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

To find the value of $\int \frac{dx}{x\sqrt{2ax - x^2}}$,the suitable substitution is

$\int \frac{x}{x^4 + x^2 + 1} dx$ is equal to

$\int {\frac{{10{x^9} + {{10}^x}{{\log }_e}10}}{{{{10}^x} + {x^{10}}}}} \;dx = $

$\int \frac{d x}{(x+100) \sqrt{x+99}}=f(x)+c \Rightarrow f(x)$

$\int \frac{1}{3-2 \cos 2 x} \,d x=$ (where $C$ is constant of integration.)

Vedclass Test Series

Exam Paper Generator

Online Exam Module