Question

Evaluate [cos-¹x{√(1-x²)}]-¹/loge{1+(sin(2x(√1-x²)/π)}

Answer 1

Answer:

[cos -1x{√(1-×2) }] ok

Answer 2

Step-by-step explanation:

∫fg′=fg−∫f′g

f =arccos(x), g′ =x/√1−x^2

f′ =−1/√1−x2, g =−√1−x2

= −√1−x2arccos (x)−∫1 dx

Now solve: ∫1 dx, apply constant rule: =x

Plug in our solved integrals:

−√1−x2arccos(x)−∫1dx =−√1−x2arccos(x)−x

Thus the answer is:

−√1−x2arccos(x)−x+C