What is $\int_{\frac{π}{8}}^{\frac{11 π}{12}} x^{2} - {cos}^{2} x + x sin x d x$ $int_(pi/8)^((11pi)/12) x^2-cos^2x+xsinxdx$ ?

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Answer 1 · 2016-06-29T09:50:32

$\approx 10$ $approx 10$

$int_(pi/8)^((11pi)/12)dx \ (x^2-cos^2x+xsinx)$

using identity $cos 2 z = 2 cos^2 z - 1 \implies cos ^2 z = (cos 2z + 1)/(2)$

$= int_(pi/8)^((11pi)/12)dx \ (x^2-(cos 2x + 1)/(2)+ color{red}{xsinx})$

first 2 terms are trivial, we do the term in red using IBP, ie

$int dx \ (u v') = uv - int dx \ (u' v)$

Here

$u = x, u' = 1$
$v' = sin x, v = -cos x$

so we have
$- x cos x - int dx \ (-cos x)$
$color{red}{= - x cos x + intdx \ (cos x)}$

so the integral becomes

$= [-x cos x] _(pi/8)^((11pi)/12) + int_(pi/8)^((11pi)/12)dx \ (x^2-(cos 2x + 1)/(2)+ cos x)$

$= [-x cos x + x^3/3-(sin 2x)/4 - x/2+ sin x ] _(pi/8)^((11pi)/12)$

crunch it in calculator ....

$approx 10$

What is int_(pi/8)^((11pi)/12) x^2-cos^2x+xsinxdx∫11π12π8x2−cos2x+xsinxdxint_(pi/8)^((11pi)/12) x^2-cos^2x+xsinxdx?