Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Ex 7.3, 1 Find the integral of sin2 (2π‘₯ + 5) ∫1β–’γ€–π’”π’Šπ’πŸ (πŸπ’™ + πŸ“) γ€— 𝒅𝒙 =∫1β–’(1 βˆ’ γ€–π‘π‘œπ‘  2〗⁑(2π‘₯ + 5))/2 𝑑π‘₯ =1/2 ∫1β–’γ€–1βˆ’cos⁑(4π‘₯+10) γ€— 𝑑π‘₯ =1/2 [∫1β–’1 𝑑π‘₯βˆ’βˆ«1β–’cos⁑(4π‘₯+10) 𝑑π‘₯] We know that 𝐜𝐨𝐬 𝟐𝜽=πŸβˆ’πŸ γ€–π’”π’Šπ’γ€—^𝟐⁑𝜽 2 sin^2 πœƒ=1βˆ’cos⁑2πœƒ sin^2 πœƒ=1/2 [1βˆ’cos⁑2πœƒ ] Replace πœƒ by (𝟐𝐱+πŸ“) sin^2 (2π‘₯+5)=(1 βˆ’ cos⁑2(2π‘₯ + 5))/2 As ∫1β–’cos⁑(π‘Žπ‘₯+𝑏) 𝑑π‘₯=sin⁑(π‘Žπ‘₯ + 𝑏)/π‘Ž+𝐢 =1/2 [π‘₯βˆ’ sin⁑(4π‘₯ + 10)/4 +𝐢] =𝒙/𝟐 βˆ’ 𝟏/πŸ– π’”π’Šπ’β‘(πŸ’π’™+𝟏𝟎)+π‘ͺ

Ask a doubt
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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.