Question

Find definite integral of :
$\int \limits_{ - \frac{\pi}{4} }^{ \frac{ \pi}{4} } \: ln(sin \: x + cos \: x)$

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Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{I = - \frac{\pi}{4} (ln \: 2)}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline{\bold{Given :}}} \\ \tt: \implies \int \limits_{ - \frac{\pi}{4} }^{ \frac{ \pi}{4} } \: ln(sin \: x + cos \: x)dx \\ \\ \red{\underline{\bold{To \: Find:}}} \\ \tt: \implies \int \limits_{ - \frac{\pi}{4} }^{ \frac{ \pi}{4} } \: ln(sin \: x + cos \: x)dx =?$

• According to given question :

$\bold{As \: we \: know \: that} \\ \tt: \implies \int \limits_{ - \frac{\pi}{4} }^{ \frac{ \pi}{4} } \: ln(sin \: x + cos \: x)dx = I -----(1)\\ \\ \green{\to} \tt \: sin \: x + cos \: x = \sqrt{2} \bigg( \frac{1}{ \sqrt{2} } sin \: x + \frac{1}{ \sqrt{2} } cos \: x) \\ \\ \green{\to} \tt \: sin \: x + cos \: x = \sqrt{2} \bigg(cos \: \frac{\pi}{4} \times sin \: x + sin \: \frac{\pi}{4} \times cos \: x\bigg)\\\\ \green{\star}\tt\:sin(A+B)=(sin\:A\:cos\:A+cos\:A\:sin\:B )\\ \\ \green{\to} \tt \: sin \: x + cos \: x = \sqrt{2} \bigg(sin \bigg(x + \frac{\pi}{4} \bigg) \bigg)-----(2)\\\\ \text{Putting\:value\:of\:(2)\:in\:(1)} \\ \\ \tt: \implies I= \int \limits_{ - \frac{\pi}{4} }^{ \frac{\pi}{4} } ln \: \sqrt{2} \: sin \bigg(x + \frac{\pi}{4} \bigg )dx \\ \\ \begin{cases} \green{\circ} \tt \: x + \frac{\pi}{4} = \theta \implies dx =d \theta \\ \\ \green{\circ} \tt \:x = - \frac{\pi}{4} \implies \theta = 0 \\ \\ \green{\circ} \tt \:x = \frac{\pi}{4} \implies \theta = \frac{\pi}{2} \end{cases}$

$\tt : \implies I= \int \limits_{0}^{ \frac{\pi}{2} } ln (\sqrt{2} sin \: \theta)d \theta\\ \\ \tt : \implies I= \int \limits_{0}^{ \frac{\pi}{2} }( ln \sqrt{2} )d \theta + \int \limits_{0}^{ \frac{\pi}{2} } \:ln( sin \: \theta) \: d \theta \\ \\ \green{ \circ } \tt \: \int \limits_{0}^{ \frac{\pi}{2} } ln(sin \: \theta)d \theta = - \frac{\pi}{2} (ln \: 2)\\ \\ \tt : \implies I= ln( \sqrt{2}) \int \limits_{0}^{ \frac{\pi}{2} }d \theta - \frac{\pi}{2} (ln \: 2) \\ \\ \tt : \implies I= ln( \sqrt{2}) \bigg[\theta \bigg]_{0}^{ \frac{\pi}{2} } - \frac{\pi}{2} (ln \: 2) \\ \\ \tt : \implies I= ln( 2^{ \frac{1}{2} } ) \bigg( \frac{\pi}{2} - 0 \bigg) - \frac{\pi}{2} (ln \: 2) \\ \\ \tt : \implies I= \frac{1}{2} ln(2) \frac{\pi}{2} - \frac{\pi}{2} (ln \: 2) \\ \\ \\ \green{\tt : \implies I = - \frac{\pi}{4} (ln \: 2)}$

Answer 2

${\bold{\underline{\underline{Answer:}}}}$

${\tt{\therefore{I = - \frac{\pi}{4} (ln \: 2)}}}$

$\sf{Given :} \\ \sf \implies\int \limits_{ - \frac{\pi}{4} }^{ \frac{ \pi}{4} } \: ln(sin \: x + cos \: x)dx \\ \\ \sf{To \: Find : } \\ \sf \implies Value \: of \: \int \limits_{ - \frac{\pi}{4} }^{ \frac{ \pi}{4} } \: ln(sin \: x + cos \: x)dx =?$

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\sf\implies Let\:I= \int \limits_{ - \frac{\pi}{4} }^{ \frac{ \pi}{4} } \: ln(sin \: x + cos \: x)dx\\ \\ \sf\implies I= \int \limits_{ - \frac{\pi}{4} }^{ \frac{ \pi}{4} } \: ln(\sqrt{2} \bigg( \frac{1}{ \sqrt{2} } sin \: x + \frac{1}{ \sqrt{2} } cos \: x) \\ \\ \sf \implies I= \int \limits_{ - \frac{\pi}{4} }^{ \frac{\pi}{4} } ln \: \sqrt{2} \: sin \bigg(x + \frac{\pi}{4} \bigg )dx \\ \\ \sf \: Replace\: x + \frac{\pi}{4} = t \implies dx+0 =dt \\ \\ \sf \:x = - \frac{\pi}{4}, so\:t = 0 \\ \\ \sf \:x = \frac{\pi}{4}, so\: t= \frac{\pi}{2}$

$\sf \implies I= \int \limits_{0}^{ \frac{\pi}{2} } ln (\sqrt{2} sin \: t)dt\\ \\ \sf\implies I= ln(2^{\frac{1}{2}})\int \limits_{0}^{ \frac{\pi}{2} }dt + \int \limits_{0}^{ \frac{\pi}{2} } \:ln( sin \: t) \: dt \\ \\ \sf \: \int \limits_{0}^{ \frac{\pi}{2} } ln(sin \: t)dt = - \frac{\pi}{2} (ln \: 2)\\ \\ \sf \implies I= \frac{1}{2}ln( 2) \bigg[t \bigg]_{0}^{ \frac{\pi}{2} } - \frac{\pi}{2} (ln \: 2) \\ \\ \sf\implies I= \frac{1}{2} ln(2) \frac{\pi}{2} - \frac{\pi}{2} (ln \: 2) \\ \\ \\ \sf \implies I = - \frac{\pi}{4} (ln \: 2)$