The value of int sqrt{1+sec x} , dx is | vedclass

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

Question

(B) Let $I = \int \sqrt{1+\sec x} \, dx$. We know that $1+\sec x = 1 + \frac{1}{\cos x} = \frac{\cos x + 1}{\cos x} = \frac{2 \cos^2 \frac{x}{2}}{\cos

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Let $I = \int \sqrt{1+\sec x} \, dx$. We know that $1+\sec x = 1 + \frac{1}{\cos x} = \frac{\cos x + 1}{\cos x} = \frac{2 \cos^2 \frac{x}{2}}{\cos x}$. So,$I = \int \frac{\sqrt{2} \cos \frac{x}{2}}{\sqrt{\cos x}} \, dx$. Using $\cos x = 1 - 2 \sin^2 \frac{x}{2}$,we get $I = \int \frac{\sqrt{2} \cos \frac{x}{2}}{\sqrt{1 - 2 \sin^2 \frac{x}{2}}} \, dx$. Let $t = \sqrt{2} \sin \frac{x}{2}$. Then $dt = \sqrt{2} \cdot \frac{1}{2} \cos \frac{x}{2} \, dx = \frac{1}{\sqrt{2}} \cos \frac{x}{2} \, dx$. This implies $\sqrt{2} \cos \frac{x}{2} \, dx = 2 \, dt$. Substituting these into the integral,we get $I = \int \frac{2 \, dt}{\sqrt{1 - t^2}} = 2 \sin^{-1}(t) + C$. Substituting back $t = \sqrt{2} \sin \frac{x}{2}$,we obtain $I = 2 \sin^{-1}(\sqrt{2} \sin \frac{x}{2}) + C$.

The value of $\int \sqrt{1+\sec x} \, dx$ is

Similar Questions

Let $I_n = \int \sec^n x \, dx$. If $5 I_6 - 4 I_4 = f(x)$, then $f\left(\frac{\pi}{4}\right)$ is equal to

If $f(x) = \int \frac{x^2 + \sin^2 x}{1 + x^2} \cdot \sec^2 x \, dx$ and $f(0) = 0$,then $f(1) = $

Let for $f(x)=7 \tan^8 x + 7 \tan^6 x - 3 \tan^4 x - 3 \tan^2 x$,$I_1 = \int_0^{\pi/4} f(x) \, dx$ and $I_2 = \int_0^{\pi/4} x f(x) \, dx$. Then $7 I_1 + 12 I_2$ is equal to:

Find the integral of the function $\tan^{3} 2x \sec 2x$.

$\int \frac{d x}{12 \cos x+5 \sin x}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module