int cos^{-1} x , dx = | vedclass

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

Question

(D) To evaluate the integral $I = \int \cos^{-1} x \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$.Let $u = \cos^

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) To evaluate the integral $I = \int \cos^{-1} x \, dx$,we use the method of integration by parts: $\int u \, dv = uv - \int v \, du$.Let $u = \cos^{-1} x$ and $dv = dx$.Then $du = -\frac{1}{\sqrt{1-x^2}} \, dx$ and $v = x$.Applying the formula: $I = x \cos^{-1} x - \int x \left( -\frac{1}{\sqrt{1-x^2}} \right) \, dx$.$I = x \cos^{-1} x + \int \frac{x}{\sqrt{1-x^2}} \, dx$.To solve $\int \frac{x}{\sqrt{1-x^2}} \, dx$,let $t = 1-x^2$,so $dt = -2x \, dx$ or $x \, dx = -\frac{1}{2} \, dt$.The integral becomes $\int \frac{-1/2}{\sqrt{t}} \, dt = -\frac{1}{2} \int t^{-1/2} \, dt = -\frac{1}{2} \left( \frac{t^{1/2}}{1/2} \right) = -\sqrt{t} = -\sqrt{1-x^2}$.Therefore,$I = x \cos^{-1} x - \sqrt{1-x^2} + c$.

$\int \cos^{-1} x \, dx =$

Similar Questions

$\int \sin ^{-1} x \, dx =$

Integrate the function: $\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$

$\int(\log x)^2 x^3 d x=\frac{x^4}{32} f(x)+C \Rightarrow f(x)=$

If $I(m, n) = \int_0^1 t^m (1 + t)^n dt$,then the expression for $I(m, n)$ in terms of $I(m + 1, n - 1)$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module