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Question Number 167328 by cortano1 last updated on 13/Mar/22

$${\int }_0^{\pi /2}\frac{\cos x+\cos {}^{5}x\sin x}{\sqrt{1+\cos {}^{2}x}}dx=?$$

Answered by MJS_new last updated on 13/Mar/22

$$\int \frac{\cos x+{\cos }^{5}x\sin x}{\sqrt{1+{\cos }^{2}x}}dx=$$$$[t=\arcsin \frac{\sin x}{\sqrt{2}}\to dx=\frac{\sqrt{1+{\cos }^{2}x}}{\cos x}dt]$$$$=$$$$...$$$$=\frac{\sqrt{2}}{4}\int (\sin 5t-\sin 3t+2\sin t+3\sqrt{2})dt=$$$$=-\frac{\sqrt{2}}{20}\cos 5t+\frac{\sqrt{2}}{12}\cos 3t-\frac{\sqrt{2}}{2}\cos t+t=$$$$...$$$$=\arcsin \frac{\sin x}{\sqrt{2}}-\frac{\sqrt{1+{\cos }^{2}x}}{15}({3\cos }^{4}x-{4\cos }^{2}x+8)+C$$

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