int frac{sin x cos x}{sqrt{1-sin ^{4} x}} d x | vedclass

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

Question

(A) Let $I = \int \frac{\sin x \cos x}{\sqrt{1-\sin ^{4} x}} dx$. Substitute $\sin^{2} x = t$. Differentiating both sides with respect to $x$,we get $

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $I = \int \frac{\sin x \cos x}{\sqrt{1-\sin ^{4} x}} dx$. Substitute $\sin^{2} x = t$. Differentiating both sides with respect to $x$,we get $2 \sin x \cos x dx = dt$,which implies $\sin x \cos x dx = \frac{1}{2} dt$. Substituting these into the integral,we get $I = \int \frac{1}{2 \sqrt{1-t^{2}}} dt$. Using the standard integral formula $\int \frac{1}{\sqrt{1-t^{2}}} dt = \sin^{-1} t + C$,we obtain $I = \frac{1}{2} \sin^{-1} t + C$. Finally,substituting $t = \sin^{2} x$ back,we get $I = \frac{1}{2} \sin^{-1}(\sin^{2} x) + C$.

$\int \frac{\sin x \cos x}{\sqrt{1-\sin ^{4} x}} d x$ is equal to

Similar Questions

If $0 < x < 1$ and $\int \frac{dx}{\sqrt{x^2-x^5}} = \frac{1}{3} \log |f(x)| + C$,then $f\left(\frac{1}{2}\right) = $

$\int \frac{1}{\cos^2 x (1 - \tan x)^2} dx = $

$\int (\sqrt{\tan x} + \sqrt{\cot x}) \, dx = $

$\int {\frac{{{e^{2x}} - 1}}{{{e^{2x}} + 1}}} \,dx = $

Evaluate the integral: $\int \frac{\sin^{-1} x}{\sqrt{1-x^2}} \, dx$

Vedclass Test Series

Exam Paper Generator

Online Exam Module