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

$$\underset{0}{\stackrel{\frac{\mathit{\pi }}{2}}{\int }}\frac{\mathbf{sin}4\mathbf{x}}{1+\mathbf{sinx}+\mathbf{cosx}}\mathbf{dx}=?$$

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

$$letI={\int }_0^{\frac{\pi }{2}}\frac{sin\left(4x\right)}{1+sinx)cosx}dxchangementx=\frac{\pi }{2}-tgive$$$$I=-{\int }_0^{\frac{\pi }{2}}\frac{sin(2\pi -4t)}{1+cost+sint}(-dt)={\int }_0^{\frac{\pi }{2}}\frac{-sin\left(4t\right)}{1+sint)cost}dt=-I\Rightarrow$$$$2I=0\Rightarrow I=0$$

Answered by MJS last updated on 18/Dec/19

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

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

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

Answered by MJS last updated on 19/Dec/19

$$\int \frac{\sin 4x}{1+\sin x+\cos x}dx=$$$$=\int (\sin 3x+\cos 3x-2\cos 2x-\sin x+\cos x)dx=$$$$=\frac{1}{3}(-\cos 3x+\sin x)-\sin 2x+\cos x+\sin x+C$$

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