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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Integral Calculus

Integrate: $Un =\int \limits_0^{\large\frac{\pi}{4}} \tan ^n x dx$, find $U_n+U_{n-2}$

\[\begin {array} {1 1} (a)\;\frac{1}{n-1} \\ (b)\;\frac{1}{n+2} \\ (c)\;\frac{1}{n-3} \\ (d)\;\frac{1}{n+4} \end {array}\]
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1 Answer

$U_n = \int _0^{\large\frac{\pi}{4}} \tan ^{n-2}x \tan ^2 x $$ dx$
$\qquad= \int \limits_0^{\large\frac{\pi}{4}} \tan ^{n-2} \sec ^2 x dx -\int \limits_0^{\large\frac{\pi}{4} } \tan ^{n-2}\;dx$
$\qquad= \int \limits_0^{\large\frac{\pi}{4}} \tan ^{n-2} x. \sec ^2 x dx -\int \limits_0^{\large\frac{\pi}{4} } \tan ^{n-2}\;dx$
$U_n +U_{n-2} =\int \limits_0^{\large\frac{\pi}{4}} \tan ^{n-2} x. \sec^2 x dx $
$\tan x =t=7\qquad=> x = 0,t=0$
$\sec^2 x dx =dt => \qquad x= \large\frac{\pi}{4},t=1$
=> $ \int \limits_0^1 t^{n-2} .dt$
$\qquad= \bigg[ \large\frac{t^{n-1}}{n-1}\bigg]_0^1$
$\qquad= \large\frac{1}{n-1}$
Hence a is the correct answer.
answered Dec 16, 2013 by meena.p
 
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