Question
$\int \frac{d x}{\tan x+\cot x+\sec x+\operatorname{cosec} x}=$(A) $\frac{1}{2}(\sin x-\cos x+x)+c$(B) $\frac{1}{2}(\sin x-\cos x-x)+c$(C) $\frac{1}{2}(\sin x+\cos x+x)+c$(D) none of these
Step 1
Step 1: We start by rewriting the integral in terms of sine and cosine: \[\int \frac{d x}{\tan x+\cot x+\sec x+\operatorname{cosec} x}=\int \frac{d x}{\frac{\sin x}{\cos x}+\frac{\cos x}{\sin x}+\frac{1}{\cos x}+\frac{1}{\sin x}}\] Show more…
Show all steps
Your feedback will help us improve your experience
Sandip Ranjan and 79 other Calculus 2 / BC educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
$\int[1 /(\tan x+\cot x+\sec x+\operatorname{cosec} x)] d x=\ldots$ (a) $(1 / 2)(\cos x-\sin x)+(x / 2)$ (b) $(1 / 2)(\sin x-\cos x)-(x / 2)$ (c) $(1 / 2)(\sin x+\cos x)+(x / 2)$ (d) $(1 / 2)(\sin x-\cos x)-(x / 2)$
$\int \frac{\sin x d x}{1+\sin x}$ is (a) $x-\tan x-\sec x+C$ (b) $x+\tan x-\sec x+C$ (c) $x+\tan x+\sec x+C$ (d) $x-\tan x+\sec x+C$
Choose the correct answer. $\int \frac{\sin ^{2} x-\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x$ is equal to (A) $\tan x+\cot x+C$ (B) $\tan x+\operatorname{cosec} x+C$ (C) $-\tan x+\cot x+C$ (D) $\tan x+\sec x+C$
Integrals
Methods of Integration
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD