((bobhans)/(β⊝β)) I = ∫ ((sin 2x)/(a cos^2 x+b sin^2 x+c))


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Question Number 108444 by bobhans last updated on 17/Aug/20

$\frac{b o b h a n s}{β ⊝ β}$ $I = \int \frac{\sin 2 x}{a co s^{2} x + b \sin^{2} x + c}$
	Answered by john santu last updated on 17/Aug/20
	$((⊸JS⊸)/★) recall → { ((cos^2 x=(1/2)+(1/2)cos 2x)),((sin^2 x=(1/2)−(1/2)cos 2x)) :} next the integral becomes I= ∫((sin 2x dx)/((1/2)(a+b)+(1/2)(a−b)cos 2x+c)) I=∫ ((2sin 2x dx)/((a+b+2c)+(a−b)cos 2x)) set cos 2x = j →−2sin 2x dx=dj I= ∫ ((−dj)/((a+b+2c)+(a−b)j)) I = (1/b)∫ ((d(a+b+2c+(a−b)j))/((a+b+2c)+(a−b)j)) I= (1/b)ln ∣a+b+2c+(a−b)j∣ + K I=(1/b)ln ∣a+b+2c+(a−b)cos 2x∣ + K$
	$\frac{⊸ J S ⊸}{★}$ $r e c a l l \to {\begin{cases} \cos^{2} x = \frac{1}{2} + \frac{1}{2} \cos 2 x \\ \sin^{2} x = \frac{1}{2} - \frac{1}{2} \cos 2 x \end{cases}$ $n e x t t h e i n t e g r a l b e c o m e s$ $I = \int \frac{\sin 2 x d x}{\frac{1}{2} (a + b) + \frac{1}{2} (a - b) \cos 2 x + c}$ $I = \int \frac{2 \sin 2 x d x}{(a + b + 2 c) + (a - b) \cos 2 x}$ $s e t \cos 2 x = j \to - 2 \sin 2 x d x = d j$ $I = \int \frac{- d j}{(a + b + 2 c) + (a - b) j}$ $I = \frac{1}{b} \int \frac{d (a + b + 2 c + (a - b) j)}{(a + b + 2 c) + (a - b) j}$ $I = \frac{1}{b} \ln ∣ a + b + 2 c + (a - b) j ∣ + K$ $I = \frac{1}{b} \ln ∣ a + b + 2 c + (a - b) \cos 2 x ∣ + K$
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