∫ (secx)/(tanx) dx #integrals #integration #trigonometry #shorts
∫ (secx)/(tanx) dx #integrals #integration #trigonometry #shorts

Misc 32 - Chapter 7 Class 12 Integrals - Part 2
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Last updated at May 29, 2023 by Teachoo

Question 2 Evaluate the definite integral ∫_0^𝜋▒(𝑥 tan𝑥 )/(sec𝑥 +〖 tan〗𝑥 ) 𝑑𝑥 Let I=∫_0^𝜋▒(𝑥 tan𝑥 )/(sec𝑥 +〖 tan〗𝑥 ) 𝑑𝑥 ∴ I=∫_0^𝜋▒((𝜋 − 𝑥) tan〖 (𝜋 − 𝑥)〗)/(sec(𝜋 − 𝑥) +〖 tan〗(𝜋 − 𝑥) ) 𝑑𝑥 I=∫_0^𝜋▒((𝜋 − 𝑥)(−tan〖 𝑥〗) )/((−sec〖 𝑥〗) + 〖( −tan〗𝑥)) 𝑑𝑥 I=∫_0^𝜋▒(−(𝜋 − 𝑥) tan𝑥 )/(−(sec𝑥 +〖 tan〗𝑥)) 𝑑𝑥 Using The Property, P4 P4 : ∫_0^𝑎▒〖𝑓(𝑥)𝑑𝑥=〗 ∫_0^𝑎▒𝑓(𝑎−𝑥)𝑑𝑥 I=∫_0^𝜋▒((𝜋 − 𝑥) tan𝑥 )/((sec𝑥 +〖 tan〗𝑥)) 𝑑𝑥 Adding (1) and (2) i.e. (1) + (2) I+I=∫_0^𝜋▒(𝑥 tan𝑥 )/(sec𝑥 +〖 tan〗𝑥 ) 𝑑𝑥+∫_0^𝜋▒(𝜋 tan𝑥 − 𝑥 tan𝑥)/(sec𝑥 +〖 tan〗𝑥 ) 𝑑𝑥 2I=∫_0^𝜋▒(𝑥 tan𝑥 + 𝜋 tan𝑥 − 𝑥 tan𝑥)/(sec𝑥 +〖 tan〗𝑥 ) 𝑑𝑥 2I=∫_0^𝜋▒(𝜋 tan𝑥)/(sec𝑥 +〖 tan〗𝑥 ) 𝑑𝑥 2I=𝜋∫_0^𝜋▒tan𝑥/(sec𝑥 +〖 tan〗𝑥 ) 𝑑𝑥 I=𝜋/2 ∫_0^𝜋▒tan𝑥/(sec𝑥 +〖 tan〗𝑥 ) 𝑑𝑥 =𝜋/2 ∫_0^𝜋▒(sin𝑥/cos𝑥 )/(1/cos𝑥 + sin𝑥/cos𝑥 ) 𝑑𝑥 =𝜋/2 ∫_0^𝜋▒sin𝑥/(1 + sin𝑥 ) 𝑑𝑥 =𝜋/2 ∫_0^𝜋▒(sin𝑥 + 1 − 1)/(1 + sin𝑥 ) 𝑑𝑥 =𝜋/2 ∫_0^𝜋▒[(1 + sin𝑥)/(1 + sin𝑥 ) −1/(1 + sin𝑥 )] 𝑑𝑥 =𝜋/2 ∫_0^𝜋▒[1 −1/(1 + sin𝑥 )] 𝑑𝑥 =𝜋/2 [∫_0^𝜋▒1 𝑑𝑥−∫_0^𝜋▒1/(1 + sin𝑥 ) 𝑑𝑥] =𝜋/2 [[𝑥]_0^𝜋−∫_0^𝜋▒1/(1 + sin𝑥 ) ((1 − sin𝑥)/(1 − sin𝑥 )) 𝑑𝑥] =𝜋/2 [[𝜋−0]−∫_0^𝜋▒(1 − sin𝑥)/(1 − sin^2𝑥 ) 𝑑𝑥] =𝜋/2 [𝜋−∫_0^𝜋▒(1 − sin𝑥)/cos^2𝑥 𝑑𝑥] =𝜋/2 [𝜋−∫_0^𝜋▒[1/cos^2𝑥 − sin𝑥/cos^2𝑥 ] 𝑑𝑥] =𝜋/2 {𝜋−∫_0^𝜋▒[sec^2𝑥−tan𝑥 sec𝑥 ] 𝑑𝑥} =𝜋/2 {𝜋−∫_0^𝜋▒sec^2𝑥 𝑑𝑥+∫_0^𝜋▒〖tan𝑥 sec𝑥 〗 𝑑𝑥} =𝜋/2 [𝜋−[tan𝑥 ]_0^𝜋+[sec𝑥 ]_0^𝜋 ] =𝜋/2 {𝜋−[tan〖(𝜋)−tan(0) 〗 ]+[sec (𝜋)−sec(0) ]} =𝜋/2 {𝜋−[0−0]+[−1−1]} =𝜋/2 {𝜋−0+[−2]} =𝝅/𝟐 (𝝅−𝟐)

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