int tan ^{-1}(sec x+tan x) d x= | vedclass

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

Question

(A) Let $I = \int 	an ^{-1}(\sec x+	an x) d x$.We know that $\sec x = \frac{1}{\cos x}$ and $	an x = \frac{\sin x}{\cos x}$,so $\sec x + 	an x = \

Vedclass · Accepted Answer

$\frac{\pi x}{4}+\frac{x^2}{4}+c$

Vedclass · Answer

(A) Let $I = \int 	an ^{-1}(\sec x+	an x) d x$.We know that $\sec x = \frac{1}{\cos x}$ and $	an x = \frac{\sin x}{\cos x}$,so $\sec x + 	an x = \frac{1+\sin x}{\cos x}$.Using half-angle formulas,$1+\sin x = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$ and $\cos x = \cos^2 \frac{x}{2} - \sin^2 \frac{x}{2} = (\cos \frac{x}{2} - \sin \frac{x}{2})(\cos \frac{x}{2} + \sin \frac{x}{2})$.Thus,$\frac{1+\sin x}{\cos x} = \frac{(\cos \frac{x}{2} + \sin \frac{x}{2})^2}{(\cos \frac{x}{2} - \sin \frac{x}{2})(\cos \frac{x}{2} + \sin \frac{x}{2})} = \frac{\cos \frac{x}{2} + \sin \frac{x}{2}}{\cos \frac{x}{2} - \sin \frac{x}{2}}$.Dividing numerator and denominator by $\cos \frac{x}{2}$,we get $\frac{1+	an \frac{x}{2}}{1-	an \frac{x}{2}} = 	an(\frac{\pi}{4} + \frac{x}{2})$.Therefore,$I = \int 	an^{-1}(	an(\frac{\pi}{4} + \frac{x}{2})) d x = \int (\frac{\pi}{4} + \frac{x}{2}) d x$.Integrating term by term,we get $I = \frac{\pi x}{4} + \frac{x^2}{4} + c$.

$\int \tan ^{-1}(\sec x+\tan x) d x=$

Similar Questions

$\int \sin^3 x \, dx$ is equal to

$\int \frac{dx}{\tan x + \cot x} = $

Evaluate the integral: $\int a^{3x + 3} dx$.

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

$\int \frac{1}{\cos x+\sqrt{3} \sin x} dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module