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1800-102-2727Cosine of a function is given by: cos θ = adjacent side / hypotenuse. The inverse of a cosine function is given by:
Θ = cos-1 (adjacent side / hypotenuse)
We use the inverse cosine function to find the unknown angles in a right-angled triangle.
We will derive the derivative of inverse cosine function using chain rule.
Let us assume y = cos-1x ⇒ cos y = x. Differentiating both sides, we get,
D (cos y) / dx = dx / dx
-sin y dy / dx = 1
dy / dx = -1 / sin y
We know, cos2 y + sin2 y = 1, we have,
sin y = √(1 - cos2 y) = √(1 - x2) [as cos y = x]
Substituting sin y = √(1 - x2) in (1), we have,
dy / dx = -1 / √(1 - x2)
Since x = -1, 1 makes the denominator √(1 - x2) equal to 0. Therefore, the derivative is not defined if x is -1 and 1.
We can conclude that the derivative of inverse cosine function is -1 / √(1 - x2), where -1 < x < 1.
We will find the integration of the inverse cosine function using integration by parts, ILATE (inverse, logarithmic, algebraic, trigonometric, and exponent).
∫cos-1 x = ∫cos-1 x · 1 dx
Using integration by parts,
∫f (x) . g (x) dx = f (x) ∫g (x) dx − ∫(f′ (x) ∫g (x) dx) dx + C
Here f (x) = cos-1 x and g (x) = 1.
∫cos-1 x · 1 dx = cos-1 x ∫1 dx - ∫ [d (cos-1 x) / dx ∫1 dx] dx + C
∫cos-1 x dx = cos-1 x . (x) - ∫ [-1 / √(1 - x²)] x dx + C
We will evaluate this integral ∫ [-1 / √(1 - x²)] x dx using substitution method. Let us assume 1 - x2 = u. Then -2x dx = du (or) x dx = -1/2 du.
∫cos-1 x dx = x cos-1x - ∫(-1/√u) (-1/2) du + C
= x cos-1 x - 1/2 ∫u-1/2 du + C
= x cos-1 x - (1/2) (u1/2 / (1/2)) + C
= x cos-1x - √u + C
= x cos-1x - √(1 - x²) + C
Therefore, ∫cos-1x dx = x cos-1x - √(1 - x²) + C