amevaa0y

2022-01-27

How can I evaluate the following integral?
$\int \frac{\mathrm{cos}\left\{x\right\}}{1+\mathrm{sin}\left\{2x\right\}}dx$
I tried the following way, but I was not able to proceed further:
$I=\int \frac{\mathrm{cos}\left\{x\right\}}{{\left(\mathrm{sin}\left\{x\right\}+\mathrm{cos}\left\{x\right\}\right)}^{2}}dx$
$=\int \frac{\mathrm{sec}\left\{x\right\}}{{\left(1+\mathrm{tan}\left\{x\right\}\right)}^{2}}dx$

Jude Carpenter

Hint:
Write numerator as
$2\mathrm{cos}x=\mathrm{cos}x+\mathrm{sin}x+\left(\mathrm{cos}x-\mathrm{sin}x\right)$
$I=\int \frac{\mathrm{cos}x+\mathrm{sin}x}{{\left\{\left(\mathrm{cos}x+\mathrm{sin}x\right\}\right)}^{2}}dx+\int \frac{\mathrm{cos}x-\mathrm{sin}x}{{\left(\mathrm{cos}x+\mathrm{sin}x\right)}^{2}}dx$
$I={I}_{1}+{I}_{2}$
${I}_{1}=\int \frac{1}{\mathrm{cos}x+\mathrm{sin}x}dx$
${I}_{1}=\int \frac{1}{\sqrt{2}\left(\mathrm{cos}x\cdot \mathrm{cos}\left\{\frac{\pi }{4}+\mathrm{sin}x\cdot \mathrm{sin}\left\{\frac{\pi }{4}\right\}\right)\right\}}dx$
${I}_{1}=\int \frac{\mathrm{sec}\left(x-\frac{\pi }{4}\right)}{\sqrt{2}}dx$
${I}_{1}=\frac{1}{\sqrt{2}}\cdot \mathrm{ln}\left[\left(\mathrm{sec}\left(x-\frac{\pi }{4}\right)+\mathrm{tan}\left(x-\frac{\pi }{4}\right)\right]+{C}_{1}$
Now
${I}_{2}=\int \frac{\mathrm{cos}x-\mathrm{sin}x}{{\left(\mathrm{cos}x+\mathrm{sin}x\right)}^{2}}dx$
$\mathrm{sin}x+\mathrm{cos}x=t⇒\left(\mathrm{cos}x-\mathrm{sin}x\right)dx=dt$
${I}_{2}=\int \frac{1}{{t}^{2}}dt$
${I}_{2}=-\frac{1}{t}+{C}_{2}=-\frac{1}{\mathrm{sin}x+\mathrm{cos}x}+{C}_{2}$

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