Question
Let $I_{n}=\int \tan ^{n} x d x,(n>1) \cdot I_{4}+I_{6}=a \tan ^{5} x+b x^{5}+C$where $C$ is a constant of integration, then the ordered [2017 JEE Main](a) $\left(-\frac{1}{5}, 0\right)$
Step 1
We are given that $I_4 + I_6 = a \tan^5 x + bx^5 + C$. Show more…
Show all steps
Your feedback will help us improve your experience
Uma Kumari and 94 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
Evaluate $\int_{0}^{2}\left[\tan ^{-1}(4-x)-\tan ^{-1} x\right] d x$ by rewriting it as a double integral and switching the order of integration.
Multiple Integrals
Double Integrals
If $I n=\int \tan n x d x$, then $I_{0}+I_{1}+2\left(I_{2}+\ldots+I_{n}\right)+I_{0}+I_{10}$ is equal to (A) $\left(\frac{\tan x}{1}+\frac{\tan ^{2} x}{2}+\ldots+\frac{\tan ^{9} x}{9}\right)$ (B) $-\left(\frac{\tan x}{1}+\frac{\tan ^{2} x}{2}+\ldots+\frac{\tan ^{9} x}{9}\right)$ (C) $\left(\frac{\cot x}{1}+\frac{\cot ^{2} x}{2}+\ldots+\frac{\cot ^{9} x}{9}\right)$ (D) $-\left(\frac{\cot x}{1}+\frac{\cot ^{2} x}{2}+\ldots+\frac{\cot ^{9} x}{9}\right)$
If $\int \tan ^{5} x d x=A \tan ^{4} x+B \tan ^{2} x+g(x)+C$, then (a) $A=\frac{1}{4}, B=-\frac{1}{2}$ (b) $g(x)=\ln |\sec x|$ (c) $g(x)=\ln |\cos x|$ (d) $A=-\frac{1}{4}, B=\frac{1}{3}$
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD