Introduction to integral of sin^-1x

In calculus, the integral is a fundamental concept that assigns numbers to functions to define displacement, area, volume, and all those functions that contain a combination of tiny elements. It is categorized into two parts, definite integral and indefinite integral. The process of integration calculates the integrals. This process is defined as finding an antiderivative of a function.

Integrals can handle almost all functions, such as trigonometric, algebraic, exponential, logarithmic, etc. This article will teach you what is integral to a trigonometric function cubic sine. You will also understand how to compute sin inverse integral by using different integration techniques.

What is the integral of sin^-1x?

The integral of sin^-1x is an antiderivative of sine function which is equal to x sin^-1x + √(1 - x²)+ c. It is also known as the reverse derivative of sine function which is a trigonometric identity.

The sine function is the ratio of opposite side to the hypotenuse of a triangle which is written as:

Sin = opposite side / hypotenuse

Integral of sin^-1x formula

The formula of integral of sin contains integral sign, coefficient of integration and the function as sine. It is denoted by ∫(sin^-1x)dx. In mathematical form, the integral of sin^-1x is:

$∫\sin^{-1}xdx=x\sin^{-1}x + \sqrt{1 - x^2}+ c$

Where c is any constant involved, dx is the coefficient of integration and ∫ is the symbol of integral. 

How to calculate the integral of sin^-1x?

The integral of sin^3x is its antiderivative that can be calculated by using different integration techniques using the integral calculator. In this article, we will discuss how to calculate integral of sine by using:

1. Integration by parts
2. Substitution method
3. Definite integral

Integral of sin inverse x by using integration by parts

The derivative of a function calculates the rate of change, and integration is the process of finding the antiderivative of a function. The integration by parts is a method of solving integral of two functions combined together. Let’s discuss calculating the integral of sin inverse x by using integration by parts finder.

Proof of integral of sin^-1x by using integration by parts

Since we know that the function sine inverse x can be written as the product of two functions. Therefore, we can calculate the integral of sin^-1x by using integration by parts. For this, suppose that:

$I = \sin^{-1}x =1.\sin^{-1}x$

Applying the integral we get,

$I = ∫1.\sin^{-1}xdx$

Since the method of integration by parts is:

$∫[f(x).g(x)]dx = f(x).∫g(x)dx - ∫[f’(x).∫g(x)dx]dx$

Now replacing f(x) and g(x) by sin x, we get,

$I = x\sin^{-1}x - ∫\left[\frac{1}{√1 – x^2}.(x)\right]dx$

It can be written as:

$I = x\sin^{-1}x - ∫\left[\frac{x}{\sqrt{1 – x^2}}\right]dx$

Now multiplying and dividing by 2, we get,

$I=x\sin^{-1}x +\frac{1}{2}∫\frac{-2x}{\sqrt{1 - x^2}} dx$

$I=x\sin^{-1}x+\frac{1}{2}∫(-2x)(1-x^2)^{-\frac{1}{2}}dx$

$I=x\sin^{-1}x+\frac{1}{2}\left[\frac{(1-x^2)^{-\frac{1}{2} + 1}}{-\frac{1}{2} + 1}\right] + C$

More simplification,

$I=x\sin^{-1}x+\frac{1}{2}\left[\frac{(1 - x^2)^{\frac{1}{2}}}{1/2}\right] + C$

$I=x\sin^{-1}x+(1 - x^2)^{\frac{1}{2}} + C$

Hence the derivation of integral of sin inverse x is:

$I=x\sin^{-1}x+\sqrt{1 - x^2} + C$

Integral of sin^-1x by using substitution method

The substitution formula calculator involves many trigonometric formulas. We can use these formulas to verify the integrals of different trigonometric functions such as sine, cosine, tangent, etc. Let’s understand how to prove the integral of sin inverse by using the substitution method.

Proof of Integral of sin^-1x by using substitution method

To proof the integral of sin^-1x by using substitution method, the following formulas can be used.

• 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

$∫\sin^{-1}xdx=∫\sin^{-1}(\sinθ)\cosθ dθ=∫θ\cosθdθ$

Now by substituting f(θ) = θ and g(θ) = cosθ in Integration by parts formula

$= θ ∫\cosθ dθ - ∫\frac{d}{dθ}(\theta)× ∫\cosθ dθ$

$= θ\sinθ - ∫1.\sinθ dθ$

$= θ \sinθ - ∫\sinθ dθ$

$= θ\sinθ + \cosθ + C$

$= θ\sinθ + \sqrt{1 - \sin^2θ} + C$

$= x\sin^{-1}x+\sqrt{1 - x^2}+C$

Hence the integration of sin^-1x is verified by using substitution method. Also, you can use our integration by trig substitution calculator to integrate sin inverse without manual calculations.

Integral of sin^-1x by using definite integral

The definite integral is a type of integral that calculates the area of a curve by using infinitesimal area elements between two points. The definite integral formula can be written as:

$∫^b_a f(x) dx = F(b) – F(a)$

Let’s understand the verification of the integral of sin^2x by using the definite integral.

Proof of integral of sin^-1x by using definite integral

To compute the integral of sin^-1x by using a definite integral, we can use the interval from 0 to 1. Let’s compute the integral of sin^-1x from 0 to 1.

Since we know the integral of sine inverse x,

$I = x\sin^{-1}x + \sqrt{1 - x^2} + C$

The definite integral of sin^-1x can be written as:

$∫^1_0 \sin^{-1}xdx=\left|x\sin^{-1}x+\sqrt{1 -x^2}\right|^1_0$

Substituting the value of limit we get,

$∫^1_0 \sin^{-1}xdx=[\sin^{-1}1 + \sqrt{1 - 1^2}] – [0 \sin^{-1}0 + \sqrt{1 - 0}]$

Since sin inverse 1 is equal to π/2, then,

$∫^1_0 \sin^{-1}xdx=\frac{π}{2}+0–1$

Therefore, the integral of sin-1x from 0 to 1 is

$∫^1_0 \sin^{-1}x dx =\frac{π}{2}–1$

Which is the calculation of the definite integral of sin^-1x. If the integral of a funciton is unbounded, use our indefinite integral calculator to eveluate such integrals.