Find the most general antiderivative or indefinite integral. ? d x ? 1 ? x 2 sin ? 1 x

Question:

Find the most general antiderivative or indefinite integral.

$$\int \frac{dx}{\sqrt{1-x^2} \sin^{-1}x}$$

Indefinite Integration in Calculus:

The given integrand is an inverse trigonometric sine function. We can use integration rules to solve this kind of problem.

We use u-substitution to get the standard form of the integrand and apply the common integral rule {eq}\displaystyle \int \dfrac{1}{u} \, \mathrm{d}u= \displaystyle \ln|u|+C {/eq} to solve.

Answer and Explanation:

We are given: {eq}\displaystyle \int \frac{dx}{\sqrt{1-x^2} \sin^{-1}x} {/eq}


Apply u-substitution:

{eq}\displaystyle u=\sin^{-1}x \rightarrow \ du= \dfrac{1}{\sqrt{1-x^2}} \ dx {/eq}

{eq}= \displaystyle \int \dfrac{1}{u} \, \mathrm{d}u {/eq}

{eq}= \displaystyle \ln|u|+C {/eq}


Substitute back {eq}\displaystyle u=\sin^{-1}x {/eq} we'll get:

{eq}= \displaystyle \ln|\sin^{-1}x |+C {/eq}


Therefore the solution is:

{eq}\displaystyle {\boxed{ \int \frac{dx}{\sqrt{1-x^2} \sin^{-1}x}=\ln|\sin^{-1}x |+C}} {/eq}.


Learn more about this topic:

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Indefinite Integral: Definition, Rules & Examples

from Calculus: Tutoring Solution

Chapter 7 / Lesson 14
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