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Home  >>  CBSE XII  >>  Math  >>  Integrals
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Choose the correct answer in the value of $\int\limits_\frac{\Large -\pi}{\Large 2}^\frac{\Large \pi}{\Large 2}(x^3+x\cos x+\tan^5x+1)dx$ is

$(a)\;0\qquad(b)\;2\qquad(c)\;\pi\qquad(d)\;1$

Can you answer this question?
 
 

1 Answer

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Toolbox:
  • (i)$\int \limits_a^bf(x)dx=F(b)-F(a)$
  • (ii)If $f(-x)=-f(x)$ it is an odd function then $ \int \limits_0^a f(x)dx=0$
  • (iii)If $f(-x)=-f(x)$ it is an odd function then $ \int \limits_0^a f(x)dx=2 \int \limits_0^a f(x)dx$
Given $I=\int\limits_\frac{\Large -\pi}{\Large 2}^\frac{\Large \pi}{\Large 2}(x^3+x\cos x+\tan^5x+1)dx$
 
On seperating the term
 
$\int\limits_\frac{\Large -\pi}{\Large 2}^\frac{\Large \pi}{\Large 2}(x^3dx+\int\limits_\frac{\Large -\pi}{\Large 2}^\frac{\Large \pi}{\Large 2}x\cos x+\int\limits_\frac{\Large -\pi}{\Large 2}^\frac{\Large \pi}{\Large 2}\tan^5x+\int\limits_\frac{\Large -\pi}{\Large 2}^\frac{\Large \pi}{\Large 2}1dx$
 
Consider $x^3;(-x)^3=-x^3$
 
Hence it is an odd function
 
Therefore $\int\limits_\frac{\Large -\pi}{\Large 2}^\frac{\Large \pi}{\Large 2} x^3dx=0$
 
Consider $ xcosx\;-x\cos(-x)=-x\cos x$
 
Hence it is an odd function
 
Therefore $\int\limits_\frac{\Large -\pi}{\Large 2}^\frac{\Large \pi}{\Large 2} x \cos x=0$
 
Consider $\int\limits_\frac{\Large -\pi}{\Large 2}^\frac{\Large \pi}{\Large 2} \tan ^5 x.\tan ^5(-x)=-\tan ^5x$
 
It is an odd function
 
Hence $\int\limits_\frac{\Large -\pi}{\Large 2}^\frac{\Large \pi}{\Large 2} \tan ^5x=0$
 
Therefore $ I=0+0+0+0+2\int \limits_0^{\frac{\pi}{2}}dx$
 
On Integrating
 
$I=[2x]_0^{\pi/2}$
 
On applying limits
 
$I.2.\frac{\pi}{2}=\pi$
 
Correct answer is C

 

answered Feb 14, 2013 by meena.p
 

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