Question

$intigral \: \sqrt{1 + \sin(2x \ \: } dx \: limit \: 0to\pi \div 4$
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Answer 1

GIVEN :–

$\\ \bf \implies \int_{0}^{ \frac{\pi}{4}} \sqrt{1 + \sin(2x)}.dx$

TO FIND :–

• Value of integration = ?

SOLUTION :–

• Let –

$\\ \bf \implies I = \int_{0}^{ \frac{\pi}{4}} \sqrt{1 + \sin(2x)}.dx$

• Using identity –

$\\ \bf \to \: \: \sin(2x) = 2 \sin(x) \cos(x)$

$\\ \bf \to \: \: \sin^{2} (x) + \cos^{2} (x) = 1$

$\\ \bf \implies I = \int_{0}^{ \frac{\pi}{4}} \sqrt{\sin^{2} (x) + \cos^{2} (x) +2\sin(x) \cos(x) }.dx$

$\\ \bf \implies I = \int_{0}^{ \frac{\pi}{4}} \sqrt{ \{\sin(x) + \cos(x) \}^{2}}.dx$

$\\ \bf \implies I = \int_{0}^{ \frac{\pi}{4}} \{ \sin(x) + \cos(x) \}.dx$

$\\ \bf \implies I = \int_{0}^{ \frac{\pi}{4}} \sin(x) .dx + \int_{0}^{ \frac{\pi}{4}} \cos(x).dx$

$\\ \bf \implies I = \bigg[ - \cos(x) \bigg]_{0}^{ \frac{\pi}{4}} + \bigg[ \sin(x) \bigg]_{0}^{ \frac{\pi}{4}}$

$\\ \bf \implies I = - \bigg[\cos \bigg(\dfrac{\pi}{4} \bigg) - \cos(0) \bigg]+ \bigg[\sin \bigg(\dfrac{\pi}{4} \bigg) - \sin(0) \bigg]$

$\\ \bf \implies I = - \bigg[ \dfrac{1}{ \sqrt{2} } - 1 \bigg]+ \bigg[\dfrac{1}{ \sqrt{2} }-0 \bigg]$

$\\ \bf \implies I = \bigg[1 - \dfrac{1}{ \sqrt{2} } \bigg]+ \bigg[\dfrac{1}{ \sqrt{2} }\bigg]$

$\\ \bf \implies I =1 - \cancel\dfrac{1}{ \sqrt{2} } + \cancel \dfrac{1}{ \sqrt{2} }$

$\\ \large\implies{ \boxed{ \bf I =1}}$