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Question Number 202224 by sonukgindia last updated on 23/Dec/23

	Answered by witcher3 last updated on 23/Dec/23

	$x \to \frac{1}{a + {bcos}^{2} (x)} = f (x), is p peridic$ $\Rightarrow \int_{k π}^{(k + 1) π} f (x) dx = \int_{0}^{π} f (x) dx; by k π + y = x$ $\Rightarrow \int_{0}^{n π} f (x) dx = \underset{k = 0}{\sum^{n - 1}} \int_{(k π)}^{(k + 1) π} f (x) dx = \underset{k = 0}{\sum^{n - 1}} \int_{0}^{π} f (x) dx$ $= n \int_{0}^{π} \frac{dx}{a + {bcos}^{2} (x)} = n (\int_{0}^{\frac{π}{2}} \frac{dx}{a + {bcos}^{2} (x)} + \int_{\frac{π}{2}}^{π} \frac{dx}{a + {bcos}^{2} (x)}_{π - x = h})$ $= 2 n \int_{0}^{\frac{π}{2}} \frac{dx}{a + {bcos}^{2} (x)} = \frac{2 n}{b} \int_{0}^{\frac{π}{2}} \frac{1}{\cos^{2} (x)} (\frac{1}{1 + \frac{a}{b} (1 + \tan^{2} (x))})$ $= \frac{2 n}{b} \int_{0}^{\frac{π}{2}} \frac{d (\tan (x))}{(\frac{b + a}{b}) (1 + {(\tan (x) . \sqrt{\frac{a}{b + a}})}^{2}}$ $You can't use 'macro parameter character #' in math mode$ $= 2 n \sqrt{\frac{1}{a (b + a)} .} \tan^{- 1} {(\sqrt{\frac{a}{b + a}} . \tan (x)]}_{0}^{\frac{π}{2}}$ $= \frac{π n}{\sqrt{a (a + b)}}$
	Answered by Calculusboy last updated on 23/Dec/23
	$Solution: multiply both numerator and denominator by (1/(cos^2 x)) I=∫_0 ^(n𝛑) ((1/(cos^2 x))/((a/(cos^2 x))+((bsin^2 x)/(cos^2 x))))dx NB: [ ((sinx)/(cosx))=tanx and (1/(cosx))=secx] I=2n∫_0 ^(𝛑/2) ((sec^2 x)/(asec^2 x+btan^2 x))dx NB: sec^2 x=1+tan^2 x I=2n∫_0 ^(𝛑/2) ((sec^2 x)/(a(1+tan^2 x)+btan^2 x))dx I=2n∫_0 ^(𝛑/2) ((sec^2 x)/(a+atan^2 x+btan^2 x))dx ⇔ I=2n∫_0 ^(𝛑/2) ((sec^2 x)/(a+tan^2 x(a+b)))dx let u=tanx du=sec^2 xdx [when x=(𝛑/2) u=∞ and when x=0 u=0] I=2n∫_0 ^∞ ((sec^2 x)/(a+u^2 (a+b)))∙(du/(sec^2 x)) I=2n∫_0 ^∞ (du/(u^2 (a+b)+a)) ⇔ I=((2n)/((a+b)))∫_0 ^∞ (du/(u^2 +((√(a/((a+b)))))^2 )) I=((2n)/((a+b)))×(1/( (√(a/((a+b)))))) ∣tan^(−1) [(u/( (√(a/((a+b))))))]∣_0 ^∞ I={((2n)/((a+b)))×((√(a+b))/( (√a)))}∣tan^(−1) [((u(√((a+b))))/( (√a)))]∣_0 ^∞ I=((2n(√((a+b))))/( (√a)(a+b)))×(𝛑/2)=((n𝛑(√((a+b))))/( (√a)(a+b)))$
	$S o l u t i o n : m u l t i p l y b o t h n u m e r a t o r a n d d e n o m i n a t o r b y \frac{1}{{c o s}^{2} x}$ $I = \int_{0}^{n π} \frac{\frac{1}{{c o s}^{2} x}}{\frac{a}{{c o s}^{2} x} + \frac{{b s i n}^{2} x}{{c o s}^{2} x}} d x N B : [\frac{s i n x}{c o s x} = t a n x a n d \frac{1}{c o s x} = s e c x]$ $I = 2 n \int_{0}^{\frac{π}{2}} \frac{{s e c}^{2} x}{{a s e c}^{2} x + {b t a n}^{2} x} d x N B : {s e c}^{2} x = 1 + {t a n}^{2} x$ $I = 2 n \int_{0}^{\frac{π}{2}} \frac{{s e c}^{2} x}{a (1 + {t a n}^{2} x) + {b t a n}^{2} x} d x$ $I = 2 n \int_{0}^{\frac{π}{2}} \frac{{s e c}^{2} x}{a + {a t a n}^{2} x + {b t a n}^{2} x} d x \Leftrightarrow I = 2 n \int_{0}^{\frac{π}{2}} \frac{{s e c}^{2} x}{a + {t a n}^{2} x (a + b)} d x$ $l e t u = t a n x d u = {s e c}^{2} x d x [w h e n x = \frac{π}{2} u = \infty a n d w h e n x = 0 u = 0]$ $I = 2 n \int_{0}^{\infty} \frac{{s e c}^{2} x}{a + u^{2} (a + b)} \cdot \frac{d u}{{s e c}^{2} x}$ $I = 2 n \int_{0}^{\infty} \frac{d u}{u^{2} (a + b) + a} \Leftrightarrow I = \frac{2 n}{(a + b)} \int_{0}^{\infty} \frac{d u}{u^{2} + {(\sqrt{\frac{a}{(a + b)}})}^{2}}$ $I = \frac{2 n}{(a + b)} \times \frac{1}{\sqrt{\frac{a}{(a + b)}}} ∣ {t a n}^{- 1} [\frac{u}{\sqrt{\frac{a}{(a + b)}}}] ∣_{0}^{\infty}$ $I = {\frac{2 n}{(a + b)} \times \frac{\sqrt{a + b}}{\sqrt{a}}} ∣ {t a n}^{- 1} [\frac{u \sqrt{(a + b)}}{\sqrt{a}}] ∣_{0}^{\infty}$ $I = \frac{2 n \sqrt{(a + b)}}{\sqrt{a} (a + b)} \times \frac{π}{2} = \frac{n π \sqrt{(a + b)}}{\sqrt{a} (a + b)}$
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