Example 20 - Chapter 7 Class 12 Integrals

Transcript

Example 20 (Method 1) Find ﷮﷮ 𝑥 sin﷮−1﷯﷮𝑥﷯﷮ ﷮1 − 𝑥﷮2﷯﷯﷯﷯ 𝑑𝑥 ﷮﷮ 𝑥 sin﷮−1﷯﷮𝑥﷯﷮ ﷮1 − 𝑥﷮2﷯﷯﷯﷯ 𝑑𝑥 Hence, we take First function :- 𝑓 𝑥﷯= sin﷮−1﷯ Second function :- g 𝑥﷯= 𝑥﷮ ﷮1 − 𝑥﷮2﷯﷯﷯ ﷮﷮ 𝑥 sin﷮−1﷯﷮𝑥﷯﷮ ﷮1 − 𝑥﷮2﷯﷯﷯﷯ 𝑑𝑥 = sin﷮−1﷯﷮𝑥﷯ ﷮﷮ 𝑥﷮ ﷮1 − 𝑥﷮2﷯﷯﷯ 𝑑𝑥﷯− ﷮﷮ 𝑑 sin﷮−1﷯﷮𝑥﷯﷯﷮𝑑𝑥﷯ ﷮﷮ 𝑥﷮ ﷮1 − 𝑥﷮2﷯﷯﷯𝑑𝑥﷯﷯﷯𝑑𝑥 Solving 𝐈﷮𝟏﷯ Let 1 − 𝑥﷮2﷯=𝑡 Differentiating both sides 𝑤.𝑟.𝑡.𝑥 −2𝑥 = 𝑑𝑡﷮𝑑𝑥﷯ 𝑑𝑥 = 𝑑𝑡﷮−2𝑥 ﷯ Putting the value of 1 − x﷮2﷯﷯ and dx in I1 I1 = ﷮﷮ 𝑥﷮ ﷮1 − 𝑥﷮2﷯﷯﷯ 𝑑𝑥﷯ I1 = ﷮﷮ 𝑥﷮ ﷮𝑡﷯﷯ 𝑑𝑡﷮−2𝑥 ﷯﷯ I1 = ﷮﷮ 1﷮−2 ﷮𝑡﷯﷯ 𝑑𝑡﷯ = −1﷮2﷯ ﷮﷮ 𝑡﷯﷮− 1﷮2﷯﷯ . 𝑑𝑡﷯= −1﷮2﷯ 𝑡﷮− 1﷮2﷯ + 1﷯﷮− 1﷮2﷯ + 1﷯﷯ = −1﷮2﷯ 𝑡﷮ 1﷮2﷯ ﷯﷮ 1﷮2﷯﷯﷯ I1 = − 𝑡﷮ 1﷮2﷯ ﷯ = − ﷮𝑡﷯ = − ﷮1 − 𝑥﷮2﷯﷯ Putting the value of I1, in eq.(1) , we get ﷮﷮ 𝑥 sin﷮−1﷯﷮𝑥﷯﷮ ﷮1 − 𝑥﷮2﷯﷯﷯﷯ 𝑑𝑥 = sin﷮−1﷯﷮𝑥﷯ ﷮﷮ 𝑥﷮ ﷮1 − 𝑥﷮2﷯﷯﷯ 𝑑𝑥﷯− ﷮﷮ 𝑑 sin﷮−1﷯﷮𝑥﷯﷯﷮𝑑𝑥﷯ ﷮﷮ sin﷮−1﷯﷮𝑥﷯. 𝑑𝑥﷯﷯﷯𝑑𝑥 = sin﷮−1﷯﷮𝑥﷯ − ﷮1 − 𝑥﷮2﷯﷯ ﷯− ﷮﷮ 𝑑 sin﷮−1﷯﷮𝑥﷯﷯﷮𝑑𝑥﷯ − ﷮1 − 𝑥﷮2﷯﷯ ﷯﷯𝑑𝑥 =− ﷮1 − 𝑥﷮2﷯﷯ sin﷮−1﷯﷮𝑥﷯+ ﷮﷮ 𝑑 sin﷮−1﷯﷮𝑥﷯﷯﷮𝑑𝑥﷯ ﷮1 − 𝑥﷮2﷯﷯﷯ . 𝑑𝑥 =− ﷮1 − 𝑥﷮2﷯﷯ sin﷮−1﷯﷮𝑥﷯+ ﷮﷮ 1﷮ ﷮1 − 𝑥﷮2﷯﷯﷯ . ﷮1 − 𝑥﷮2﷯﷯﷯ . 𝑑𝑥 =− ﷮1 − 𝑥﷮2﷯﷯ sin﷮−1﷯﷮𝑥﷯+ ﷮﷮1﷯. 𝑑𝑥 =𝒙− ﷮𝟏 − 𝒙﷮𝟐﷯﷯ 𝒔𝒊𝒏﷮−𝟏﷯﷮𝒙﷯+𝑪 Example 20 (Method 2) Find ﷮﷮ 𝑥 sin﷮−1﷯﷮𝑥﷯﷮ ﷮1 − 𝑥﷮2﷯﷯﷯﷯ 𝑑𝑥 ﷮﷮ 𝑥 sin﷮−1﷯﷮𝑥﷯﷮ ﷮1 − 𝑥﷮2﷯﷯﷯﷯ 𝑑𝑥 Let t = 𝑠𝑖𝑛﷮−1﷯(𝑥) 𝑑𝑡﷮𝑑𝑥﷯= 1﷮ ﷮1 − 𝑥﷮2﷯﷯﷯ dt = 𝑑𝑥﷮ ﷮1 − 𝑥﷮2﷯﷯﷯ Substituting, = ﷮﷮ sin﷮𝑡×𝑡 ﷯× 𝑑𝑥﷮ ﷮1 − 𝑥﷮2﷯﷯﷯﷯ = ﷮﷮ sin﷮𝑡×𝑡 ﷯𝑑𝑡﷯ Hence we take First function :-f(x) = t Second function :- g(x) = sin t ﷮﷮ sin﷮𝑡×𝑡 𝑑𝑡=𝑡 ﷮﷮ sin﷮𝑡 𝑑𝑡 − ﷮﷮ 𝑑(𝑡)﷮𝑑𝑡﷯﷯﷯ ﷯﷯ ﷯ ﷮﷮ sin﷮𝑡 𝑑𝑡 ﷯﷯ 𝑑𝑡 = t (−cost) − ﷮﷮ − cos﷮𝑡﷯﷯﷯ 𝑑𝑡 = − t cost + ﷮﷮ cos﷮𝑡﷯﷯𝑑𝑡 = −t cost + sin t + C Now, Hence putting the values. ﷮﷮ 𝑥 𝑠𝑖𝑛﷮−1﷯𝑥﷮ ﷮1− 𝑥﷮2﷯﷯﷯﷯ 𝑑𝑥=𝑥− ﷮1− 𝑥﷮2﷯﷯ 𝑠𝑖𝑛﷮−1﷯𝑥 +C