integral of sin^-1(cos x)dx
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I = sin-¹(cosx ).dx
use integration by part, if f and g are two function and f is 1st function and g is 2nd .
f.dg = f.g - g.df use this concept .
f = sin-¹(cosx)
df = 1/√(1 - cos²x) (-sinx).dx
df = -dx
I = sin-¹(cosx).x - x{-dx}
= x²/2 + xsin-¹(cosx) + Q
use integration by part, if f and g are two function and f is 1st function and g is 2nd .
f.dg = f.g - g.df use this concept .
f = sin-¹(cosx)
df = 1/√(1 - cos²x) (-sinx).dx
df = -dx
I = sin-¹(cosx).x - x{-dx}
= x²/2 + xsin-¹(cosx) + Q
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