int sqrt{1 + csc x} , dx equals | vedclass

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

Question

(D) Let $I = \int \sqrt{1 + \frac{1}{\sin x}} \, dx = \int \sqrt{\frac{\sin x + 1}{\sin x}} \, dx$. Multiply numerator and denominator by $\sqrt{1 - \

Vedclass · Accepted Answer

Both $(A)$ and $(C)$

Vedclass · Answer

(D) Let $I = \int \sqrt{1 + \frac{1}{\sin x}} \, dx = \int \sqrt{\frac{\sin x + 1}{\sin x}} \, dx$. Multiply numerator and denominator by $\sqrt{1 - \sin x}$: $I = \int \frac{\sqrt{(1 + \sin x)(1 - \sin x)}}{\sqrt{\sin x(1 - \sin x)}} \, dx = \int \frac{\cos x}{\sqrt{\sin x - \sin^2 x}} \, dx$. Let $t = \sin x$,then $dt = \cos x \, dx$. $I = \int \frac{dt}{\sqrt{t - t^2}} = \int \frac{dt}{\sqrt{\frac{1}{4} - (t^2 - t + \frac{1}{4})}} = \int \frac{dt}{\sqrt{(\frac{1}{2})^2 - (t - \frac{1}{2})^2}}$. Using $\int \frac{du}{\sqrt{a^2 - u^2}} = \sin^{-1}(\frac{u}{a}) + c$: $I = \sin^{-1}\left(\frac{t - 1/2}{1/2}\right) + c = \sin^{-1}(2t - 1) + c = \sin^{-1}(2 \sin x - 1) + c$. Since $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$,$\sin^{-1}(2 \sin x - 1) = \frac{\pi}{2} - \cos^{-1}(2 \sin x - 1)$. Also,$\cos^{-1}(1 - 2 \sin x) = \pi - \cos^{-1}(2 \sin x - 1)$. By simplifying the trigonometric identities,the integral results in $2 \sin^{-1} \sqrt{\sin x} + c$ and $\cos^{-1}(1 - 2 \sin x) + c$ (up to a constant). Thus,the correct option is $(D)$.

$\int \sqrt{1 + \csc x} \, dx$ equals

Similar Questions

Let $g:(0, \infty) \rightarrow R$ be a differentiable function such that $\int \left( \frac{x(\cos x - \sin x)}{e^x + 1} + \frac{g(x)(e^x + 1 - x e^x)}{(e^x + 1)^2} \right) dx = \frac{x g(x)}{e^x + 1} + c$ for all $x > 0$,where $c$ is an arbitrary constant. Then:

If $f(x) = g(x)$,then the value of $\int {f'(x) \cdot g(x)} \, dx$ is

$\int (\sin^{-1} x + \cos^{-1} x) \, dx = $

If $\int \frac{dx}{\cos^3 x \sqrt{2 \sin 2x}} = (\tan x)^A + C(\tan x)^B + K$,where $K$ is a constant of integration,then the value of $5(A+B+C)$ is equal to

$y=\int \cos \left\{2 \tan ^{-1} \sqrt{\frac{1-x}{1+x}}\right\} d x$ is an equation of a family of

Vedclass Test Series

Exam Paper Generator

Online Exam Module