Ex 7.4

Chapter 7 Class 12 Integrals
Serial order wise

### Transcript

Ex 7.4, 15 Integrate the function 1/√((𝑥 − 𝑎)(𝑥 − 𝑏)) ∫1▒1/√((𝑥 − 𝑎) (𝑥 − 𝑏)) 𝑑𝑥 =∫1▒1/√(𝑥(𝑥 − 𝑎) − 𝑎(𝑥 − 𝑏)) 𝑑𝑥 =∫1▒1/√(𝑥^2 − 𝑏𝑥 − 𝑎𝑥 + 𝑎𝑏) 𝑑𝑥 =∫1▒1/√(𝑥^2 − 𝑥(𝑎 + 𝑏) + 𝑎𝑏) 𝑑𝑥 =∫1▒1/√(𝑥^2 − 2(𝑥)((𝑎 + 𝑏)/2) + 𝑎𝑏) 𝑑𝑥 =∫1▒1/√(𝑥^2 − 2(𝑥)((𝑎 + 𝑏)/2) + ((𝑎 + 𝑏)/2)^2− ((𝑎 + 𝑏)/2)^2+ 𝑎𝑏) 𝑑𝑥 =∫1▒1/√((𝑥 − (𝑎 + 𝑏)/2)^2 − ((𝑎 + 𝑏)/2)^2+ 𝑎𝑏) 𝑑𝑥 =∫1▒1/√((𝑥 − (𝑎 + 𝑏)/2)^2 − ((𝑎^2 + 𝑏^2+ 2𝑎𝑏)/4) + 𝑎𝑏) 𝑑𝑥 =∫1▒1/√((𝑥 − (𝑎 + 𝑏)/2)^2+ (− 𝑎^2 − 𝑏^2 − 2𝑎𝑏 + 4𝑎𝑏)/4) 𝑑𝑥 =∫1▒1/√((𝑥 − (𝑎 + 𝑏)/2)^2 + (− 𝑎^2 − 𝑏^2 + 2𝑎𝑏)/4) 𝑑𝑥 =∫1▒1/(√((𝑥 − (𝑎 + 𝑏)/2)^2 − ((𝑎^2 + 𝑏^2 − 2𝑎𝑏)/4) ) ) 𝑑𝑥 =∫1▒1/(√((𝑥 − (𝑎 + 𝑏)/2)^2 − ((𝑎 − 𝑏)/2)^2 ) ) 𝑑𝑥 =𝑙𝑜𝑔⁡|𝑥 − (𝑎 + 𝑏)/2 +√((𝑥 − (𝑎 + 𝑏)/2)^2− ((𝑎 − 𝑏)/2)^2 )|+𝐶 =𝑙𝑜𝑔⁡|𝑥− (𝑎 + 𝑏)/2 +√(𝑥^2+((𝑎 + 𝑏)/2)^2−2(𝑥)((𝑎 + 𝑏)/2)−((𝑎 − 𝑏)/2)^2 )|+𝐶 It is of form ∫1▒𝑑𝑥/√(𝑥^2 − 𝑎^2 ) =𝑙𝑜𝑔⁡|𝑥+√(𝑥^2−𝑎^2 )|+𝐶 ∴ Replacing 𝑥 by (𝑥− (𝑎 + 𝑏)/2) and a by ((𝑎 − 𝑏)/2) , we get =𝑙𝑜𝑔⁡|𝑥− (𝑎 + 𝑏)/2 +√(𝑥^2−2(𝑥)((𝑎 + 𝑏)/2)+((𝑎 + 𝑏)/2)^2−((𝑎 − 𝑏)/2)^2 )|+𝐶 =𝑙𝑜𝑔⁡|𝑥− (𝑎 + 𝑏)/2 +√(𝑥^2−𝑥(𝑎+𝑏)+(𝑎^2 + 𝑏^2 + 2𝑎𝑏)/4−(𝑎^2 + 𝑏^2 − 2𝑎𝑏)/4)|+𝐶 =𝑙𝑜𝑔⁡|𝑥− (𝑎 + 𝑏)/2 +√(𝑥^2−𝑥(𝑎+𝑏)+2𝑎𝑏/4+2𝑎𝑏/4) |+𝐶 =𝑙𝑜𝑔⁡|𝑥− (𝑎 + 𝑏)/2 +√(𝑥^2−𝑥(𝑎+𝑏)+4𝑎𝑏/4) |+𝐶 =𝑙𝑜𝑔⁡|𝑥− (𝑎 + 𝑏)/2 +√(𝑥^2−𝑥(𝑎+𝑏)+𝑎𝑏) |+𝐶 =𝑙𝑜𝑔⁡|𝑥− (𝑎 + 𝑏)/2 +√(𝑥^2−𝑎𝑥−𝑏𝑥+𝑎𝑏) |+𝐶 =𝑙𝑜𝑔⁡|𝑥− (𝑎 + 𝑏)/2 +√(𝑥(𝑥−𝑎)−𝑏(𝑥−𝑎) ) |+𝐶 =𝒍𝒐𝒈⁡|𝒙− (𝒂 + 𝒃)/𝟐 +√((𝒙−𝒂)(𝒙−𝒃) ) |+𝑪