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Kason Palmer

Answered question

2022-03-20

Solve trigonometric integral

\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{\sin^{2014} x}{\sin^{2014} x + \cos^{2014} x} d x

uqhekekocj8f

Beginner2022-03-21Added 8 answers

Consider the integral

I = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{\sin^{2 a} x}{\sin^{2 a} x + \cos^{2 a} x}

This may also be seen as

= \int_{0}^{\frac{π}{2}} \frac{\sin^{2 a} x}{\sin^{2 a} x + \cos^{2 a} x} + \int_{- \frac{π}{2}}^{0} \frac{\sin^{2 a} x}{\sin^{2 a} x + \cos^{2 a} x}

= \int_{0}^{\frac{π}{2}} \frac{\sin^{2 a} x}{\sin^{2 a} x + \cos^{2 a} x} d x - \int_{\frac{π}{2}}^{0} \frac{\sin^{2 a} x}{\sin^{2 a} x + \cos^{2 a} x} d x

= 2, \int_{0}^{\frac{π}{2}} \frac{\sin^{2 a} x}{\sin^{2 a} x + \cos^{2 a} x} d x

where

x \to - x

was made in the second integral. Now make the substitution

x = t - \frac{π}{2}

to obtain

I = 2 \int_{- \frac{π}{2}}^{0} \frac{\cos^{2 a} t}{\sin^{2 a} t + \cos^{2 a} t} d t

Now let

t \to - x

to obtain

I = 2 \int_{0}^{\frac{π}{2}} \frac{\cos^{2 a} x}{\sin^{2 a} x + \cos^{2 a} x} d x

Adding the two integral expressions leads to

= 2 \int_{0}^{\frac{π}{2}} \frac{\sin^{2 a} x + \cos^{2 a} x}{\sin^{2 a} x + \cos^{2 a} x} d x = 2 \int_{0}^{\frac{π}{2}} d x = π .

It can now be started that

\int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{\sin^{2 a} x}{\sin^{2 a} x + \cos^{2 a} x} d x = \frac{π}{2}

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