Question

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✌ 100 points ​

Answer 1

Question :-

$(1) \sf \int \red{ \sqrt{ \sin \phi} } \cos \phi \: \green{ d \phi \: }$

Of limit 0 to π/2

i will take limits after some steps -

Solution :-

$\leadsto \sf \: let \: \sin \phi \: = \: t \\ \\ \sf \green{ differentiate \: with} \: \red{ respect \: to \: t \:} \\ \\ \leadsto \sf \cos \phi \: = \frac{dt}{d \phi} \\ \\ \leadsto \sf \cos \phi \: d \phi \: = dt \\ \\ \therefore \\ \\ \rightarrow \sf \: \: \int \red{ \sqrt{t} } \: dt \\ \\ \to \: \int \sf {t}^{ \frac{1}{2} } \: dt \\ \\ \to \sf \frac{ {t}^{ \frac{1}{2} + 1} }{ \frac{1}{2} + 1} + c \\ \\ \sf \: now \: changing \: limits \: \\( \star \sf \: upper \: limit) if \: \phi \: = \frac{ \pi}{2} \\ \\ \to \sf \sin \frac{ \pi}{2} = t \\ \\ \to \boxed{ \sf t = 1} \\ \\ \star ( \sf \: lower \: limit \: ) \\ \\ \to \: \phi \: = 0 \\ \\ \to \: \sin 0 = t \\ \to \boxed{ \sf t = 0} \\ \\ \sf \: now \: taking \: limits \: \\ \\ \to \: \frac{ {1}^{ \frac{3}{2} } }{ \frac{3}{2} } - \frac{ {0}^{ \frac{3}{2} } }{ \frac{3}{2} } \\ \\ \to \: \frac{2}{3} \:$

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$(2) \: \int \sf \frac{ \sin x}{1 + { \cos}^{2}x } \: dx$

of limit 0 to π/2

Solution :-

We have

$\to \int \sf \frac{ \sin x}{1 + { \cos}^{2}x } \: dx \\ \: \\ \sf put \: \: \sf \cos x \: = t \\ \\ \sf differentiate \: with \: respect \:to \: t \\ \\ \to \sf - \sin x \: dx = dt \\ \\ \therefore \sf changing \: limits \: \\ \\ \star \sf \: upper \: limit \: \\ \to \: x = \frac{ \pi}{2} \\ \\ \to \: \cos \frac{ \pi}{2} = t \: = 0 \\ \\ \star \sf \: lower \: limit \: \\ \\ \to \: x = 0 \\ \\ \to \sf \cos 0 = t = 1 \\ \\ \therefore \: \\ \to \sf - \int \: \frac{1}{1 + {t}^{2} } \: dt \\ \\ \to \: - { \tan}^{ - 1} t \\ \\ \sf taking \: limit \: \\ \\ \to \: - \{ \: { \tan}^{ - 1} 1 - { \tan}^{ - 1} 0 \} \\ \\ \to \: - \frac{ \pi}{4} \\ \\ \large \sf{ \red{ BRAINLIST } }$

Hope it helps you .

Answer 2

Explanation:

hello dude refer to pic above mate.

have a good day dued ✌