0-pi-4-sinx-cosx-16-9sin2x-dx-

Question Number 54224 by rahul 19 last updated on 31/Jan/19

\int_{0}^{\frac{π}{4}} \frac{\sin x + \cos x}{16 + 9 \sin 2 x} d x = ?

Commented by Meritguide1234 last updated on 01/Feb/19

\Rightarrow \int_{0}^{π / 4} \frac{sinx + cosx}{25 - 9 {(sinx - cosx)}^{2}} dx

put sinx - cosx = t rest easy

Commented by rahul 19 last updated on 01/Feb/19

there is a little mistake in the denominator! �� Anyways, I've understood .

Commented by maxmathsup by imad last updated on 01/Feb/19

l e t I = \int_{0}^{\frac{π}{4}} \frac{s i n x + c o s x}{16 + 9 s i n (2 x)} d x \Rightarrow I = \int_{0}^{\frac{π}{4}} \frac{s i n (x + \frac{π}{4})}{16 + 9 s i n (2 x)} d x

=_{x + \frac{π}{4} = t} \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{s i n (t)}{16 + 9 s i n (2 t - \frac{π}{2})} d t = \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{s i n (t)}{16 - 9 c o s (2 t)} d t

= \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{s i n t d t}{16 - 9 (2 {c o s}^{2} t - 1)} = \int_{\frac{π}{4}}^{\frac{π}{2}} \frac{s i n t d t}{25 - 18 {c o s}^{2} t} =_{c o s t = u} \int_{\frac{1}{\sqrt{2}}}^{0} \frac{- d u}{25 - 18 u^{2}}

= - \int_{0}^{\frac{1}{\sqrt{2}}} \frac{d u}{18 u^{2} - 25} = - \int_{0}^{\frac{1}{\sqrt{2}}} \frac{d u}{{(3 \sqrt{2} u)}^{2} - 25} =_{3 \sqrt{2} u = 5 α} - \int_{0}^{\frac{3}{5}} \frac{1}{25 (u^{2} - 1)} \frac{5 d α}{3 \sqrt{2}}

= - \frac{1}{15 \sqrt{2}} \int_{0}^{\frac{3}{5}} (\frac{1}{u - 1} - \frac{1}{u + 1}) d u = \frac{1}{30 \sqrt{2}} {[l n ∣ \frac{u + 1}{u - 1} ∣]}_{0}^{\frac{3}{5}}

= \frac{1}{30 \sqrt{2}} l n ∣ \frac{\frac{3}{5} + 1}{\frac{3}{5} - 1} ∣ = \frac{1}{30 \sqrt{2}} l n (\frac{8}{2}) = \frac{2 l n (2)}{30 \sqrt{2}} = \frac{l n (2)}{15 \sqrt{2}}

Commented by maxmathsup by imad last updated on 01/Feb/19

e r r o r f r o m t h e f i r s t l i n e

I = \int_{0}^{\frac{π}{4}} \frac{\sqrt{2} s i n (x + \frac{π}{4})}{16 + 9 s i n (2 x)} d x \Rightarrow I = \sqrt{2} \frac{l n (2)}{15 \sqrt{2}} \Rightarrow I = \frac{l n (2)}{15} .

Answered by tanmay.chaudhury50@gmail.com last updated on 31/Jan/19

\int_{0}^{\frac{π}{4}} \frac{d (s i n x - c o s x)}{16 + 9 \times 2 s i n x c o s x}

\int_{0}^{\frac{π}{4}} \frac{d (s i n x - c o s x)}{25 - 9 (1 - 2 s i n x c o s x)}

\int_{0}^{\frac{π}{4}} \frac{d (s i n x - c o s x)}{25 - 9 {(s i n x - c o s x)}^{2}}

\frac{1}{9} \int_{0}^{\frac{π}{4}} \frac{d (s i n x - c o s x)}{{(\frac{5}{3})}^{2} - {(s i n x - c o s x)}^{2}}

\frac{1}{9} \times \frac{1}{2 (\frac{5}{3})} ∣ l n (\frac{\frac{5}{3} + s i n x - c o s x}{\frac{5}{3} - s i n x + c o s x}) ∣_{0}^{\frac{π}{4}}

= \frac{1}{30} [l n (1) - l n (\frac{\frac{5}{3} - 1}{\frac{5}{3} + 1})]

= - \frac{1}{30} l n (\frac{1}{4}) = \frac{1}{30} l n 4 = \frac{l n 2}{15} r a h u l p l s c h e c k \dots p l s u p l o a d a n s w e r o f

p r e v i o u s 8 i n t t e g a l s y o u p o s r e d \dots

Commented by rahul 19 last updated on 01/Feb/19

thank you sir!

Answered by Prithwish sen last updated on 31/Jan/19

= \int_{0}^{\frac{π}{4}} \frac{\sin (\frac{π}{4} - x) + \cos (\frac{π}{4} - x)}{16 + 9 \sin (\frac{π}{2} - 2 x)} dx

= \int_{0}^{\frac{π}{4}} \frac{\sqrt{2} cosx}{16 + 9 \cos 2 x} dx

= \sqrt{2} \int_{0}^{\frac{π}{4}} \frac{cosx}{16 + 9 - {18 \sin}^{2} x} dx

= \sqrt{2} \int_{0}^{\frac{π}{4}} \frac{cosx}{25 - {18 \sin}^{2} x} dx

putting, sinx = t

∴ cosx dx = dt, x \to \frac{π}{4} \Rightarrow t \to \frac{1}{\sqrt{2}}, x \to 0 \Rightarrow t \to 0

= \sqrt{2} \int_{0}^{\frac{1}{\sqrt{2}}} \frac{dt}{25 - {18 t}^{2}}

= \frac{\sqrt{2}}{18} \int_{0}^{\frac{1}{\sqrt{2}}} \frac{dt}{{(\frac{5}{3 \sqrt{2}})}^{2} - t^{2}}

= \frac{1}{30} \ln ∣ \frac{5 + 3 \sqrt{2} t}{5 - 3 \sqrt{2} t} ∣_{0_{}}^{\frac{1}{\sqrt{2}}}

= \frac{1}{30_{}} \ln 4

= \frac{1}{15_{}} \ln 2

Commented by rahul 19 last updated on 01/Feb/19

thank you sir!