Ex 7.9, 17
Last updated at Dec. 8, 2016 by Teachoo
Transcript
Ex7.9, 17 0 𝜋4 2 sec2𝑥+ 𝑥3+2 𝑑𝑥 Step 1 :- Let F 𝑥= 2 sec2𝑥+ 𝑥3+2 𝑑𝑥 =2 sec2𝑥 𝑑𝑥+ 𝑥3 𝑑𝑥+ 2 𝑑𝑥 =2 tan𝑥+ 𝑥44+2𝑥 Hence F 𝑥=2 tan𝑥+ 𝑥44+2𝑥 Step 2 :- 0 𝜋4 2𝑠𝑒 𝑐2𝑥+ 𝑥3+2𝑑𝑥=𝐹 𝜋4−𝐹 0 = 2𝑡𝑎𝑛 𝜋4+ 𝜋444+2 𝜋4− 2𝑡𝑎𝑛 0+ 044+2 ×0 =2𝑡𝑎𝑛 𝜋4+ 𝜋4 44 × 14+ 𝜋2− 2×0+0+0 =2 ×1+ 𝜋4 45+ 𝜋2−0 =𝟐+ 𝝅𝟒𝟏𝟎𝟐𝟒+ 𝝅𝟐
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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.