int sin^{-1} x , dx is equal to | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) To evaluate $I = \int \sin^{-1} x \, dx$,we use integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = \sin^{-1} x$ and $dv = dx$. The

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) To evaluate $I = \int \sin^{-1} x \, dx$,we use integration by parts: $\int u \, dv = uv - \int v \, du$. Let $u = \sin^{-1} x$ and $dv = dx$. Then $du = \frac{1}{\sqrt{1 - x^2}} \, dx$ and $v = x$. Applying the formula: $I = x \sin^{-1} x - \int x \cdot \frac{1}{\sqrt{1 - x^2}} \, dx$. To solve the integral $\int \frac{x}{\sqrt{1 - x^2}} \, dx$,let $t = 1 - x^2$,so $dt = -2x \, dx$,which means $x \, dx = -\frac{1}{2} dt$. Substituting this: $\int \frac{x}{\sqrt{1 - x^2}} \, dx = \int \frac{-1/2}{\sqrt{t}} \, dt = -\frac{1}{2} \int t^{-1/2} \, dt = -\frac{1}{2} \cdot (2t^{1/2}) = -\sqrt{t} = -\sqrt{1 - x^2}$. Therefore,$I = x \sin^{-1} x - (-\sqrt{1 - x^2}) + c = x \sin^{-1} x + \sqrt{1 - x^2} + c$.

$\int \sin^{-1} x \, dx$ is equal to

Similar Questions

Prove that $\int_{0}^{1} x e^{x} d x = 1$.

$\int \frac{x \cdot \log x}{\left(\sqrt{x^2-1}\right)^3} d x=$

The positive integer $n \leq 5$ for which $\int_0^1 e^x(x-1)^n dx = 16-6e$ is

Find $\int \log x \, dx$.

If $f(x)$ and $g(x)$ are integrable functions, then which of the following is equal to $[\int f(x) dx][\int g(x) dx]$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module