The value of ∫-(π)/2(π)/2(x^2cos⁡x)/(1+e^x) dx is equal to

Correct answer: (A) (π^2)/4-2. Let I= ∫ − x/2 x/2 (x^2 cosx)/(1+e^x) dx Replacing x by -x I= ∫ −(π)/2 (π)/2 ( (x^2 cosx)/(1+e^(−x)) )dx ⇒2I= ∫ −(π)/2 (π)/2 ( 1/(1+e^x) +1/(1+e^(−x)) )x^2 cosx dx ⇒2I= ∫ −(π)/2 (π)/2 x^2 cosx dx , (even function) I = ∫0(π)/2x^2cos⁡x dx=x^2sin⁡x0(π)/2-2∫0(π)/2x.sin⁡x dx (Using integration by parts, taking x^2 =u,cos x dx = dv ) =(π^2)/4-2-x.cos⁡x0(π)/2+ ∫0(π)/21.cos⁡xdx=(π^2)/4-20+1=(π^2)/4-2

Q23 / 54 · JEE Advanced 2016 (Paper 2) · Mathematics

The value of $\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{x^{2} \cos x}{1 + e^{x}} d x$ is equal to

Correct answer
A
$\frac{π^{2}}{4} - 2$
B
$\frac{π^{2}}{4} + 2$
C
$π^{2} - e^{\frac{π}{2}}$
D
$π^{2} + e^{\frac{π}{2}}$

Solution

Let

I = \int_{- \frac{x}{2}}^{\frac{x}{2}} \frac{x^{2} \cos x}{1 + e^{x}} d x

Replacing

x

- x

I = \int_{- \frac{π}{2}}^{\frac{π}{2}} (\frac{x^{2} \cos x}{1 + e^{- x}}) d x

\Rightarrow 2 I = \int_{- \frac{π}{2}}^{\frac{π}{2}} (\frac{1}{1 + e^{x}} + \frac{1}{1 + e^{- x}}) x^{2} \cos x d x

\Rightarrow 2 I = \int_{- \frac{π}{2}}^{\frac{π}{2}} x^{2} \cos x d x

, (even function)

I

= \int_{0}^{\frac{π}{2}} x^{2} \cos x d x = {(x^{2} \sin x)}_{0}^{\frac{π}{2}} - 2 \int_{0}^{\frac{π}{2}} x . \sin x d x

(Using integration by parts, taking

x^{2} = u, \cos x d x = d v

)

= \frac{π^{2}}{4} - 2 [- {(x . \cos x)}_{0}^{\frac{π}{2}} + \int_{0}^{\frac{π}{2}} 1 . \cos x d x] = \frac{π^{2}}{4} - 2 [0 + 1] = (\frac{π^{2}}{4} - 2)

Hence, the correct answer is option: 'A'.

The value of ∫-π2π2x2cos⁡x1+ex dx is equal to

Solution

Relevant Questions

The value of $\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{x^{2} \cos x}{1 + e^{x}} d x$ is equal to