logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Integral Calculus
0 votes

$I_n =\int \limits_0^{z/4} \tan ^n x \;dx$ find $I_{n-1}+I_{n+1}$

\[\begin {array} {1 1} (a)\;\frac{2}{n} \\ (b)\;\frac{1}{n} \\ (c)\;\frac{\pi}{n} \\ (d)\;\frac{\pi}{4n} \end {array}\]
Can you answer this question?
 
 

1 Answer

0 votes
$I_{n-1} =\int \limits_0^{\pi/4} \tan ^{n-1} x \;dx$
$I_{n+1} =\int \limits_0^{\pi/4} \tan ^{n+1} x \;dx$
$I_{n-1} =\int \limits_0^{\pi/4} \large\frac{\tan ^n x}{\tan ^ x}$$ \;dx$
$I_{n-1}+I_{n+1} =\int \limits_0^{\pi/4} \tan ^n \bigg[\large\frac{1}{\tan x}$$+ \tan n\bigg]$$dx$
$\qquad=\int \limits_0^{\pi/4} \large\frac{\tan ^n n}{\tan n}$$( \sec^2 x) dx$
$\tan n =t$
$\sec^2 x dx=dt$
$\qquad\qquad= \int \limits_0^1 t^{n-1} dt$
$\qquad\qquad= \large\frac{t^n}{n} \bigg]_0^1$
$\qquad\qquad =\large\frac{1}{n}$
Hence b is the correct answer.
answered Dec 17, 2013 by meena.p
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...