Chapter 7 Class 12 Integrals
Concept wise

Misc 10 - Integrate sin8 x - cos8 x / 1 - 2 sin2 x cos2 x Misc 10 - Chapter 7 Class 12 Integrals - Part 2 Misc 10 - Chapter 7 Class 12 Integrals - Part 3

You saved atleast 2 minutes of distracting ads by going ad-free. Thank you :)

You saved atleast 2 minutes by viewing the ad-free version of this page. Thank you for being a part of Teachoo Black.


Transcript

Misc 10 Integrate the function (γ€–sin^8 π‘₯γ€—β‘βˆ’ cos^8⁑π‘₯)/(1 βˆ’ 2 sin^2⁑〖π‘₯ cos^2⁑π‘₯ γ€— ) ∫1β–’(γ€–sin^8 π‘₯γ€—β‘βˆ’ cos^8⁑π‘₯)/(1 βˆ’ 2 sin^2⁑〖π‘₯ cos^2⁑π‘₯ γ€— ) =∫1β–’((sin^4 π‘₯)^2β‘γ€–βˆ’ γ€— (cos^4 π‘₯)^2)/(1 βˆ’ 2 sin^2⁑〖π‘₯ cos^2⁑π‘₯ γ€— ) =∫1β–’((sin^4 π‘₯ + cos^4⁑π‘₯ )⁑(sin^4⁑π‘₯ βˆ’ cos^4⁑π‘₯ )𝑑π‘₯)/(1 βˆ’ 2 sin^2⁑〖π‘₯ cos^2⁑π‘₯ γ€— ) =∫1β–’((sin^4 π‘₯ + cos^4⁑π‘₯ )⁑〖 ((sin^2 π‘₯)^2 βˆ’ (cos^2 π‘₯)^2 )γ€— 𝑑π‘₯)/(1 βˆ’ 2 sin^2⁑〖π‘₯ cos^2⁑π‘₯ γ€— ) =∫1β–’((sin^4 π‘₯ + cos^4⁑π‘₯ )⁑(sin^2⁑π‘₯ + cos^2⁑π‘₯ ) (sin^2⁑π‘₯ βˆ’ cos^2⁑π‘₯ )𝑑π‘₯)/(1 βˆ’ 2 sin^2⁑〖π‘₯ cos^2⁑π‘₯ γ€— ) =∫1β–’((sin^4 π‘₯ + cos^4⁑π‘₯ )⁑(sin^2⁑π‘₯ + cos^2⁑π‘₯ ) (sin^2⁑π‘₯ βˆ’ cos^2⁑π‘₯ )𝑑π‘₯)/(1 βˆ’ 2 sin^2⁑〖π‘₯ cos^2⁑π‘₯ γ€— ) =∫1β–’(γ€–(sin^4 π‘₯ + cos^4⁑π‘₯ ) (1)〗⁑〖 (sin^2⁑π‘₯ βˆ’ cos^2⁑π‘₯ )γ€— 𝑑π‘₯)/(1 βˆ’ 2 sin^2⁑〖π‘₯ cos^2⁑π‘₯ γ€— ) Adding & Subtracting 2 sin^2⁑π‘₯ cos^2⁑π‘₯ =∫1β–’((sin^4 π‘₯ + cos^4⁑π‘₯ + 2 sin^2⁑π‘₯ cos^2⁑π‘₯ βˆ’ 2 sin^2⁑cos^2⁑π‘₯ )⁑〖 (sin^2⁑π‘₯ βˆ’ cos^2⁑π‘₯ )γ€— 𝑑π‘₯)/(1 βˆ’ 2 sin^2⁑〖π‘₯ cos^2⁑π‘₯ γ€— ) =∫1β–’((((sin^2⁑π‘₯ )^2+ (cos^2⁑π‘₯ )^2 + 2 sin^2⁑π‘₯ cos^2⁑π‘₯ )βˆ’2 sin^2⁑π‘₯ cos^2⁑π‘₯ )⁑(sin^2⁑π‘₯ βˆ’ cos^2⁑π‘₯ )𝑑π‘₯)/(1 βˆ’ 2 sin^2⁑〖π‘₯ cos^2⁑π‘₯ γ€— ) =∫1β–’(γ€–((sin^2⁑π‘₯ + cos^2⁑π‘₯ )^2 βˆ’ 2 sin^2⁑π‘₯ cos^2⁑π‘₯ ) 〗⁑(sin^2⁑π‘₯ βˆ’ cos^2⁑π‘₯ )𝑑π‘₯)/(1 βˆ’ 2 sin^2⁑〖π‘₯ cos^2⁑π‘₯ γ€— ) =∫1β–’(γ€–(1^2 βˆ’ 2 sin^2⁑π‘₯ cos^2⁑π‘₯ ) 〗⁑(sin^2⁑π‘₯ βˆ’ cos^2⁑π‘₯ )𝑑π‘₯)/(1 βˆ’ 2 sin^2⁑〖π‘₯ cos^2⁑π‘₯ γ€— ) =∫1β–’(γ€–(1 βˆ’ 2 sin^2⁑π‘₯ cos^2⁑π‘₯ ) 〗⁑(sin^2⁑π‘₯ βˆ’ cos^2⁑π‘₯ )𝑑π‘₯)/(1 βˆ’ 2 sin^2⁑〖π‘₯ cos^2⁑π‘₯ γ€— ) =∫1β–’(sin^2⁑π‘₯βˆ’cos^2⁑π‘₯ ) 𝑑π‘₯ =βˆ’βˆ«1β–’(cos^2⁑π‘₯βˆ’sin^2⁑π‘₯ ) 𝑑π‘₯ =βˆ’βˆ«1β–’cos⁑2π‘₯ . 𝑑π‘₯ =(βˆ’πŸ)/𝟐 π¬π’π§β‘πŸπ’™+π‘ͺ (sin^2⁑π‘₯ + cos^2⁑π‘₯=1" " ) (Using cos 2πœƒ=γ€–π‘π‘œπ‘ γ€—^2 πœƒβˆ’γ€–π‘ π‘–π‘›γ€—^2 πœƒ)

Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo