0-pi-4-x-2-x-sin-x-cos-x-2-dx-

Question Number 118111 by bemath last updated on 15/Oct/20

\underset{0}{\int^{π / 4}} \frac{x^{2}}{{(x \sin x + \cos x)}^{2}} dx = ?

Answered by Lordose last updated on 15/Oct/20

Ω = \int_{0}^{\frac{π}{4}} \frac{x^{2}}{{(xsinx + cosx)}^{2}} dx

Ω = \int_{0}^{\frac{π}{4}} \frac{x^{2} \sin^{2} x + x^{2} \cos^{2} x}{{(xsinx + cosx)}^{2}} dx

Ω = \int_{0}^{\frac{π}{4}} \frac{x^{2} \sin^{2} x + xsinxcosx}{{(xsinx + cosx)}^{2}} dx + \int_{0}^{\frac{π}{4}} \frac{x^{2} \cos^{2} x - xsinxcosx}{{(xsinx + cosx)}^{2}} dx

Ω = \int_{0}^{\frac{π}{4}} \frac{xsinx (xsinx + cosx)}{{(xsinx + cosx)}^{2}} dx + \int_{0}^{\frac{π}{4}} \frac{- xcosx (- xcosx + sinx)}{{(xsinx + cosx)}^{2}} dx

Ω = Φ + Λ

resolving Λ by IBP

u = sinx - xcosx \Rightarrow du = xsinxdx

dv = \frac{- xcosx}{{(xsinx + cosx)}^{2}} dx \Rightarrow v = \frac{1}{xsinx + cosx}

Λ =∣ \frac{sinx - xcosx}{xsinx + cosx} ∣_{0}^{\frac{π}{4}} - \int_{0}^{\frac{π}{4}} \frac{xsinx}{xsinx + cosx} dx

Λ =∣ \frac{sinx - xcosx}{xsinx + cosx} ∣_{0}^{\frac{π}{4}} - Φ

Ω = Φ + ∣ \frac{sinx - xcosx}{xsinx + cosx} ∣_{0}^{\frac{π}{4}} - Φ

Ω =∣ \frac{sinx - xcosx}{xsinx + cosx} ∣_{0}^{\frac{π}{4}}

Ω = \frac{4 - π}{4 + π}

Commented by bemath last updated on 15/Oct/20

g a v e k u d o s

Answered by Ar Brandon last updated on 15/Oct/20

\frac{d}{dx} (xsinx + cosx) = xcosx + sinx - sinx = xcosx

I = \int_{0}^{\frac{π}{4}} \frac{x^{2}}{{(xsinx + cosx)}^{2}} dx = \int_{0}^{\frac{π}{4}} \frac{xcosx}{{(xsinx + cosx)}^{2}} \cdot \frac{x}{cosx} dx

= {\frac{x}{cosx} \int \frac{xcosx}{{(xsinx + cosx)}^{2}} dx}_{0}^{\frac{π}{4}} - \int_{0}^{\frac{π}{4}} {\frac{d}{dx} (\frac{x}{cosx}) \cdot \int \frac{xcosx}{{(xsinx + cosx)}^{2}} dx}

= - {[\frac{x}{cosx} \cdot \frac{1}{xsinx + cosx}]}_{0}^{\frac{π}{4}} + \int_{0}^{\frac{π}{4}} {(\frac{cosx + xsinx}{\cos^{2} x}) (\frac{1}{xsinx + cosx})} dx

= - {[\frac{x}{cosx} \cdot \frac{1}{xsinx + cosx}]}_{0}^{\frac{π}{4}} + \int_{0}^{\frac{π}{4}} \frac{dx}{\cos^{2} x}

= {[tanx - \frac{x}{cosx} \cdot \frac{1}{xsinx + cosx}]}_{0}^{\frac{π}{4}} = [1 - \frac{2 π}{4 \sqrt{2}} \cdot \frac{8}{π \sqrt{2} + 4 \sqrt{2}}]

= 1 - \frac{16 π}{8 π + 32} = \frac{8 π + 32 - 16 π}{8 π + 32} = \frac{32 - 8 π}{32 + 8 π} = \frac{4 - π}{4 + π}

Commented by bemath last updated on 15/Oct/20

s a n t u y y

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