Ex 7.2, 33 - Chapter 7 Class 12 Integrals

Transcript

Ex 7.2, 33 1﷮1 − 𝑡𝑎𝑛﷮𝑥﷯﷯ 1﷮1 − 𝑡𝑎𝑛﷮𝑥﷯﷯ The given function cannot be integrated directly, so we have to simplify it. Step 1: Simplify the given function ﷮﷮ 1﷮1− tan﷮𝑥﷯﷯﷯ 𝑑𝑥 = ﷮﷮ 1﷮1 − sin﷮𝑥﷯﷮ cos﷮𝑥﷯﷯﷯﷯ 𝑑𝑥 = ﷮﷮ 1﷮ cos﷮𝑥﷯ − sin﷮𝑥﷯﷮ cos﷮𝑥﷯﷯﷯﷯ 𝑑𝑥 = ﷮﷮ cos﷮𝑥﷯﷮ cos﷮𝑥﷯ − sin﷮𝑥﷯﷯﷯ 𝑑𝑥 = ﷮﷮ 2 cos﷮𝑥﷯﷮2 cos﷮𝑥﷯ − sin﷮𝑥﷯﷯﷯﷯ 𝑑𝑥 = ﷮﷮ cos﷮𝑥﷯ + cos﷮𝑥﷯﷮2 cos﷮𝑥﷯ − sin﷮𝑥﷯﷯﷯﷯ 𝑑𝑥 = ﷮﷮ cos﷮𝑥﷯ + cos﷮𝑥﷯ + sin﷮𝑥﷯ − sin﷮𝑥﷯﷮2 cos﷮𝑥﷯ − sin﷮𝑥﷯﷯﷯﷯ 𝑑𝑥 = ﷮﷮ cos﷮𝑥﷯ − sin﷮𝑥﷯ + cos﷮𝑥﷯ + sin﷮𝑥﷯﷮2 cos﷮𝑥﷯ − sin﷮𝑥﷯﷯﷯﷯ 𝑑𝑥 = 1﷮2﷯ ﷮﷮ cos﷮𝑥﷯ − sin﷮𝑥﷯ + cos﷮𝑥﷯ + sin﷮𝑥﷯﷮ cos﷮𝑥﷯ − sin﷮𝑥﷯﷯﷯﷯ 𝑑𝑥 = 1﷮2﷯ ﷮﷮ cos﷮𝑥﷯ + sin﷮𝑥﷯﷮ cos﷮𝑥﷯ − sin﷮𝑥﷯﷯+ cos﷮𝑥﷯ − sin﷮𝑥﷯﷮ cos﷮𝑥﷯ − sin﷮𝑥﷯﷯ ﷯﷯ 𝑑𝑥 𝑑𝑥= 𝑑𝑡﷮− sin﷮𝑥﷯ − cos﷮𝑥﷯﷯﷯ Thus, our equation becomes I1 = ﷮﷮ cos﷮𝑥﷯ + sin﷮𝑥﷯﷮ cos﷮𝑥﷯ − sin﷮𝑥﷯﷯﷯ . 𝑑𝑥 I1 = ﷮﷮ cos﷮𝑥﷯ − sin﷮𝑥﷯﷮𝑡﷯﷯ . 𝑑𝑡﷮− sin﷮𝑥﷯ − cos﷮𝑥﷯﷯﷯ I1 = ﷮﷮ 𝑑𝑡﷮−𝑡﷯﷯ I1 = −1 ﷮﷮ 𝑑𝑡﷮𝑡﷯﷯ I1 = − log﷮ 𝑡﷯﷯+𝐶2 I1 = − log﷮ cos﷮𝑥﷯− sin﷮𝑥﷯﷯﷯+𝐶2 Putting the value of I1 in (1) ∴ ﷮﷮ 1﷮1 + tan﷮𝑥﷯﷯ ﷯= 𝑥﷮2﷯+ 1﷮2﷯ ﷮﷮ cos﷮𝑥﷯ + sin﷮𝑥﷯﷮ cos﷮𝑥﷯ − sin﷮𝑥﷯﷯﷯ 𝑑𝑥+ 𝐶﷮1﷯ = 𝑥﷮2﷯+− 1﷮2﷯ log﷮ cos﷮𝑥﷯− sin﷮𝑥﷯﷯﷯+ 𝐶2﷮2﷯+𝐶1 = 𝒙﷮𝟐﷯ − 𝟏﷮𝟐﷯ 𝒍𝒐𝒈﷮ 𝒄𝒐𝒔﷮𝒙﷯− 𝒔𝒊𝒏﷮𝒙﷯﷯﷯+𝑪