Slide9.JPG

Slide10.JPG

Go Ad-free

Transcript

Ex 7.10, 4 By using the properties of definite integrals, evaluate the integrals : ∫_0^(πœ‹/2)β–’(cos^5⁑π‘₯ 𝑑π‘₯)/(sin^5⁑π‘₯ + cos^5⁑π‘₯ ) Let I=∫_0^(πœ‹/2)β–’γ€–cos^5⁑π‘₯/(sin^5⁑π‘₯ + cos^5⁑π‘₯ ) 𝑑π‘₯γ€— I= ∫_0^(πœ‹/2)β–’γ€–(cos^5 (πœ‹/2 βˆ’ π‘₯))/(〖𝑠𝑖𝑛〗^5 (πœ‹/2 βˆ’ π‘₯) + γ€–π‘π‘œπ‘ γ€—^5 (πœ‹/2 βˆ’ π‘₯) ) 𝑑π‘₯γ€— ∴ I = ∫_0^(πœ‹/2)β–’γ€– sin^5⁑π‘₯/(cos^5⁑π‘₯ + sin^5⁑π‘₯ ) 𝑑π‘₯γ€— Adding (1) and (2) i.e. (1) + (2) I+I=(γ€–π‘π‘œπ‘ γ€—^5 π‘₯)/(〖𝑠𝑖𝑛〗^5 π‘₯ + γ€–π‘π‘œπ‘ γ€—^5 π‘₯) 𝑑π‘₯+∫_0^(πœ‹/2)β–’γ€–sin^5⁑π‘₯/(cos^5⁑π‘₯ + sin^5⁑π‘₯ ) 𝑑π‘₯γ€— 2I=∫_0^(πœ‹/2)β–’γ€–[(γ€–π‘π‘œπ‘ γ€—^5 π‘₯ + 〖𝑠𝑖𝑛〗^5 π‘₯)/(γ€–π‘π‘œπ‘ γ€—^5 π‘₯ + 〖𝑠𝑖𝑛〗^5 π‘₯)] 𝑑π‘₯γ€— 2I= ∫_0^(πœ‹/2)β–’γ€– 𝑑π‘₯γ€— I=1/2 ∫_0^(πœ‹/2)β–’γ€– 𝑑π‘₯γ€— I=1/2 [π‘₯]_0^(πœ‹/2) I=1/2 [πœ‹/2βˆ’0] 𝑰=𝝅/πŸ’

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