Question

$\sf{{How \: to \: evaluate \: the \: integral \: from \: 0 \: to \: \frac{\pi}{4} \: of}{ \: \: \frac{1 \: + \: {cos}^{2} x }{ {cos}^{2} x } dx \: \: ?}}$

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Answer 1

$\sf\underline{ \underline{\: Solution: }{}}$

$\sf{{ ⟼ \: {∫}^{ \frac{\pi}{4} } _{0} \frac{1 + {cos}^{2}(x) }{ {cos}^{2}(x) } }{ dx}}$

$\sf{{⟼ \: {∫}^{ \frac{\pi}{4} } _{0} \frac{2 \: + \: {cos}^{2}(x) \: - 1 }{ {cos}^{2}(x) }dx }{ \: }}$

Factorize :

$\sf{{⟼ \: {∫}^{ \frac{\pi}{4} } _{0} \frac{ - (1 \: - \: {cos}^{2} (x)) \: + \: 2}{ {cos}^{2}(x) } dx }{ \: }}$

$\sf{{⟼ \: {∫}^{ \frac{\pi}{4} } _{0} \frac{ - \: {sin}^{2}(x) + 2 }{ {cos}^{2}(x) } dx }{ \: }}$

$\sf{{⟼ \: - {∫}^{ \frac{\pi}{4} } _{0} \frac{ {sin}^{2}(x) }{ {cos}^{2}(x) }dx + \: 2{∫}^{ \frac{\pi}{4} } _{0} \frac{1}{ {cos}^{2} }dx }{ \: }}$

$\sf{{⟼ \: - {∫}^{ \frac{\pi}{4} } _{0} \: 1 \: + \: {tan}^{2} (x) - 1dx \: + \: 2{∫}^{ \frac{\pi}{4} } _{0} \frac{1}{ {cos}^{2}(x) } dx \: }{ \: }}$

$\sf{{ ⟼ \: - (tan(x) - x){⌉}^{ \frac{\pi}{4} } _{0} \:+ \: 2 (tan(x)){⌉}^{ \frac{\pi}{4} } _{0} \: }{ \: }}$

$⟼ \: - 1 + \: \frac{\pi}{4} + 2$

$⟼ \: 1 \: + \: \frac{\pi}{4}$

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Hope that helps! :)

Answer 2

Answer:

hey!!

actually wanted your help a bit

do u know (itzofficialAshu)?