Math, asked by shreyaSingh2022, 1 month ago


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Integrate. [cos^-1x (√1-x^2)]^-1 / logₑ{1+(sin(2x√1-x^2)/π}

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Answers

Answered by latabara97
2

Let I=∫

1−x

2

xcos

−1

x

dx

I=

2

−1

1−x

2

−2x

⋅cos

−1

xdx

Taking cos

−1

x as first function and (

1−x

2

−2x

) as second function and integrating by parts, we obtain I=

2

−1

[cos

−1

x∫

1−x

2

−2x

dx−∫{(

dx

d

cos

−1

x)∫

1−x

2

−2x

dx}dx]

=

2

−1

[cos

−1

x⋅2

1−x

2

−∫

1−x

2

−1

⋅2

1−x

2

dx]

=

2

−1

[2

1−x

2

cos

−1

x+∫2dx]

=

2

−1

[2

1−x

2

cos

−1

x+2x]+C

=−[

1−x

2

cos

−1

x+x]+C

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