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

Take a fresh quiz. Then take another.
Every attempt is a new AI-adaptive Teachoo quiz with 5 questions, selected from your answers, mistakes, and progress.
Remove Ads

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 is an IIT Kanpur graduate and has been teaching for 16+ years. At Teachoo, he breaks down Maths, Science and Computer Science into simple steps so students understand concepts deeply and score with confidence.

Many students prefer Teachoo Black for a smooth, ad-free learning experience.