∫ (dx/(√(Acos x+B)))


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Question Number 104459 by bemath last updated on 21/Jul/20

$\int \frac{d x}{\sqrt{A \cos x + B}}$
Commented by Dwaipayan Shikari last updated on 21/Jul/20
$∫(dx/(√(Acosx+B)))=∫((A(−sinx))/(−Asinx)).(1/(√(Acosx+B)))dx {Acosx+B=t^2 ∫((2tdt)/(t(−Asinx)))=−(2/A)∫(1/(sinx))dt {cosx=(t^2 −B).(1/A) −(2/A)∫ (1/(√(A^2 −(t^2 −B)^2 )))dt {sinx=(√(1−(((t^2 −B)/A))^2 )) −(2/A)∫((2tdt)/(2t(√(A^2 −u^2 )))) {t^2 −B=u −(1/A)∫(du/(√(u+B))).(1/(√(A^2 −u^2 )))....continue$
$\int \frac{d x}{\sqrt{A c o s x + B}} = \int \frac{A (- s i n x)}{- A s i n x} . \frac{1}{\sqrt{A c o s x + B}} d x {A c o s x + B = t^{2}$ $\int \frac{2 t d t}{t (- A s i n x)} = - \frac{2}{A} \int \frac{1}{s i n x} d t {c o s x = (t^{2} - B) . \frac{1}{A}$ $- \frac{2}{A} \int \frac{1}{\sqrt{A^{2} - {(t^{2} - B)}^{2}}} d t {s i n x = \sqrt{1 - {(\frac{t^{2} - B}{A})}^{2}}$ $- \frac{2}{A} \int \frac{2 t d t}{2 t \sqrt{A^{2} - u^{2}}} {t^{2} - B = u$ $- \frac{1}{A} \int \frac{d u}{\sqrt{u + B}} . \frac{1}{\sqrt{A^{2} - u^{2}}} . . . . c o n t i n u e$
	Answered by bobhans last updated on 21/Jul/20

	$I = \int \frac{d x}{\sqrt{A \cos x + B}} . l e t x = π - t$ $I = \int \frac{- d t}{\sqrt{- A \cos t + B}} = \int \frac{- d x}{\sqrt{- A \cos x + B}}$ $2 I = \int \frac{1}{\sqrt{A \cos x + B}} - \frac{1}{\sqrt{- A \cos x + B}} d x$ $2 I = \int \frac{\sqrt{B - A \cos x} - \sqrt{B + A \cos x}}{\sqrt{B^{2} - A^{2} \cos^{2} x}} d x$
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