Integrating ∫0π4sin⁡x+cos⁡xsin4x+cos2xdx

Question

Eliza Flores · Accepted Answer

Hints:1) Divide it into two integrals :∫−dcos⁡{x}(1−cos2{x})2+cos2{x}+∫dsin⁡{x}sin4{x}+1−sin2{x}Then use rational functions2) Use tan⁡{x2}=t then it&#039;s easy to get rational functions in numerator and denominator.

Leanna Boone · Accepted Answer

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

[\sin (x) + \cos (x)] d x = - d [\cos (x) - \sin (x)]

and

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

so

\int_{0}^{\frac{π}{4}} \frac{\sin (x) + \cos (x)}{{\sin (x)}^{4} + {\cos (x)}^{2}} d x = 4 \int_{0}^{1} \frac{d u}{4 - {(1 - u^{2})}^{2}} = 4 \int_{0}^{1} \frac{d u}{(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 = \cos (x) - \sin (x) = \sqrt{2} \cos (x + \frac{π}{4})

, then plug in

x = \cos^{- 1} (\frac{u}{\sqrt{2}}) - \frac{π}{4}

and do the algebra. It's not fun, but it gets there in the end.

Integrating \int_{0}^{π/4} \frac{\sin x+\cos x}{\sin^4 x+\cos^2 x} dx

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