JEE-Adv Paper-2: Can you solve it?

Calculus Level 3

$\large \int_{-\pi/2}^{\pi /2} \dfrac{ x^2 \cos x}{1 + e^x} \, dx = \, ?$

\frac{\pi^2}{4}-2

\frac{\pi^2}{4}+2

\pi^2+e^{{\pi^2}/ { 2}}

\pi^2-e^{ {\pi} / {2}}

1 solution

Aditya Narayan Sharma
May 25, 2016

$\displaystyle f(x) =\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2\cos x}{1+e^x}dx$

$\displaystyle 2f(x) = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2\cos x\cancel{(1+e^x)}}{\cancel{1+e^x}}dx$ Using $\displaystyle[\int_{a}^{b}f(x)dx=\int_{a}^{b}f(a+b-x)dx]$

Since $\displaystyle g(x)=x^2\cos x = g(-x)$ , we have $\displaystyle 2f(x)=2\int_{0}^{\frac{\pi}{2}} x^2\cos x \ dx$

Using IBP twice with first function $=x^2$ ,second function $=\cos x$ we get,

$\displaystyle f(x)=[x^2\sin x+2x\cos x-2\sin x]_{0}^{\frac{\pi}{2}}=\boxed{\frac{\pi^2}{4}-2}$

Could you elaborate a bit on step 2? I think I've seen this technique used before but I've never managed to figure out how it works.

Thicky Bushi - 5 years ago

$\int_a^bf(x)=\int_a^bf(a+b-x)$

Using the above property, and adding it to the give equation, u get the 2nd step

Sparsh Sarode - 5 years ago

I think I'm having problems understanding that property. Is there a name for it? Do you know where I can read more about it?

Thicky Bushi - 5 years ago

@Thicky Bushi – Just search for "Properties of definite integrals" . You would get lot of websites. This is one of the elementary properties

Aditya Narayan Sharma - 5 years ago

@Thicky Bushi – The property is also known as KING'S Rule in integration.

Anupam Khandelwal - 5 years ago

@Thicky Bushi – It involves properties of definite integrals

Rishi K - 5 years ago

Yes as sparsh stated, $\displaystyle f(x)=\int_{0}^{\frac{\pi}{2}} \frac{x^2cosx}{1+e^{-x}}dx$

So adding two two different representations of $f(x)$ brings the second step as $1+e^x$ gets cancelled

Aditya Narayan Sharma - 5 years ago

JEE-Adv Paper-2: Can you solve it?

1 solution

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