Example 6 - Chapter 7 Class 12 Integrals

Transcript

Example 6 Find the following integrals: (i) ∫1▒〖sin^3⁡𝑥 cos^2⁡𝑥 〗 𝑑𝑥 ∫1▒〖sin^3⁡𝑥 cos^2⁡𝑥 〗 𝑑𝑥 Let cos 𝑥=𝑡 Differentiating both sides 𝑤.𝑟.𝑡.𝑥. −sin⁡𝑥=𝑑𝑡/𝑑𝑥 𝑑𝑥=(−𝑑𝑡)/sin⁡𝑥 Now are equation becomes ∫1▒〖sin^3⁡𝑥 cos^2⁡𝑥 〗 𝑑𝑥 Putting value of 𝑐𝑜𝑠⁡𝑥 and 𝑑𝑥 = ∫1▒sin^3⁡𝑥 .𝑡^2. 𝑑𝑥 = ∫1▒sin^3⁡𝑥 .𝑡^2. 𝑑𝑡/(−sin⁡𝑥 ) = ∫1▒sin^3⁡𝑥/(−sin⁡𝑥 ) 𝑡^2. 𝑑𝑡 = –∫1▒sin^2⁡𝑥 𝑡^2. 𝑑𝑡 = – ∫1▒(1−cos^2⁡𝑥 ) 𝑡^2. 𝑑𝑡 = – ∫1▒(1−𝑡^2 ) 𝑡^2. 𝑑𝑡 = – ∫1▒(𝑡^2−𝑡^4 ) 𝑑𝑡 = ∫1▒(−𝑡^2+𝑡^4 ) 𝑑𝑡 = ∫1▒〖−𝑡^2 〗. 𝑑𝑡 + ∫1▒𝑡^4 . 𝑑𝑡 (∴ sin^2⁡𝑥=1−cos^2⁡𝑥) = (〖−𝑡〗^2+1)/(2 + 1)+𝑡^(4 + 1)/(4 + 1)+𝐶 = (−𝑡^3)/3 +𝑡^5/5 +𝐶 Putting back value of t = cos x = (−𝟏)/𝟑 〖𝒄𝒐𝒔〗^𝟑⁡𝒙 +𝟏/𝟓 〖𝒄𝒐𝒔〗^𝟓⁡𝒙 +𝑪 Example 6 Find the following integrals (ii) ∫1▒sin⁡𝑥/sin⁡(𝑥 + 𝑎) 𝑑𝑥 Let 𝑥+𝑎=𝑡 Differentiate both sides 𝑤.𝑟.𝑡.𝑥. 1=𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡 Hence, our equation becomes ∫1▒sin⁡𝑥/sin⁡(𝑥 + 𝑎) 𝑑𝑥 Putting the value of (𝑥⁡+ 𝑎) and 𝑑𝑥 = ∫1▒sin⁡𝑥/sin⁡𝑡 𝑑𝑥" " = ∫1▒𝒔𝒊𝒏⁡(𝒕 − 𝒂)/sin⁡𝑡 𝑑𝑡" " = ∫1▒(𝒔𝒊𝒏⁡𝒕 𝐜𝐨𝐬⁡𝒂 − 𝐬𝐢𝐧⁡𝒂 𝐜𝐨𝐬⁡𝒕)/sin⁡𝑡 𝑑𝑡 = ∫1▒(sin⁡〖𝑡 cos⁡𝑎 〗/sin⁡𝑡 − (sin⁡𝑎 cos⁡𝑡)/sin⁡𝑡 ) 𝑑𝑡 = ∫1▒sin⁡〖𝑡 cos⁡𝑎 〗/sin⁡𝑡 . 𝑑𝑡−∫1▒(sin⁡𝑎 cos⁡𝑡)/sin⁡𝑡 . 𝑑𝑡 = ∫1▒cos⁡𝑎 . 𝑑𝑡−∫1▒〖sin⁡𝑎 co𝑡⁡𝑡 〗 . 𝑑𝑡 = cos 𝑎∫1▒1. 𝑑𝑡−sin⁡𝑎 ∫1▒𝒄𝒐𝒕⁡𝒕 . 𝒅𝒕 = cos 𝑎 (𝑡)−sin⁡𝑎 𝒍𝒐𝒈⁡|𝒔𝒊𝒏⁡𝒕 |+ 𝐶 = 𝑡.cos⁡𝑎−𝑠𝑖𝑛⁡𝑎 𝑙𝑜𝑔⁡|𝑠𝑖𝑛⁡𝑡 |+ 𝐶 Putting back value of t = x + a (Using x + a = t ∴ x = t – a) (█(𝑈𝑠𝑖𝑛𝑔 sin⁡(𝑎−𝑏)=@sin⁡𝑎 cos⁡𝑏−sin⁡𝑏 cos⁡𝑎 )) ( ∫1▒co𝑡⁡𝑥 𝑑𝑥=log⁡|sin⁡𝑥 | ) = ∫1▒𝒔𝒊𝒏⁡(𝒕 − 𝒂)/sin⁡𝑡 𝑑𝑡" " = ∫1▒𝒔𝒊𝒏⁡〖𝒕 𝐜𝐨𝐬⁡𝒂 −𝐬𝐢𝐧⁡𝒂 𝐜𝐨𝐬⁡𝒕 〗/sin⁡𝑡 𝑑𝑡 = ∫1▒(sin⁡〖𝑡 cos⁡𝑎 〗/sin⁡𝑡 − (sin⁡𝑎 cos⁡𝑡)/sin⁡𝑡 ) 𝑑𝑡 = ∫1▒sin⁡〖𝑡 cos⁡𝑎 〗/sin⁡𝑡 . 𝑑𝑡−∫1▒(sin⁡𝑎 cos⁡𝑡)/sin⁡𝑡 . 𝑑𝑡 = ∫1▒cos⁡𝑎 . 𝑑𝑡−∫1▒〖sin⁡𝑎 co𝑡⁡𝑡 〗 . 𝑑𝑡 = cos 𝑎∫1▒1. 𝑑𝑡−sin⁡𝑎 ∫1▒𝒄𝒐𝒕⁡𝒕 . 𝒅𝒕 = cos 𝑎 (𝑡)−sin⁡𝑎 [𝒍𝒐𝒈⁡|𝒔𝒊𝒏⁡𝒕 | ]+𝐶1 = 𝑡.cos⁡𝑎−sin⁡𝑎 [log⁡|sin⁡𝑡 | ]+ 𝐶1 Putting back value of t = x + a = (𝑥+𝑎) cos⁡𝑎−sin⁡𝑎 [log⁡|sin⁡(𝑥+𝑎) | ]+ 𝐶1 (Using x + a = t x = t – a) (█(𝑈𝑠𝑖𝑛𝑔 sin⁡(𝑎−𝑏)=@sin⁡𝑎 cos⁡𝑏−sin⁡𝑏 cos⁡𝑎 )) ( ∫1▒co𝑡⁡𝑥 𝑑𝑥=log⁡|sin⁡𝑥 | ) = (𝑥+𝑎) cos⁡𝑎−sin⁡𝑎 log⁡〖 |sin⁡(𝑥+𝑎) |〗+ 𝐶 = 𝑥 cos⁡𝑎+〖𝑎 cos〗⁡𝑎−sin⁡𝑎 log⁡|sin⁡(𝑥+𝑎) |+𝐶 = 𝑥 cos⁡𝑎−sin⁡𝑎 log⁡|sin⁡(𝑥+𝑎) |+〖𝒂 𝒄𝒐𝒔〗⁡𝒂+𝑪 = 𝒙 𝒄𝒐𝒔⁡𝒂−𝐬𝐢𝐧⁡〖 𝒂〗 𝒍𝒐𝒈⁡|𝒔𝒊𝒏⁡(𝒙+𝒂) |+𝐂𝟏 (𝑊ℎ𝑒𝑟𝑒 𝐶1=𝑎 cos⁡𝑎+𝐶) Example 6 Find the following integrals (iii) ∫1▒1/(1 + tan⁡𝑥 ) 𝑑𝑥 The given function cannot be integrated by direct substitution, so we have to simplify the given function . Simplifying the given function and integrating. ∫1▒1/(1 + tan⁡𝑥 ) .𝑑𝑥 = ∫1▒1/(1 + sin⁡𝑥/cos⁡𝑥 ) .𝑑𝑥 = ∫1▒1/((cos⁡𝑥 + sin⁡𝑥)/cos⁡𝑥 ) .𝑑𝑥 (𝑈𝑠𝑖𝑛𝑔 tan⁡𝑥=sin⁡𝑥/cos⁡𝑥 ) = ∫1▒cos⁡𝑥/(sin⁡𝑥 + cos⁡𝑥 ) .𝑑𝑥 Multiplying and dividing by 2 = ∫1▒(2 cos⁡𝑥)/(2 (sin⁡𝑥 + cos⁡𝑥 ) ) .𝑑𝑥 = ∫1▒(cos⁡𝑥 + cos⁡𝑥)/(2 (sin⁡𝑥 + cos⁡𝑥 ) ) .𝑑𝑥 Adding and subtracting sin⁡𝑥 in the numerator = ∫1▒(cos⁡𝑥 + cos⁡𝑥 + sin⁡𝑥 − sin⁡𝑥)/(2 (sin⁡𝑥 + cos⁡𝑥 ) ) .𝑑𝑥 = 1/2 ∫1▒(sin⁡𝑥 + cos⁡𝑥 + cos⁡𝑥 − sin⁡𝑥)/(sin⁡𝑥 + cos⁡𝑥 ) .𝑑𝑥 = 1/2 ∫1▒[(sin⁡𝑥 + cos⁡𝑥)/(sin⁡𝑥 + cos⁡𝑥 )+〖cos⁡𝑥 − sin〗⁡𝑥/(sin⁡𝑥 + cos⁡𝑥 )] 𝑑𝑥 = 1/2 ∫1▒[1+〖cos⁡𝑥 − sin〗⁡𝑥/(sin⁡𝑥 + cos⁡𝑥 )] 𝑑𝑥 = 1/2 [∫1▒〖1.𝑑𝑥+∫1▒〖cos⁡𝑥 − sin〗⁡𝑥/(sin⁡𝑥 + cos⁡𝑥 )〗 𝑑𝑥] = 1/2 [𝑥+𝐶1+∫1▒〖cos⁡𝑥 − sin〗⁡𝑥/(sin⁡𝑥 + cos⁡𝑥 ) 𝑑𝑥] Take, I1 =∫1▒〖cos⁡𝑥 − sin〗⁡𝑥/(sin⁡𝑥 + cos⁡𝑥 ) .𝑑𝑥 Let sin⁡𝑥 + cos⁡𝑥=𝑡 Differentiate both sides 𝑤.𝑟.𝑡.𝑥. cos⁡𝑥−sin⁡𝑥=𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡/(cos⁡𝑥 − sin⁡𝑥 ) Putting value of (𝑠𝑖𝑛⁡𝑥+𝑐𝑜𝑠⁡𝑥 ) and 𝑑𝑥 in I1 . I1=∫1▒〖cos⁡𝑥 − sin〗⁡𝑥/(sin⁡𝑥 + cos⁡𝑥 ) .𝑑𝑥 I1 = ∫1▒〖cos⁡𝑥 − sin〗⁡𝑥/𝑡 .𝑑𝑡/(cos⁡𝑥 − sin⁡𝑥 ) I1 =∫1▒1/𝑡 .𝑑𝑡 Let sin⁡𝑥 + cos⁡𝑥=𝑡 Differentiate both sides 𝑤.𝑟.𝑡.𝑥. cos⁡𝑥−sin⁡𝑥=𝑑𝑡/𝑑𝑥 𝑑𝑥=𝑑𝑡/(cos⁡𝑥 − sin⁡𝑥 ) Putting value of (𝑠𝑖𝑛⁡𝑥+𝑐𝑜𝑠⁡𝑥 ) and 𝑑𝑥 in I1 . I1=∫1▒〖cos⁡𝑥 − sin〗⁡𝑥/(sin⁡𝑥 + cos⁡𝑥 ) .𝑑𝑥 I1 = ∫1▒〖cos⁡𝑥 − sin〗⁡𝑥/𝑡 .𝑑𝑡/(cos⁡𝑥 − sin⁡𝑥 ) I1 =∫1▒1/𝑡 .𝑑𝑡 (𝑈𝑠𝑖𝑛𝑔 ∫1▒〖1/𝑥 .〗 𝑑𝑥=log⁡〖 |𝑥|〗+𝐶) (𝑈𝑠𝑖𝑛𝑔 𝑡=sin⁡𝑥+cos⁡𝑥 ) (𝑊ℎ𝑒𝑟𝑒 𝐶=𝐶1/2+𝐶2/2)