∫_0 ^( α) ((tan θ)/(√(a^2 cos^2 θ−b^2 sin^2 θ))) dθ = ?


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Question Number 35949 by ajfour last updated on 26/May/18

$\int_{0}^{α} \frac{\tan θ}{\sqrt{a^{2} \cos^{2} θ - b^{2} \sin^{2} θ}} d θ = ?$
Commented by abdo mathsup 649 cc last updated on 26/May/18

$l e t p u t I_{α} = \int_{0}^{α} \frac{t a n θ}{\sqrt{a^{2} {c o s}^{2} θ - b^{2} {s i n}^{2} θ}} d θ$ $w e h s v e t h e f o r m u l a 1 + {t a n}^{2} θ = \frac{1}{{c o s}^{2} θ} \Rightarrow$ ${c o s}^{2} θ = \frac{1}{1 + {t a n}^{2} θ} a n d {s i n}^{2} θ = 1 - \frac{1}{1 + {t a n}^{2} θ} \Rightarrow$ $a^{2} {c o s}^{2} θ - b^{2} {s i n}^{2} θ$ $= \frac{a^{2}}{1 + {t a n}^{2} θ} - b^{2} \frac{{t a n}^{2} θ}{1 + {t a n}^{2} θ} = \frac{a^{2} - b^{2} {t a n}^{2} θ}{1 + {t a n}^{2} θ} \Rightarrow$ $I_{α} = \int_{0}^{α} \frac{t a n θ \sqrt{1 + {t a n}^{2} θ}}{\sqrt{a^{2} - b^{2} {t a n}^{2} θ}} d θ c h a n g e m e n t t a n θ = t$ $g i v e I_{α} = \int_{0}^{t a n α} \frac{t \sqrt{1 + t^{2}}}{\sqrt{a^{2} - b^{2} t^{2}}} \frac{d t}{1 + t^{2}}$ $= \int_{0}^{t a n (α)} \frac{t}{\sqrt{1 + t^{2}}} \frac{d t}{\sqrt{a^{2} - b^{2} t^{2}}} b y p a r t s$ $u^{^{'}} = \frac{t}{\sqrt{1 + t^{2}}} a n d v = \frac{1}{\sqrt{a^{2} - b^{2} t^{2}}}$ $I_{α} = {[\frac{\sqrt{1 + t^{2}}}{\sqrt{a^{2} - b^{2} t^{2}}}]}_{0}^{t a n (α)} - \int_{0}^{t a n (α)} \sqrt{1 + t^{2}} (- \frac{1}{2} - 2 b^{2} t) {(a^{2} - b^{2} t^{2})}^{- \frac{3}{2}} d t$ $= \frac{\sqrt{1 + {t a n}^{2} α}}{\sqrt{a^{2} - b^{2} {t a n}^{2} α}} - \frac{1}{∣ a ∣} - b^{2} \int_{0}^{t a n (α)} \sqrt{1 + t^{2}} {(a^{2} - b^{2} t^{2})}^{- \frac{3}{2}} d t$ $l e t J = \int_{0}^{t a n (α)} \sqrt{1 + t^{2}} {(a^{2} - b^{2} t^{2})}^{- \frac{3}{2}} d t$ $= c \int_{0}^{t a n (α)} \frac{\sqrt{1 + t^{2}}}{(a^{2} - b^{2} t^{2}) \sqrt{a^{2} - b^{2} t^{2}}} d t . . . b e c o n t i n u e d . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 26/May/18
	$∫_0 ^α ((tanθ.secθ)/(√(a^2 −b^2 tan^2 θ)))dθ ∫_0 ^α ((tanθsecθ)/(√(a^2 −b^2 sec^2 θ+b^2 )))dθ (1/b)∫((d(secθ))/(√(((a^2 +b^2 )/b^2 )−sec^2 θ))) now use formula ∫(dx/(√(A^2 −X^2 ))) (1/b)×∣sin^(−1) (((secθ)/(√((a^2 +b^2 )/b^2 ))))∣_0 ^α (1/b)×{sin^(−1) (((secα)/(√((a^2 +b^2 )/b^2 ))))−sin^(−1) ((1/((√(a^2 +b^2 ))/b^2 )))}$
	$\int_{0}^{α} \frac{t a n θ . s e c θ}{\sqrt{a^{2} - b^{2} {t a n}^{2} θ}} d θ$ $\int_{0}^{α} \frac{t a n θ s e c θ}{\sqrt{a^{2} - b^{2} {s e c}^{2} θ + b^{2}}} d θ$ $\frac{1}{b} \int \frac{d (s e c θ)}{\sqrt{\frac{a^{2} + b^{2}}{b^{2}} - {s e c}^{2} θ}}$ $n o w u s e f o r m u l a \int \frac{d x}{\sqrt{A^{2} - X^{2}}}$ $\frac{1}{b} \times ∣ {s i n}^{- 1} (\frac{s e c θ}{\sqrt{\frac{a^{2} + b^{2}}{b^{2}}}}) ∣_{0}^{α}$ $\frac{1}{b} \times {{s i n}^{- 1} (\frac{s e c α}{\sqrt{\frac{a^{2} + b^{2}}{b^{2}}}}) - {s i n}^{- 1} (\frac{1}{\frac{\sqrt{a^{2} + b^{2}}}{b^{2}}})}$
	Commented by ajfour last updated on 26/May/18

	$T h a n k s .$
	Commented by tanmay.chaudhury50@gmail.com last updated on 26/May/18

	$i t s o k . . .$
	Commented by ajfour last updated on 26/May/18

	$i a p p l i e d y o u r t e c h n i q u e a g a i n$ $a n d c o r r e c t e d m y a n s w e r t o$ $Q . 35940 . T h a n k s a g a i n .$
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