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Question Number 62981 by hovea cw last updated on 27/Jun/19

Commented by mathmax by abdo last updated on 27/Jun/19
$2) let A =∫_0 ^(2π) ((x∣sin(2x)∣)/(3+sin^2 x)) dx changement x =π +t give A =∫_(−π) ^π (((π+t)∣sin(2t)∣)/(3+sin^2 t)) dt = ∫_(−π) ^0 (((π+t)∣sin(2t)∣)/(3+sin^2 t)) dt +∫_0 ^π (((π+t)∣sin(2t)∣)/(3+sin^2 t)) dt ∫_(−π) ^0 (((π+t)∣sin(2t)∣)/(3+sin^2 t)) dt =_(t=−u) −∫_0 ^π (((π−u)∣sin(2u)∣)/(3+sin^2 u))(−du) ⇒ A = ∫_0 ^π (((π+t+π−t)∣sin(2t)∣)/(3+sin^2 t)) dt =2π ∫_0 ^π ((∣sin(2t)∣)/(3+sin^2 t))dt =_(2t =x) 2π ∫_0 ^(2π) ((∣sin(x)∣)/(3+sin^2 ((x/2)))) (dx/2) =π { ∫_0 ^π ((sinx)/(3+sin^2 ((x/2))))dx +∫_π ^(2π) ((∣sinx∣)/(3+sin^2 ((x/2))))dx} ∫_π ^(2π) ((∣sinx∣)/(3+sin^2 ((x/2))))dx =_(x =π +α) ∫_0 ^π ((sin(α))/(3+sin^2 ((π/2)+(α/2)))) dα =∫_0 ^π ((sin(α))/(3+cos^2 ((α/2))))dα ⇒ A =π { ∫_0 ^π ((sin(x))/(3+sin^2 ((x/2))))dx +∫_0 ^π ((sinx)/(3+cos^2 ((x/2))))dx} =π ∫_0 ^π sinx(7/((3+sin^2 ((x/2)))(3+cos^2 ((x/2))))dx =7π ∫_0 ^π ((sinx)/((3+((1−cosx)/2))(3+((1+cosx)/2))))dx =28π∫_0 ^π sinx{(1/(7−cosx)) +(1/(7+cosx))}dx =28π { ∫_0 ^π ((sinx)/(7−cosx))dx +∫_0 ^π ((sinx)/(7 +cosx))dx} =28π{ [ln∣7−cosx∣]_0 ^π −[ln∣7+cosx∣]_0 ^π } =28π{ ln(8)−ln(6)−ln(6) +ln(8)} =28π{2ln(8)−2ln(6)} =56π{ln(8)−ln(6)} =56π{3ln(2)−ln(2)−ln(3)} ⇒ A =56π{2ln(2)−ln(3)} .$
$2) l e t A = \int_{0}^{2 π} \frac{x ∣ s i n (2 x) ∣}{3 + {s i n}^{2} x} d x c h a n g e m e n t x = π + t g i v e$ $A = \int_{- π}^{π} \frac{(π + t) ∣ s i n (2 t) ∣}{3 + {s i n}^{2} t} d t = \int_{- π}^{0} \frac{(π + t) ∣ s i n (2 t) ∣}{3 + {s i n}^{2} t} d t + \int_{0}^{π} \frac{(π + t) ∣ s i n (2 t) ∣}{3 + {s i n}^{2} t} d t$ $\int_{- π}^{0} \frac{(π + t) ∣ s i n (2 t) ∣}{3 + {s i n}^{2} t} d t =_{t = - u} - \int_{0}^{π} \frac{(π - u) ∣ s i n (2 u) ∣}{3 + {s i n}^{2} u} (- d u) \Rightarrow$ $A = \int_{0}^{π} \frac{(π + t + π - t) ∣ s i n (2 t) ∣}{3 + {s i n}^{2} t} d t = 2 π \int_{0}^{π} \frac{∣ s i n (2 t) ∣}{3 + {s i n}^{2} t} d t$ $=_{2 t = x} 2 π \int_{0}^{2 π} \frac{∣ s i n (x) ∣}{3 + {s i n}^{2} (\frac{x}{2})} \frac{d x}{2} = π {\int_{0}^{π} \frac{s i n x}{3 + {s i n}^{2} (\frac{x}{2})} d x + \int_{π}^{2 π} \frac{∣ s i n x ∣}{3 + {s i n}^{2} (\frac{x}{2})} d x}$ $\int_{π}^{2 π} \frac{∣ s i n x ∣}{3 + {s i n}^{2} (\frac{x}{2})} d x =_{x = π + α} \int_{0}^{π} \frac{s i n (α)}{3 + {s i n}^{2} (\frac{π}{2} + \frac{α}{2})} d α = \int_{0}^{π} \frac{s i n (α)}{3 + {c o s}^{2} (\frac{α}{2})} d α \Rightarrow$ $A = π {\int_{0}^{π} \frac{s i n (x)}{3 + {s i n}^{2} (\frac{x}{2})} d x + \int_{0}^{π} \frac{s i n x}{3 + {c o s}^{2} (\frac{x}{2})} d x}$ $= π \int_{0}^{π} s i n x \frac{7}{(3 + {s i n}^{2} (\frac{x}{2})) (3 + {c o s}^{2} (\frac{x}{2})} d x = 7 π \int_{0}^{π} \frac{s i n x}{(3 + \frac{1 - c o s x}{2}) (3 + \frac{1 + c o s x}{2})} d x$ $= 28 π \int_{0}^{π} s i n x {\frac{1}{7 - c o s x} + \frac{1}{7 + c o s x}} d x$ $= 28 π {\int_{0}^{π} \frac{s i n x}{7 - c o s x} d x + \int_{0}^{π} \frac{s i n x}{7 + c o s x} d x}$ $= 28 π {{[l n ∣ 7 - c o s x ∣]}_{0}^{π} - {[l n ∣ 7 + c o s x ∣]}_{0}^{π}}$ $= 28 π {l n (8) - l n (6) - l n (6) + l n (8)} = 28 π {2 l n (8) - 2 l n (6)}$ $= 56 π {l n (8) - l n (6)} = 56 π {3 l n (2) - l n (2) - l n (3)} \Rightarrow$ $A = 56 π {2 l n (2) - l n (3)} .$
	Answered by Hope last updated on 27/Jun/19

	$1) \int \sqrt{t a n x + 1} d x$ $t^{2} = 1 + t a n x \to 2 t d t = {s e c}^{2} x d x$ $\frac{2 t d t}{1 + {(t^{2} - 1)}^{2}} = d x$ $\int \frac{t \times 2 t d t}{t^{4} - 2 t^{2} + 2}$ $\int \frac{\frac{2}{t^{2}}}{t^{2} + \frac{2}{t^{2}} - 2} d t$ $\frac{1}{\sqrt{2}} \int \frac{(1 + \frac{\sqrt{2}}{t^{2}}) - (1 - \frac{\sqrt{2}}{t^{2}})}{(t^{2} + \frac{2}{t^{2}}) - 2} d t$ $\frac{1}{\sqrt{2}} [\int \frac{d (t - \frac{\sqrt{2}}{t})}{{(t - \frac{\sqrt{2}}{t})}^{2} + 2 \sqrt{2} - 2} - \int \frac{d (t + \frac{\sqrt{2}}{t})}{{(t + \frac{\sqrt{2}}{t})}^{2} - 2 \sqrt{2} - 2}$ $= \frac{1}{\sqrt{2}} [\frac{1}{\sqrt{2 \sqrt{2} - 2}} \times {t a n}^{- 1} (\frac{t - \frac{\sqrt{2}}{t})}{\sqrt{2 \sqrt{2} - 2}}) - \frac{1}{2 (\sqrt{2 \sqrt{2} + 2}} l n (\frac{t + \frac{\sqrt{2}}{t}) - \sqrt{2 \sqrt{2} + 2}}{(t + \frac{\sqrt{2}}{t}) + \sqrt{2 \sqrt{2} + 2}})$ $p u t t = \sqrt{1 + t a n x}$
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