Example 6 - Class 12

Transcript

Example 6 Find the following integrals: (i) ﷮﷮ sin﷮3﷯﷮𝑥﷯ cos﷮2﷯﷮𝑥﷯﷯ 𝑑𝑥 Let cos 𝑥=𝑡 Differentiate both sides 𝑤.𝑟.𝑡.𝑥. − sin﷮𝑥﷯= 𝑑𝑡﷮𝑑𝑥﷯ 𝑑𝑥= −𝑑𝑡﷮ sin﷮𝑥﷯﷯ Now are equation becomes ﷮﷮ sin﷮3﷯﷮𝑥﷯ cos﷮2﷯﷮𝑥﷯﷯ 𝑑𝑥 Put the value of 𝑐𝑜𝑠⁡𝑥 and 𝑑𝑥 = ﷮﷮ sin﷮3﷯﷮𝑥﷯﷯. 𝑡﷮2﷯. 𝑑𝑥 = ﷮﷮ sin﷮3﷯﷮𝑥﷯﷯. 𝑡﷮2﷯. 𝑑𝑡﷮− sin﷮𝑥﷯﷯ = ﷮﷮ sin﷮3﷯﷮𝑥﷯﷮− sin﷮𝑥﷯﷯﷯ 𝑡﷮2﷯. 𝑑𝑡 = – ﷮﷮ sin﷮2﷯﷮𝑥﷯﷯ 𝑡﷮2﷯. 𝑑𝑡 = – ﷮﷮ 1− cos﷮2﷯﷮𝑥﷯﷯﷯ 𝑡﷮2﷯. 𝑑𝑡 = – ﷮﷮ 1− 𝑡﷮2﷯﷯﷯ 𝑡﷮2﷯. 𝑑𝑡 = – ﷮﷮ 𝑡﷮2﷯− 𝑡﷮4﷯﷯﷯𝑑𝑡 = ﷮﷮ − 𝑡﷮2﷯+ 𝑡﷮4﷯﷯﷯𝑑𝑡 = ﷮﷮− 𝑡﷮2﷯﷯. 𝑑𝑡 + ﷮﷮ 𝑡﷮4﷯﷯. 𝑑𝑡 = −𝑡﷮2﷯+1﷮2 + 1﷯+ C﷮1﷯+ 𝑡﷮4 + 1﷯﷮4 + 1﷯+ C﷮2﷯ = − 𝑡﷮3﷯﷮3﷯ +𝐶1+ 𝑡﷮5﷯﷮5﷯ +𝐶2 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) ﷮﷮ 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 ﷮﷮ sin﷮𝑥﷯﷮ sin﷮ 𝑥 + 𝑎﷯﷯﷯﷯ 𝑑𝑥 Putting the value of 𝑥﷮+﷯𝑎﷯ and 𝑑𝑥 = ﷮﷮ sin﷮𝑥﷯﷮ sin﷮𝑡﷯﷯﷯ 𝑑𝑥 = ﷮﷮ 𝒔𝒊𝒏﷮ 𝒕 − 𝒂﷯﷯﷮ sin﷮𝑡﷯﷯﷯ 𝑑𝑡 = ﷮﷮ 𝒔𝒊𝒏﷮𝒕 𝐜𝐨𝐬﷮𝒂﷯ − 𝐬𝐢𝐧﷮𝒂﷯ 𝐜𝐨𝐬﷮𝒕﷯﷯﷮ sin﷮𝑡﷯﷯﷯ 𝑑𝑡 = ﷮﷮ sin﷮𝑡 cos﷮𝑎﷯﷯﷮ sin﷮𝑡﷯﷯ − sin﷮𝑎﷯ cos﷮𝑡﷯﷮ sin﷮𝑡﷯﷯﷯﷯ 𝑑𝑡 = ﷮﷮ sin﷮𝑡 cos﷮𝑎﷯﷯﷮ sin﷮𝑡﷯﷯﷯. 𝑑𝑡− ﷮﷮ sin﷮𝑎﷯ cos﷮𝑡﷯﷮ sin﷮𝑡﷯﷯﷯ . 𝑑𝑡 = ﷮﷮ cos﷮𝑎﷯﷯. 𝑑𝑡− ﷮﷮ sin﷮𝑎﷯ co𝑡﷮𝑡﷯﷯ . 𝑑𝑡 = cos 𝑎 ﷮﷮1﷯. 𝑑𝑡− sin﷮𝑎﷯ ﷮﷮ 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 = 𝑥 cos﷮𝑎﷯+ 𝑎 cos﷮𝑎﷯− sin﷮𝑎﷯﷯ log﷮ sin﷮ 𝑥+𝑎﷯﷯﷯﷯+𝐶3 = 𝑥 cos﷮𝑎﷯− sin﷮ 𝑎﷯﷯ log﷮ sin﷮ 𝑥+𝑎﷯﷯﷯﷯+C Example 6 Find the following integrals (iii) ﷮﷮ 1﷮1 + tan﷮𝑥﷯﷯﷯ 𝑑𝑥 ﷮﷮ 1﷮1 + tan﷮𝑥﷯﷯﷯ 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 + tan﷮𝑥﷯﷯﷯ .𝑑𝑥 = ﷮﷮ 1﷮1 + sin﷮𝑥﷯﷮ cos﷮𝑥﷯﷯﷯﷯ .𝑑𝑥 = ﷮﷮ 1﷮1 + cos﷮𝑥﷯ + sin﷮𝑥﷯﷮ cos﷮𝑥﷯﷯﷯﷯ .𝑑𝑥 = ﷮﷮ cos﷮𝑥﷯﷮ sin﷮𝑥﷯ + cos﷮𝑥﷯﷯﷯ .𝑑𝑥 Multiplying and dividing by 2 = ﷮﷮ 2 cos﷮𝑥﷯﷮2 sin﷮𝑥﷯ + cos﷮𝑥﷯﷯﷯﷯ .𝑑𝑥 = ﷮﷮ cos﷮𝑥﷯ + cos﷮𝑥﷯﷮2 sin﷮𝑥﷯ + cos﷮𝑥﷯﷯﷯﷯ .𝑑𝑥 Adding and subtracting sin⁡𝑥 in the numerator = ﷮﷮ cos﷮𝑥﷯ + cos﷮𝑥﷯ + sin﷮𝑥﷯ − sin﷮𝑥﷯﷮2 sin﷮𝑥﷯ + cos﷮𝑥﷯﷯﷯﷯ .𝑑𝑥 = 1﷮2﷯ ﷮﷮ cos﷮𝑥﷯ + cos﷮𝑥﷯ + sin﷮𝑥﷯ − sin﷮𝑥﷯﷮ sin﷮𝑥﷯ + cos﷮𝑥﷯﷯﷯ .𝑑𝑥 = 1﷮2﷯ ﷮﷮ sin﷮𝑥﷯ + cos﷮𝑥﷯﷮ sin﷮𝑥﷯ + cos﷮𝑥﷯﷯+ cos﷮𝑥﷯ − sin﷮𝑥﷯﷮ sin﷮𝑥﷯ + cos﷮𝑥﷯﷯﷯﷯ 𝑑𝑥 = 1﷮2﷯ ﷮﷮ 1+ cos﷮𝑥﷯ − sin﷮𝑥﷯﷮ sin﷮𝑥﷯ + cos﷮𝑥﷯﷯﷯﷯ 𝑑𝑥 = 1﷮2﷯ ﷮﷮1.𝑑𝑥+ ﷮﷮ cos﷮𝑥﷯ − sin﷮𝑥﷯﷮ sin﷮𝑥﷯ + cos﷮𝑥﷯﷯﷯﷯ 𝑑𝑥﷯ = 1﷮2﷯ 𝑥+𝐶1+ ﷮﷮ cos﷮𝑥﷯ − sin﷮𝑥﷯﷮ sin﷮𝑥﷯ + cos﷮𝑥﷯﷯﷯ 𝑑𝑥﷯ Take, I1 = ﷮﷮ cos﷮𝑥﷯ − sin﷮𝑥﷯﷮ sin﷮𝑥﷯ + cos﷮𝑥﷯﷯﷯ .𝑑𝑥 This can be solved by using the method of substitution 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﷯﷯ Put the value of sin﷮x﷯+ cos﷮x﷯﷯ and dx in I1 . I1= ﷮﷮ cos﷮𝑥﷯ − sin﷮𝑥﷯﷮ sin﷮𝑥﷯ + cos﷮𝑥﷯﷯﷯ .𝑑𝑥 I1 = ﷮﷮ cos﷮𝑥﷯ − sin﷮𝑥﷯﷮𝑡﷯﷯ . 𝑑𝑡﷮ cos﷮𝑥﷯ − sin﷮𝑥﷯﷯ I1 = ﷮﷮ 1﷮𝑡﷯﷯ .𝑑𝑡 I1 = log﷮ 𝑡﷯﷯+𝐶2 I1 = log﷮ sin﷮𝑥﷯+ cos﷮𝑥﷯﷯﷯+𝐶2 Put the value of I1 in eq. (1) ﷮﷮ 1﷮1 + tan﷮𝑥﷯﷯﷯ = 1﷮2﷯ 𝑥+𝐶1+ ﷮﷮ cos﷮𝑥﷯ − sin﷮𝑥﷯﷮ sin﷮𝑥﷯ + cos﷮𝑥﷯﷯﷯ 𝑑𝑥﷯ = 1﷮2﷯ 𝑥+𝐶1+ log﷮ sin﷮𝑥﷯+ cos﷮𝑥﷯﷯﷯+𝐶2﷯ = 𝑥﷮2﷯ + 𝐶1﷮2﷯ + 1﷮2﷯ log﷮ sin﷮𝑥﷯+ cos﷮𝑥﷯﷯﷯+ 𝐶2﷮2﷯ = 𝑥﷮2﷯ + 1﷮2﷯ log﷮ sin﷮𝑥﷯+ cos﷮𝑥﷯﷯﷯+𝐶