0}^ { \frac{\pi}{4} }( \cos(x) )$

$\left \lceil - \cos( \frac{\pi}{4} ) + \cos(0) \right \rceil + \left \lceil \sin( \frac{\pi}{4} ) - \sin(0) \right \rceil \\ \\ \cancel{ - \frac{1}{ \sqrt{2} }} + 1 + \cancel{ \frac{1}{ \sqrt{2} }} + 0 \\ \\ \huge \boxed{1}$