Misc 19 - Integrate sin-1 root x - cos-1 root x - CBSE - Integration using trigo identities - Inv Trigo formulae

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  1. Class 12
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Misc 19 Integrate the function sin﷮−1﷯﷮ ﷮𝑥﷯﷯ − cos﷮−1﷯﷮ ﷮𝑥﷯﷯﷮ sin﷮−1﷯﷮ ﷮𝑥﷯﷯ + cos﷮−1﷯﷮ ﷮𝑥﷯﷯﷯ , 𝑥∈ 0, 1﷯ Let 𝐼 = ﷮﷮ sin﷮−1﷯﷮ ﷮𝑥﷯﷯ − cos﷮−1﷯﷮ ﷮𝑥﷯﷯﷮ sin﷮−1﷯﷮ ﷮𝑥﷯﷯ + cos﷮−1﷯﷮ ﷮𝑥﷯﷯﷯﷯ 𝑑𝑥 We can write as sin﷮−1﷯﷮ ﷮𝑥﷯﷯ − cos﷮−1﷯﷮ ﷮𝑥﷯﷯﷮ sin﷮−1﷯﷮ ﷮𝑥﷯﷯ + cos﷮−1﷯﷮ ﷮𝑥﷯﷯﷯ = sin﷮−1﷯﷮ ﷮𝑥﷯﷯ − 𝜋﷮2﷯ − 𝑠𝑖𝑛﷮−1﷯﷮ ﷮𝑥﷯﷯﷯﷮ 𝜋﷮2﷯﷯ = 1﷮ 𝜋﷮2﷯﷯﷯ sin﷮−1﷯﷮ ﷮𝑥﷯﷯ − 𝜋﷮2﷯ − 𝑠𝑖𝑛﷮−1﷯﷮𝑥﷯﷯ = 2﷮𝜋﷯ 2 sin﷮−1﷯﷮ ﷮𝑥﷯﷯ − 𝜋﷮2﷯﷯ = 2﷮𝜋﷯ ×2 sin﷮−1﷯﷮ ﷮𝑥﷯﷯− 2﷮𝜋﷯× 𝜋﷮2﷯ = 4﷮𝜋﷯ sin﷮−1﷯﷮ ﷮𝑥﷯﷯−1 Integrating 𝑤.𝑟.𝑡.𝑥 ﷮﷮ sin﷮−1﷯﷮ ﷮𝑥﷯﷯ − cos﷮−1﷯﷮ ﷮𝑥﷯﷯﷮ sin﷮−1﷯﷮ ﷮𝑥﷯﷯ + cos﷮−1﷯﷮ ﷮𝑥﷯﷯﷯﷯ 𝑑𝑥= ﷮﷮ 4﷮𝜋﷯ sin﷮−1﷯﷮ ﷮𝑥﷯﷯−1﷯﷯𝑑𝑥 = ﷮﷮ 4﷮𝜋﷯ sin﷮−1﷯﷮ ﷮𝑥﷯﷯﷯𝑑𝑥− ﷮﷮𝑑𝑥﷯ = 4﷮𝜋﷯ ﷮﷮ sin﷮−1﷯﷮ ﷮𝑥﷯﷯﷯𝑑𝑥−𝑥+𝐶1 Hence I = 4﷮𝜋﷯ 𝐼1−𝑥+𝐶1 Solving 𝐈﷮𝟏﷯ 𝐼1 = ﷮﷮ sin﷮−1﷯﷮ ﷮𝑥﷯﷯﷯𝑑𝑥 Put ﷮𝑥﷯=𝑡 𝑥= 𝑡﷮2﷯ Differentiating 𝑤.𝑟.𝑡.𝑥 𝑑𝑥﷮𝑑𝑥﷯ = 𝑑 𝑡﷮2﷯﷮𝑑𝑥﷯ 1 = 2𝑡 𝑑𝑡﷮𝑑𝑥﷯ 𝑑𝑥 = 2𝑡 𝑑𝑡 Therefore ﷮﷮ sin﷮−1﷯﷮ ﷮𝑥﷯﷯﷯𝑑𝑥= ﷮﷮ sin﷮−1﷯﷮ 𝑡﷯﷯﷯.2𝑡 𝑑𝑡 =2 ﷮﷮ sin﷮−1﷯﷮𝑡﷯﷯.𝑡 𝑑𝑡 =2 ﷮﷮𝑡 sin﷮−1﷯﷮𝑡 ﷯﷯𝑑𝑡 Hence we take First function :- 𝑓 𝑡﷯= sin﷮−1﷯﷮𝑡 ﷯ Second function :- g 𝑡﷯=𝑡 =2[ sin﷮−1﷯﷮𝑡 ﷯ ﷮﷮𝑡﷯𝑑𝑡− ﷮﷮ 𝑑﷮𝑑𝑡﷯ sin﷮−1﷯﷮𝑡﷯﷯ ﷮﷮𝑡 𝑑𝑡﷯﷯﷯𝑑𝑡 =2 sin﷮−1﷯﷮𝑡 ﷯ 𝑡﷮2﷯﷮2﷯ − ﷮﷮ 1﷮ ﷮1 − 𝑡﷮2﷯﷯﷯﷯ × 𝑡﷮2﷯﷮2﷯ 𝑑𝑡+𝐶﷯ =2× 𝑡﷮2﷯﷮2﷯ sin﷮−1﷯﷮𝑡﷯−2× ﷮﷮ 1﷮2﷯ × 𝑡﷮2﷯﷮ ﷮1 − 𝑡﷮2﷯﷯﷯﷯ 𝑑𝑡+𝐶 = 𝑡﷮2﷯ sin﷮−1﷯﷮𝑡﷯− ﷮﷮ 𝑡﷮2﷯﷮ ﷮1 − 𝑡﷮2﷯﷯﷯﷯ 𝑑𝑡+𝐶 Solving ﷮﷮ − 𝒕﷮𝟐﷯﷮ ﷮𝟏 − 𝒕﷮𝟐﷯﷯﷯﷯ 𝒅𝒕 We can write − 𝑡﷮2﷯﷮ ﷮1 − 𝑡﷮2﷯﷯﷯ = − 𝑡﷮2﷯ + 1 − 1﷮ ﷮1 − 𝑡﷮2﷯﷯﷯ = 1 − 𝑡﷮2﷯ − 1﷮ ﷮1 − 𝑡﷮2﷯﷯﷯ = 1 − 𝑡﷮2﷯﷮ ﷮1 − 𝑡﷮2﷯﷯﷯ − 1﷮ ﷮1 − 𝑡﷮2﷯﷯﷯ = ﷮1 − 𝑡﷮2﷯﷯ − 1﷮ ﷮1 − 𝑡﷮2﷯﷯﷯ Integrating 𝑤.𝑟.𝑡.𝑥 ﷮﷮ − 𝑡﷮2﷯﷮1 − 𝑡﷮2﷯﷯﷯ dt = ﷮﷮ ﷮1 − 𝑡﷮2﷯﷯ − 1﷮ ﷮1 − 𝑡﷮2﷯﷯﷯﷯ ﷯ 𝑑𝑡 = ﷮﷮ ﷮ 1﷮2﷯ − 𝑡﷮2﷯﷯﷯ 𝑑𝑡− ﷮﷮ 1﷮ ﷮ 1﷮2﷯ − 𝑡﷮2﷯﷯﷯﷯ 𝑑𝑡 = 𝑡﷮2﷯ ﷮ 1﷮2﷯ − 𝑡﷮2﷯﷯+ 1﷮2﷯﷮2﷯ sin﷮−1﷯﷮ 𝑡﷮1﷯﷯− sin﷮−1﷯﷮ 𝑡﷮1﷯﷯ = 𝑡﷮2﷯ ﷮1 − 𝑡﷮2﷯﷯+ 1﷮2﷯ sin﷮−1﷯﷮𝑡﷯− sin﷮−1﷯﷮𝑡﷯ = 𝑡﷮2﷯ ﷮1 − 𝑡﷮2﷯﷯ − 1﷮2﷯ sin﷮−1﷯﷮𝑡﷯ Hence we can write 𝐼1 = 𝑡﷮2﷯ sin﷮−1﷯﷮𝑡﷯+ ﷮﷮ − 𝑡﷮2﷯﷮ ﷮1 − 𝑡﷮2﷯﷯﷯﷯ 𝑑𝑡 = 𝑡﷮2﷯ sin﷮𝑡﷯+ 𝑡﷮2﷯ ﷮1 − 𝑡﷮2﷯﷯− 1﷮2﷯ sin﷮−1﷯﷮𝑡﷯ Putting 𝑡 = ﷮𝑥﷯ 𝐼1 = ﷮𝑥﷯﷯﷮2﷯ sin﷮ ﷮𝑥﷯﷯+ ﷮𝑥﷯﷮2﷯ ﷮1 − ﷮𝑥﷯﷯﷮2﷯﷯− 1﷮2﷯ sin﷮−1﷯﷮ ﷮𝑥﷯﷯ = 𝑥 sin﷮ ﷮𝑥﷯﷯+ ﷮𝑥﷯﷮2﷯ ﷮1−𝑥﷯− 1﷮2﷯ sin﷮−1﷯﷮ ﷮𝑥﷯﷯ Hence 𝐼 = 4﷮𝜋﷯ 𝐼﷮1 ﷯−𝑥+ C﷮1﷯ 𝐼 = 4﷮𝜋﷯ 𝑥 𝑠𝑖𝑛﷮−1﷯ ﷮𝑥﷯+ ﷮𝑥﷯﷮2﷯ ﷮1−𝑥﷯− 1﷮2﷯ 𝑠𝑖𝑛﷮−1﷯ ﷮𝑥﷯﷯−𝑥+ C﷮1﷯ 𝐼 = 4﷮𝜋﷯ 𝑥 𝑠𝑖𝑛﷮−1﷯ ﷮𝑥﷯+ ﷮𝑥 − 𝑥﷮2﷯﷯﷮2﷯ − 1﷮2﷯ 𝑠𝑖𝑛﷮−1﷯ ﷮𝑥﷯﷯−𝑥+ C﷮1﷯ 𝐼 = 4﷮𝜋﷯𝑥 𝑠𝑖𝑛﷮−1﷯ ﷮𝑥﷯+ 2﷮𝜋﷯ ﷮𝑥 − 𝑥﷮2﷯﷯− 2﷮𝜋﷯ 𝑠𝑖𝑛﷮−1﷯ ﷮𝑥﷯−𝑥+ C﷮1﷯ 𝐼 = 𝑠𝑖𝑛﷮−1﷯ ﷮𝑥﷯ 4𝜋﷮𝜋﷯− 2𝜋﷮𝜋﷯﷯+ 2 ﷮𝜋 − 𝜋﷮2﷯﷯﷮𝜋﷯−𝑥+ C﷮1﷯ 𝐼 = 𝟐𝝅﷮𝝅﷯ 𝟐𝒙−𝟏﷯ 𝒔𝒊𝒏﷮−𝟏﷯ ﷮𝒙﷯+ 𝟐 ﷮𝝅 − 𝝅﷮𝟐﷯﷯﷮𝝅﷯−𝒙+ 𝑪﷮𝟏﷯

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