int {xsqrt {frac{{1 - {x^2}}}{{1 + {x^2}}}} } | vedclass

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

Question

(A) Let $I = \int x\sqrt {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \;dx$. Multiply the numerator and denominator by $\sqrt{1-x^2}$: $I = \int \frac{x(1-x^2)}{

Vedclass · Accepted Answer

$\frac{1}{2}[{\sin ^{ - 1}}({x^2}) + \sqrt {1 - {x^4}} ] + c$

Vedclass · Answer

(A) Let $I = \int x\sqrt {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \;dx$. Multiply the numerator and denominator by $\sqrt{1-x^2}$: $I = \int \frac{x(1-x^2)}{\sqrt{(1-x^2)(1+x^2)}} \;dx = \int \frac{x - x^3}{\sqrt{1-x^4}} \;dx$. Split the integral: $I = \int \frac{x}{\sqrt{1-x^4}} \;dx - \int \frac{x^3}{\sqrt{1-x^4}} \;dx$. For the first part,let $x^2 = t$,then $2x \;dx = dt$,so $x \;dx = \frac{1}{2} dt$: $\int \frac{x}{\sqrt{1-x^4}} \;dx = \frac{1}{2} \int \frac{dt}{\sqrt{1-t^2}} = \frac{1}{2} \sin^{-1}(t) = \frac{1}{2} \sin^{-1}(x^2)$. For the second part,let $1-x^4 = u$,then $-4x^3 \;dx = du$,so $x^3 \;dx = -\frac{1}{4} du$: $\int \frac{x^3}{\sqrt{1-x^4}} \;dx = -\frac{1}{4} \int u^{-1/2} \;du = -\frac{1}{4} (2\sqrt{u}) = -\frac{1}{2} \sqrt{1-x^4}$. Combining these results: $I = \frac{1}{2} \sin^{-1}(x^2) - (-\frac{1}{2} \sqrt{1-x^4}) + c = \frac{1}{2} [\sin^{-1}(x^2) + \sqrt{1-x^4}] + c$.

$\int {x\sqrt {\frac{{1 - {x^2}}}{{1 + {x^2}}}} } \;dx = $

Similar Questions

If $\int \frac{5 \tan x}{\tan x-2} dx=ax+b \log |\sin x-2 \cos x|+c$,then $a+b=$

For $x > 0$,if $\int (\log x)^5 dx$ is equal to $x[A(\log x)^5 + B(\log x)^4 + C(\log x)^3 + D(\log x)^2 + E(\log x) + F] + \text{constant}$,then $A + B + C + D + E + F$ is equal to

If $\int f(x) \cos x \, dx = \frac{1}{2} [f(x)]^2 + C$ and $f(0) = 0$,then $f'(0) = $

For $n \ge 2$,if $I_n = \int (\sin x + \cos x)^n dx$,then $nI_n - 2(n-1)I_{n-2} = $

Let $I_n = \int \tan^n x dx, (n > 1)$. If $I_4 + I_6 = a \tan^5 x + b x^5 + C$,where $C$ is the constant of integration,then the ordered pair $(a, b)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module