1) ∫_0 ^( (π/4)) ((sin x+cos x)/(cos^2 x+sin^4 x))dx = ?


Question and Answers Forum

All Questions Topic List
Integration Questions
Previous in All Question Next in All Question
Previous in Integration Next in Integration
Question Number 59040 by rahul 19 last updated on 03/May/19

$1) \int_{0}^{\frac{π}{4}} \frac{\sin x + \cos x}{\cos^{2} x + \sin^{4} x} d x = ?$
	Answered by tanmay last updated on 03/May/19

	$1) \int \frac{s i n x d x}{{c o s}^{2} x + {(1 - {c o s}^{2} x)}^{2}} + \int \frac{c o s x d x}{1 - {s i n}^{2} x + {s i n}^{4} x}$ $\int \frac{- d a}{a^{2} + 1 - 2 a^{2} + a^{4}} + \int \frac{d b}{1 - b^{2} + b^{4}}$ $= - I_{1} + I_{2}$ $c a l c u l a t i o n o f - I_{1}$ $(- 1) \int \frac{\frac{1}{a^{2}}}{\frac{1}{a^{2}} - 1 + a^{2}} d a$ $(- \frac{1}{2}) \int \frac{(1 + \frac{1}{a^{2}}) - (1 - \frac{1}{a^{2}})}{a^{2} + \frac{1}{a^{2}} - 1} d a$ $= \frac{1}{2} [\int \frac{1 - \frac{1}{a^{2}}}{a^{2} + \frac{1}{a^{2}} - 1} d a - \int \frac{1 + \frac{1}{a^{2}}}{a^{2} + \frac{1}{a^{2}} - 1} d a]$ $= \frac{1}{2} [\int \frac{d (a + \frac{1}{a})}{{(a + \frac{1}{a})}^{2} - 3} - \int \frac{d (a - \frac{1}{a})}{{(a - \frac{1}{a})}^{2} + 1}]$ $= \frac{1}{2} [\frac{1}{2 \sqrt{3}} l n (\frac{a + \frac{1}{a} - \sqrt{3}}{a + \frac{1}{a} + \sqrt{3}}) - {t a n}^{- 1} (a - \frac{1}{a})]$ $=∣ \frac{1}{2} [\frac{1}{2 \sqrt{3}} l n (\frac{c o s x + s e c x - \sqrt{3}}{c o s x + s e c x + \sqrt{3}}) - {t a n}^{- 1} (c o s x - s e c x)] ∣_{0}^{\frac{π}{4}}$ $n o w c a l c u l a t i o n o f I_{2}$ $\int \frac{d b}{b^{4} - b^{2} + 1}$ $= \frac{1}{2} \int \frac{\frac{2}{b^{2}}}{b^{2} + \frac{1}{b^{2}} - 1} d b$ $= \frac{1}{2} \int \frac{(1 + \frac{1}{b^{2}}) - (1 - \frac{1}{b^{2}})}{b^{2} + \frac{1}{b^{2}} - 1} d b$ $= \frac{1}{2} [\int \frac{d (b - \frac{1}{b})}{{(b - \frac{1}{b})}^{2} + 1} - \int \frac{d (b + \frac{1}{b})}{{(b + \frac{1}{b})}^{2} - 3}]$ $= \frac{1}{2} [{t a n}^{- 1} (b - \frac{1}{b}) - \frac{1}{2 \sqrt{3}} l n (\frac{b + \frac{1}{b} - \sqrt{3}}{b + \frac{1}{b} + \sqrt{3}})]$ $= \frac{1}{2} ∣ [{t a n}^{- 1} (s i n x - c o s e c x) - \frac{1}{2 \sqrt{3}} l n (\frac{s i n x + c o s e c x - \sqrt{3}}{s i n x + c o s e c x + \sqrt{3}})] ∣_{0}^{\frac{π}{4}}$ $R a h u l p l s c h e c k . . . l e n t h y p r o b l e m . . . m i s t a k e i f a n y$
	Commented by rahul 19 last updated on 04/May/19

	$t h a n k Y o u S i r .$
	Answered by MJS last updated on 03/May/19

	$\underset{0}{\int^{\frac{π}{4}}} \frac{\cos x}{\cos^{2} x + \sin^{4} x} d x =$ $[t = \sin x \to d x = \frac{d t}{\cos x}]$ $= \underset{0}{\int^{\frac{\sqrt{2}}{2}}} \frac{d t}{t^{4} - t^{2} + 1} = \underset{0}{\int^{\frac{\sqrt{2}}{2}}} \frac{d t}{(t^{2} - \sqrt{3} t + 1) (t^{2} + \sqrt{3} t + 1)}$ $\underset{0}{\int^{\frac{π}{4}}} \frac{\sin x}{\cos^{2} x + \sin^{4} x} d x =$ $[u = \cos x \to d x = - \frac{d u}{\sin x}]$ $= - \underset{1}{\int^{\frac{\sqrt{2}}{2}}} \frac{d u}{u^{4} - u^{2} + 1} \Rightarrow we only have to solve the 1^{st} one$ $\int \frac{d t}{(t^{2} - \sqrt{3} t + 1) (t^{2} + \sqrt{3} t + 1)} =$ $= \int \frac{\frac{1}{2} - \frac{\sqrt{3}}{6} t}{t^{2} - \sqrt{3} t + 1} d t + \int \frac{\frac{1}{2} + \frac{\sqrt{3}}{6} t}{t^{2} + \sqrt{3} t + 1} d t =$ $= - \frac{\sqrt{3}}{6} \int \frac{t - \sqrt{3}}{t^{2} - \sqrt{3} t + 1} d t + \frac{\sqrt{3}}{6} \int \frac{t + \sqrt{3}}{t^{2} + \sqrt{3} t + 1} d t =$ $= \frac{1}{2} \arctan (2 t - \sqrt{3}) - \frac{\sqrt{3}}{12} \ln (t^{2} - \sqrt{3} t + 1) + \frac{1}{2} \arctan (2 t + \sqrt{3}) + \frac{\sqrt{3}}{12} \ln (t^{2} + \sqrt{3} t + 1)$ $now please finish it, Sir Rahul, I have to go$ $to bed . . .$
	Commented by MJS last updated on 03/May/19

	$I get, after ‘ ‘ {cleaning}^{″} the roots$ $\frac{\sqrt{3}}{6} \ln (2 + \sqrt{3}) + \frac{π}{4}$
	Commented by rahul 19 last updated on 04/May/19

	$t h a n k y o u s i r .$
	Answered by MJS last updated on 04/May/19

	$I found another path$ $\int \frac{\sin x + \cos x}{\cos^{2} x + \sin^{4} x} d x = 8 \sqrt{2} \int \frac{\sin (x + \frac{π}{4})}{7 + \cos 4 x} d x =$ $[t = \cos (x + \frac{π}{4}) \to d x = - \frac{d t}{\sin (x + \frac{π}{4})}]$ $= 4 \sqrt{2} \int \frac{d t}{4 t^{4} - 4 t^{2} + 1} = 4 \sqrt{2} \int \frac{d t}{(2 t^{2} - 3) (2 t^{2} + 1)} =$ $= \frac{\sqrt{6}}{6} \int \frac{d t}{\sqrt{2} t - \sqrt{3}} - \frac{\sqrt{6}}{6} \int \frac{d t}{\sqrt{2} t + \sqrt{3}} - \sqrt{2} \int \frac{d t}{2 t^{2} + 1} =$ $= \frac{\sqrt{3}}{6} \ln \frac{\sqrt{2} t - \sqrt{3}}{\sqrt{2} t + \sqrt{3}} - \arctan \sqrt{2} t =$ $= \frac{\sqrt{3}}{6} \ln \frac{{(2 t - \sqrt{6})}^{2}}{2 (2 t^{2} - 3)} - \arctan \sqrt{2} t =$ $= \frac{\sqrt{3}}{3} \ln (2 t - \sqrt{6}) - \frac{\sqrt{3}}{6} \ln (2 t^{2} - 3) - \frac{\sqrt{3}}{6} \ln 2 - \arctan \sqrt{2} t =$ $= \frac{\sqrt{3}}{3} \ln ∣ 2 t - \sqrt{6} ∣ - \frac{\sqrt{3}}{6} \ln ∣ 2 t^{2} - 3 ∣ - \arctan \sqrt{2} t + C$ $the borders [0 ⩽ x ⩽ \frac{π}{4}] \to [\frac{\sqrt{2}}{2} ⩾ t ⩾ 0]$ $\Rightarrow \underset{0}{\int^{\frac{π}{4}}} \frac{\sin x + \cos x}{\cos^{2} x + \sin^{4} x} d x = \frac{π}{4} - \frac{\sqrt{3}}{6} \ln (2 - \sqrt{3})$
	Commented by MJS last updated on 04/May/19

	$\frac{π}{4} - \frac{\sqrt{3}}{6} \ln (2 - \sqrt{3}) = \frac{π}{4} + \frac{\sqrt{3}}{6} \ln (2 + \sqrt{3}) \approx 1.16557$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum