Ex 7.9, 8 - Chapter 7 Class 12 Integrals (Term 2)

Transcript

Ex 7.9, 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| (Put 𝑡=𝑐𝑜𝑠𝑒𝑐 𝑥+𝑐𝑜𝑡 𝑥) =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)] (log a − log b = log 𝑎/𝑏)