Explained: "Evaluating ∫−π2π2sin⁡(x)+cos⁡(x)sin4(x)−4dx"

Gerald Ritter

Answered question

2022-01-28

Evaluating

\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{\sin (x) + \cos (x)}{\sin^{4} (x) - 4} d x

Brynn Ortiz

Beginner2022-01-29Added 12 answers

Following a suggestion from the comments, you can reduce your integral to

\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{\cos x}{\sin^{4} (x) - 4} d x

Now, do the substitution

\sin x = t

and

\cos x d x = d t

votaren10

Beginner2022-01-30Added 11 answers

\int \frac{\sin x}{\sin^{4} x - 4} d x = - \int \frac{1}{{(1 - u^{2})}^{2} - 4} d u

\int \frac{\cos x}{\sin^{4} x - 4} d x = \int \frac{d u}{u^{4} - 4} = 0.25 \int \frac{d u}{u^{2} - 2} - 0.25 \int \frac{d u}{u^{2} + 2}

the first one is zero over a symmetric interval and for the second we have

I = 0.25 \int_{- 1}^{1} \frac{d u}{u^{2} - 2} - 0.25 \int_{- 1}^{1} \frac{d u}{u^{2} + 2} = \frac{1}{4 \sqrt{2}} \ln \frac{\sqrt{2} - 1}{\sqrt{2} + 1} - \frac{\sqrt{2}}{4} \tan^{- 1} \frac{1}{\sqrt{2}}

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?