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Question Number 122922 by bemath last updated on 21/Nov/20

$$\underset{0}{\stackrel{\pi /4}{\int }}\frac{\cos x+\sin x}{16\sin 2x+9}dx$$

Answered by liberty last updated on 21/Nov/20

$$\varsigma =\underset{0}{\stackrel{\pi /4}{\int }}\frac{\cos x+\sin x}{16\sin 2x+9}dx$$$$[let16\sin 2x+9=t\Rightarrow dt=32\cos 2xdx]$$$$\Rightarrow dt=32(\cos x+\sin x)(\cos x-\sin x)dx$$$$\Rightarrow dt=32(\cos x+\sin x)\sqrt{1-\sin 2x}dx$$$$\Rightarrow dt=32(\cos x+\sin x)\sqrt{1-\left(\frac{t-9}{16}\right)}dx$$$$\Rightarrow (\cos x+\sin x)dx=\frac{dt}{32\sqrt{\frac{25-t}{16}}}=\frac{dt}{8\sqrt{25-t}}$$$$\varsigma ={\int }_9^{25}\frac{dt}{8t\sqrt{25-t}}=\frac{1}{8}{\int }_9^{25}\frac{dt}{t\sqrt{25-t}}$$$$[let\sqrt{25-t}=s\Rightarrow t=25-s^2]$$$$\varsigma =\frac{1}{8}{\int }_4^0\frac{-2sds}{s(25-s^2)}=\frac{1}{4}{\int }_0^4\frac{ds}{25-s^2}$$$$[\frac{-1}{(s+5)(s-5)}=\frac{A}{s+5}+\frac{B}{s-5}]$$$$[A=\frac{-1}{s-5}{\mid }_{s=-5}=\frac{1}{10}\wedge B=\frac{-1}{s+5}{\mid }_{s=5}=-\frac{1}{10}]$$$$\varsigma =\frac{1}{40}({\int }_0^4\frac{ds}{s+5}-{\int }_0^4\frac{ds}{s-5})$$$$\varsigma =\frac{1}{40}{(\ln \mid \frac{s+5}{s-5}\mid )}_0^4=\frac{\ln \left(9\right)}{40}.▴$$

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