Integral of Sin Inverse

Integral of sin inverse can be calculated using different formulas of integration. Integration is a reverse process of differentiation. An integral is nothing but an anti-derivative. Sin inverse integral is written as ∫sin^-1x dx = x sin^-1x + √(1 - x²) + C, where ∫ is the sign of integration, dx denotes that the integration of sin inverse is with respect to x, and C is the constant of integration.

Let us learn the process of finding the integral of sin inverse x using different methods and determine the definite integral of sin inverse.

The integral of sin inverse is given by x sin^-1x + √(1 - x²) + C, where C is the constant of integration. Mathematically, the sin inverse integral is written as ∫arcsin x dx = ∫sin^-1x dx = x sin^-1x + √(1 - x²) + C. Integral of sin inverse x is also called the antiderivative of sin inverse x. Integration of sin inverse can be done using different methods such as integration by parts and substitution method followed by integration by parts.

Sin Inverse Integration Formula

The formula for the integral of arcsin is given by, ∫sin^-1x dx = x sin^-1x + √(1 - x²) + C, where C is the constant of integration.

Now, that we know that the integration of sin inverse is ∫sin^-1x dx = x sin^-1x + √(1 - x²) + C, we will prove this using integration by parts. We will use the following formulas and facts to prove the integration of sin inverse.

• The formula for integration by parts is ∫f(x)g(x)dx = f(x) ∫g(x)dx - ∫[d(f(x))/dx × ∫g(x) dx] dx.
• Note that sin^-1x can be written as sin^-1x = sin^-1x.1.
• We have f(x) = sin^-1x, g(x) = 1
• d(sin^-1x)/dx = 1/√(1 - x²)

Using these formulas and facts, we have

∫sin^-1x dx = ∫sin^-1x.1 dx

= sin^-1x ∫1dx - ∫[d(sin^-1x)/dx × ∫1 dx] dx

= x sin^-1x - ∫[1/√(1 - x²) × x] dx

= x sin^-1x - ∫x/√(1 - x²) dx

= x sin^-1x + (1/2) ∫-2x/√(1 - x²) dx [Multiplying and dividing by 2]

= x sin^-1x + (1/2) ∫(-2x)(1 - x²)^-1/2 dx

= x sin^-1x + (1/2) [(1 - x²)^{-1/2 + 1}/ (-1/2 + 1)] + C {Using formula ∫[f(x)]ⁿ f'(x) dx = [f(x)]^{n + 1}/(n + 1) + C}

= x sin^-1x + (1/2) [(1 - x²)^1/2/ (1/2)] + C

= x sin^-1x + (1 - x²)^1/2 + C

= x sin^-1x + √(1 - x²) + C

We have proved the integral of sin inverse using the method integration by parts. Now, we will prove the sin inverse integral by substitution method of integration followed by the method of integration by parts. To determine the integration of sin inverse x, we will substitute x with sin θ. We will use the following formulas to determine the integral of sin inverse:

• Assume x = sinθ ⇒ sin^-1x = sin^-1(sinθ) = θ
• dx = cosθ dθ
• Integration by Parts: ∫f(x)g(x)dx = f(x) ∫g(x)dx - ∫[d(f(x))/dx × ∫g(x) dx] dx
• sin²θ + cos²θ = 1 ⇒ cosθ = √(1 - sin²θ)

Using the above formulas, we have

∫sin^-1x dx = ∫sin^-1(sinθ) cosθ dθ

= ∫θ cosθ dθ

= θ ∫cosθ dθ - ∫[d(θ)/dθ × ∫cosθ dθ] {By substituting f(θ) = θ and g(θ) = cosθ in Integration by parts formula}

= θ sinθ - ∫1.sinθ dθ

= θ sinθ - ∫sinθ dθ

= θ sinθ + cosθ + C

= θ sinθ + √(1 - sin²θ) + C

= x sin^-1x + √(1 - x²) + C

Hence, we have proved that the integral of arcsin is x sin^-1x + √(1 - x²) + C using the substitution method followed by integration by parts.

We know that the integral of sin inverse is given by, ∫sin^-1x dx = x sin^-1x + √(1 - x²) + C. We will use this formula to determine the definite integral of sin inverse with limits from 0 to 1.

\(\begin{align}\int_{0}^{1}\sin^{-1}x &=\left [ x\sin^{-1}x+\sqrt{1-x^2}+C \right ]_0^1\\&=1\sin^{-1}(1)+\sqrt{1-1^2}+C - (0\sin^{-1}(0)+\sqrt{1-0^2}+C)\\&=\dfrac{\pi}{2}+0+C-0-1-C\\&=\dfrac{\pi}{2}-1 \end{align}\)

Hence the definite integral of sin inverse x with limits from 0 to 1 is given by π/2 - 1.

Important Notes on Integral of Sin Inverse

• The integral of sin inverse is also called the integral of arcsin.
• The integral of sin inverse is x sin^-1x + √(1 - x²) + C
• The derivative of sin inverse is 1/√(1 - x²), -1 < x < 1

Related Topics on Integral of Sin Inverse

• Derivative of Sin Inverse
• Integral of Sin x
• Integration

What is Integral of Sin Inverse in Calculus?

The integral of sin inverse is given by ∫sin^-1x dx = x sin^-1x + √(1 - x²) + C, where C is the constant of integration.

How to Find the Sin Inverse Integral?

Integration of sin inverse can be done using different methods such as integration by parts and substitution method followed by integration by parts.

What is the Integral of Sin Inverse Whole Square?

The Integral of Sin Inverse Whole Square is given by ∫(sin^-1x)²dx = (sin^-1x)² + 2(sin^-1x)√(1 - x²) - 2x + C, where C is the constant of integration.

Is the Derivative of Sin Inverse Equal to the Integral of Arcsin?

No, the derivative of sin inverse is not equal to the arcsin integration. The derivative of sin inverse is given by, d(sin^-1x)/dx = 1/√(1 - x²) whereas the integral of sin inverse is ∫sin^-1x dx = x sin^-1x + √(1 - x²) + C

What is the Integral of Sin Inverse from 0 to 1?

The definite integral of sin inverse x with limits from 0 to 1 is given by π/2 - 1.