pi-2-pi-2-x-2-ln-pi-x-pi-x-cos-x-dx-

Question Number 121174 by benjo_mathlover last updated on 05/Nov/20

\underset{- π / 2}{\int^{π / 2}} (x^{2} + \ln (\frac{π + x}{π - x})) \cos x dx ?

Answered by TANMAY PANACEA last updated on 05/Nov/20

I = \int_{\frac{- π}{2}}^{\frac{π}{2}} {{(0 - x)}^{2} + l n (\frac{π + (0 - x)}{π - (0 - x)})} c o s (0 - x) d x

I = \int_{\frac{- π}{2}}^{\frac{π}{2}} {x^{2} + l n (\frac{π - x}{π + x})} c o s x d x

2 I = \int_{\frac{- π}{2}}^{\frac{π}{2}} 2 x^{2} \times c o s x + c o s x {l n (\frac{π + x}{π - x}) + l n (\frac{π - x}{π + x})} d x

2 I = \int_{\frac{- π}{2}}^{\frac{π}{2}} 2 x^{2} \times c o s x + c o s x \times l n 1 d x

I = \int_{\frac{- π}{2}}^{\frac{π}{2}} x^{2} c o s x d x

J = \int x^{2} e^{i x} d x

= x^{2} \frac{e^{i x}}{i} - \int 2 x \times \frac{e^{i x}}{i} d x

= \frac{x^{2} e^{i x} \times i}{- 1} + 2 i \int {x e}^{i x} d x

= - {i x}^{2} (c o s x + i s i n x) + 2 i [x \times \frac{e^{i x}}{i} - \int \frac{e^{i x}}{i} d x]

= - {i x}^{2} (c o s x + i s i n x) + 2 x (c o s x + i s i n x) - 2 \times \frac{e^{i x}}{i}

= (- {i x}^{2} c o s x) + x^{2} s i n x + 2 x c o s x + i \times 2 x c o s x + 2 i (c o s x + i s i n x)

= r e a l p a r t + i m p a r t

r e a l p s r t = (x^{2} s i n x + 2 x c o s x - 2 s i n x)

I =∣ x^{2} s i n x + 2 x c o s x - 2 s i n x ∣_{\frac{- π}{2}}^{\frac{π}{2}}

= \frac{π^{2}}{4} \times 1 - \frac{π^{2}}{4} \times (- 1) + 2 \times \frac{π}{2} \times 0 - 2 \times \frac{- π}{2} \times 0 - 2 (1 + 1)

= \frac{π^{2}}{2} - 4

Commented by TANMAY PANACEA last updated on 05/Nov/20

y e s y e s i c o u l d n o t s e e \dots

Commented by benjo_mathlover last updated on 06/Nov/20

yess \dots

Answered by Ar Brandon last updated on 05/Nov/20

I = \int_{- \frac{π}{2}}^{\frac{π}{2}} {x^{2} + \ln (\frac{π + x}{π - x})} cosxdx

I = \int_{- \frac{π}{2}}^{\frac{π}{2}} {x^{2} + \ln (\frac{π - x}{π + x})} cosxdx

I + I = \int_{- \frac{π}{2}}^{\frac{π}{2}} {{2 x}^{2} + \ln (\frac{π + x}{π - x}) + \ln (\frac{π - x}{π + x})} cosxdx

\Rightarrow 2 I = \int_{- \frac{π}{2}}^{\frac{π}{2}} {2 x}^{2} cosxdx = 2 {[x^{2} \int cosxdx - \int {\frac{{dx}^{2}}{dx} \cdot \int cosxdx}]}_{- \frac{π}{2}}^{\frac{π}{2}}

\Rightarrow I = {[x^{2} sinx - 2 \int xsinxdx]}_{- \frac{π}{2}}^{\frac{π}{2}}

= {[x^{2} sinx + 2 {xcosx - \int cosxdx}]}_{- \frac{π}{2}}^{\frac{π}{2}}

= {[x^{2} sinx + 2 xcosx - 2 sinx]}_{- \frac{π}{2}}^{\frac{π}{2}} = {\frac{π^{2}}{4} - 2} + {\frac{π^{2}}{4} - 2}

= \frac{π^{2}}{2} - 4

Commented by Ar Brandon last updated on 05/Nov/20

Commented by Ar Brandon last updated on 05/Nov/20

\int_{a}^{b} f (x) dx = \int_{a}^{b} f (a + b - x) dx

Answered by Dwaipayan Shikari last updated on 05/Nov/20

I n t e g r a l i s s y m m e t r i c

\int_{- \frac{π}{2}}^{\frac{π}{2}} l o g (\frac{π + x}{π - x}) c o s x = 0

\int_{- \frac{π}{2}}^{\frac{π}{2}} x^{2} c o s x d x

= {[x^{2} s i n x]}_{- \frac{π}{2}}^{\frac{π}{2}} - \int_{- \frac{π}{2}}^{\frac{π}{2}} 2 x s i n x d x

= \frac{π^{2}}{2} + 2 {[x c o s x]}_{- \frac{π}{2}}^{\frac{π}{2}} - 2 \int_{- \frac{π}{2}}^{\frac{π}{2}} c o s x

= \frac{π^{2}}{2} - 2.2 = \frac{π^{2}}{2} - 4

Answered by Bird last updated on 05/Nov/20

\int_{- \frac{π}{2}}^{\frac{π}{2}} (x^{2} + l n (\frac{π + x}{π - x})) c o s x d x

= \int_{- \frac{π}{2}}^{\frac{π}{2}} x^{2} c o s x d x + \int_{- \frac{π}{2}}^{\frac{π}{2}} l n (\frac{π + x}{π - x}) c o s x d x (\to= 0 o d d f u n c t i o n u n d e r i n t e g r a l)

= \int_{- \frac{π}{2}}^{\frac{π}{2}} x^{2} c o s x d x = {[x^{2} s i n x]}_{- \frac{π}{2}}^{\frac{π}{2}}

- \int_{- \frac{π}{2}}^{\frac{π}{2}} 2 x s i n x d x

= \frac{π^{2}}{2} - 2 \int_{- \frac{π}{2}}^{\frac{π}{2}} x s i n x d x w e h a v e

\int_{- \frac{π}{2}}^{\frac{π}{2}} x s i n x d x = 2 \int_{0}^{\frac{π}{2}} x s i n x d x

= 2 {{[- x c o s x]}_{0}^{\frac{π}{2}} + \int_{0}^{\frac{π}{2}} c o s x d x}

= 2 {[s i n x]}_{0}^{\frac{π}{2}} = 2 \Rightarrow

I = \frac{π^{2}}{2} - 4

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