Rylan Sullivan

2022-04-25

Integrating ${\int}_{0}^{\frac{\pi}{4}}\frac{\mathrm{sin}x+\mathrm{cos}x}{{\mathrm{sin}}^{4}x+{\mathrm{cos}}^{2}x}dx$

Eliza Flores

Beginner2022-04-26Added 16 answers

Hints:

1) Divide it into two integrals :

$\int \frac{-d\mathrm{cos}\left\{x\right\}}{{(1-{\mathrm{cos}}^{2}\left\{x\right\})}^{2}+{\mathrm{cos}}^{2}\left\{x\right\}}+\int \frac{d\mathrm{sin}\left\{x\right\}}{{\mathrm{sin}}^{4}\left\{x\right\}+1-{\mathrm{sin}}^{2}\left\{x\right\}}$

Then use rational functions

2) Use$\mathrm{tan}\left\{\frac{x}{2}\right\}=t$ then it's easy to get rational functions in numerator and denominator.

1) Divide it into two integrals :

Then use rational functions

2) Use

Leanna Boone

Beginner2022-04-27Added 11 answers

Use your trig identities to create a u-substitution. We have

$[\mathrm{sin}\left(x\right)+\mathrm{cos}\left(x\right)]dx=-d[\mathrm{cos}\left(x\right)-\mathrm{sin}\left(x\right)]$

and

${\mathrm{sin}\left(x\right)}^{4}+{\mathrm{cos}\left(x\right)}^{2}=1-{\mathrm{sin}\left(x\right)}^{2}{\mathrm{cos}\left(x\right)}^{2}=1-\frac{1}{4}{(1-{[\mathrm{cos}\left(x\right)-\mathrm{sin}\left(x\right)]}^{2})}^{2},$

so

$\int}_{0}^{\frac{\pi}{4}}\frac{\mathrm{sin}\left(x\right)+\mathrm{cos}\left(x\right)}{{\mathrm{sin}\left(x\right)}^{4}+{\mathrm{cos}\left(x\right)}^{2}}dx=4{\int}_{0}^{1}\frac{du}{4-{(1-{u}^{2})}^{2}}=4{\int}_{0}^{1}\frac{du}{(3-{u}^{2})(1+{u}^{2})$

and I'm guessing you know where to go from here.

Note that if you didn't know those trig identities for the denominator, you could still use$u=\mathrm{cos}\left(x\right)-\mathrm{sin}\left(x\right)=\sqrt{2}\mathrm{cos}(x+\frac{\pi}{4})$ , then plug in $x={\mathrm{cos}}^{-1}\left(\frac{u}{\sqrt{2}}\right)-\frac{\pi}{4}$ and do the algebra. It's not fun, but it gets there in the end.

and

so

and I'm guessing you know where to go from here.

Note that if you didn't know those trig identities for the denominator, you could still use

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