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


Transcript

Ex 7.8, 8 ∫_(𝜋/6)^(𝜋/4)▒〖𝑐𝑜𝑠𝑒𝑐 𝑥〗⁡𝑑𝑥 Let F(𝑥)=∫1▒〖𝑐𝑜𝑠𝑒𝑐 𝑥 . 𝑑𝑥〗 Multiplying and Dividing by 𝑐𝑜𝑠𝑒𝑐 𝑥+𝑐𝑜𝑡 𝑥 F(𝑥)=∫1▒(𝑐𝑜𝑠𝑒𝑐 𝑥 (𝑐𝑜𝑠𝑒𝑐 𝑥 + 𝑐𝑜𝑡 𝑥))/(𝑐𝑜𝑠𝑒𝑐 𝑥 + 𝑐𝑜𝑡 𝑥) 𝑑𝑥 Let c𝑜𝑠𝑒𝑐 𝑥+𝑐𝑜𝑡 𝑥=𝑡 Differentiating w.r.t. 𝑥 𝑑/𝑑𝑥 (𝑐𝑜𝑠𝑒𝑐 𝑥+𝑐𝑜𝑡 𝑥)=𝑑𝑡/𝑑𝑥 −𝑐𝑜𝑠𝑒𝑐^2 𝑥−𝑐𝑜𝑠𝑒𝑐 𝑥 𝑐𝑜𝑡 𝑥=𝑑𝑡/𝑑𝑥 −𝑐𝑜𝑠𝑒𝑐 𝑥(𝑐𝑜𝑠𝑒𝑐 𝑥+𝑐𝑜𝑡 𝑥)=𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡/(−𝑐𝑜𝑠𝑒𝑐 𝑥 (𝑐𝑜𝑠𝑒𝑐 𝑥 + 𝑐𝑜𝑡 𝑥) ) Therefore, ∫1▒(𝑐𝑜𝑠𝑒𝑐 𝑥(𝑐𝑜𝑠𝑒𝑐 𝑥 + 𝑐𝑜𝑡 𝑥))/(𝑐𝑜𝑠𝑒𝑐 𝑥 + 𝑐𝑜𝑡 𝑥) 𝑑𝑥 =∫1▒(𝑐𝑜𝑠𝑒𝑐 𝑥(𝑐𝑜𝑠𝑒𝑐 𝑥 + 𝑐𝑜𝑡 𝑥))/𝑡. 𝑑𝑡/(−𝑐𝑜𝑠𝑒𝑐 𝑥(𝑐𝑜𝑠𝑒𝑐 𝑥 + 𝑐𝑜𝑡 𝑥) ) =−∫1▒𝑑𝑡/𝑡 =−log⁡〖 |𝑡|〗 =−log⁡〖 |𝑐𝑜𝑠𝑒𝑐 𝑥+𝑐𝑜𝑡 𝑥|〗 Hence, F(𝑥)=−log⁡|𝑐𝑜𝑠𝑒𝑐 𝑥+cot⁡𝑥 | Now, ∫_(𝜋/6)^(𝜋/4)▒〖𝑐𝑜𝑠𝑒𝑐 𝑥=𝐹(𝜋/4)−𝐹(𝜋/6) 〗 =−𝑙𝑜𝑔|𝑐𝑜𝑠𝑒𝑐(𝜋/4)+𝑐𝑜𝑡(𝜋/4)|− (−log⁡|𝑐𝑜𝑠𝑒𝑐(𝜋/6)+ 𝑐𝑜𝑡(𝜋/6)| ) =−𝑙𝑜𝑔|√2+1|+𝑙𝑜𝑔|2+√3| =−𝑙𝑜𝑔|2+√3|− 𝑙𝑜𝑔|√2+1| =𝑙𝑜𝑔|(2 + √3)/(√2 + 1)| =𝑙𝑜𝑔|(2 + √3)/(√2 + 1)×(2 − √3)/(2 − √3)| =𝑙𝑜𝑔[((2)^2 − (√3)^2)/((√2 + 1) × (2 − √3) )] =𝑙𝑜𝑔[(4 − 3)/(√2 + 1)(2 − √3) ] =𝑙𝑜𝑔[(1 × (√2 − 1))/((2 − √3)(√2 + 1) ×(√2 − 1) )] =𝑙𝑜𝑔[(√2 − 1)/(2 − √3)[(√2)^2− 1^2 ] ] =𝒍𝒐𝒈[(√𝟐 − 𝟏)/(𝟐 − √𝟑)]

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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.