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Question Number 75848 by behi83417@gmail.com last updated on 18/Dec/19

∫_0 ^( (𝛑/2)) ((sin4x)/(1+sinx+cosx))dx=?

$$\underset{\mathrm{0}} {\overset{\:\:\:\:\:\:\:\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} {\int}}\frac{\boldsymbol{\mathrm{sin}}\mathrm{4}\boldsymbol{\mathrm{x}}}{\mathrm{1}+\boldsymbol{\mathrm{sinx}}+\boldsymbol{\mathrm{cosx}}}\boldsymbol{\mathrm{dx}}=? \\ $$

Commented by mathmax by abdo last updated on 18/Dec/19

let I=∫_0 ^(Ο€/2) ((sin(4x))/(1+sinx )cosx))dx changement x=(Ο€/2)βˆ’t give I =βˆ’βˆ«_0 ^(Ο€/2) ((sin(2Ο€βˆ’4t))/(1+cost +sint))(βˆ’dt) =∫_0 ^(Ο€/2) ((βˆ’sin(4t))/(1+sint )cost))dt =βˆ’I β‡’ 2I=0 β‡’I =0

$${let}\:{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{sin}\left(\mathrm{4}{x}\right)}{\left.\mathrm{1}+{sinx}\:\right){cosx}}{dx}\:{changement}\:{x}=\frac{\pi}{\mathrm{2}}βˆ’{t}\:{give} \\ $$$${I}\:=βˆ’\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{sin}\left(\mathrm{2}\piβˆ’\mathrm{4}{t}\right)}{\mathrm{1}+{cost}\:+{sint}}\left(βˆ’{dt}\right)\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{βˆ’{sin}\left(\mathrm{4}{t}\right)}{\left.\mathrm{1}+{sint}\:\right){cost}}{dt}\:=βˆ’{I}\:\Rightarrow \\ $$$$\mathrm{2}{I}=\mathrm{0}\:\Rightarrow{I}\:=\mathrm{0} \\ $$

Answered by MJS last updated on 18/Dec/19

sin 4x has a period of (Ο€/2) β‡’ weβ€²re integrating over a whole period β‡’ answer is 0

$$\mathrm{sin}\:\mathrm{4}{x}\:\mathrm{has}\:\mathrm{a}\:\mathrm{period}\:\mathrm{of}\:\frac{\pi}{\mathrm{2}}\:\Rightarrow\:\mathrm{we}'\mathrm{re}\:\mathrm{integrating} \\ $$$$\mathrm{over}\:\mathrm{a}\:\mathrm{whole}\:\mathrm{period}\:\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{0} \\ $$

Commented by behi83417@gmail.com last updated on 19/Dec/19

e^x cellent dear proph. thanks in advance sir.

$$\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{x}}} \boldsymbol{\mathrm{cellent}}\:\boldsymbol{\mathrm{dear}}\:\boldsymbol{\mathrm{proph}}. \\ $$$$\boldsymbol{\mathrm{than}}\mathrm{k}\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{in}}\:\boldsymbol{\mathrm{advance}}\:\boldsymbol{\mathrm{sir}}. \\ $$

Answered by MJS last updated on 19/Dec/19

∫((sin 4x)/(1+sin x +cos x))dx= =∫(sin 3x +cos 3x βˆ’2cos 2x βˆ’sin x +cos x)dx= =(1/3)(βˆ’cos 3x +sin x)βˆ’sin 2x +cos x +sin x +C

$$\int\frac{\mathrm{sin}\:\mathrm{4}{x}}{\mathrm{1}+\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}}{dx}= \\ $$$$=\int\left(\mathrm{sin}\:\mathrm{3}{x}\:+\mathrm{cos}\:\mathrm{3}{x}\:βˆ’\mathrm{2cos}\:\mathrm{2}{x}\:βˆ’\mathrm{sin}\:{x}\:+\mathrm{cos}\:{x}\right){dx}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left(βˆ’\mathrm{cos}\:\mathrm{3}{x}\:+\mathrm{sin}\:{x}\right)βˆ’\mathrm{sin}\:\mathrm{2}{x}\:+\mathrm{cos}\:{x}\:+\mathrm{sin}\:{x}\:+{C} \\ $$

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