Resolved: "Calculating with Mathematica, one can have∫0π2sin3tsin3t+cos3tdt=π4"

Shayla Massey

Answered question

2022-03-17

Calculating with Mathematica, one can have

\int_{0}^{\frac{π}{2}} \frac{\sin^{3} t}{\sin^{3} t + \cos^{3} t} d t = \frac{π}{4}

PCCNQN4XKhjx

Beginner2022-03-18Added 8 answers

Use the Calculus identity that

f (x) = f (a - x)

and let

I = \int_{0}^{\frac{π}{2}} \frac{\sin^{3} t}{\sin^{3} t + \cos^{3} t} d t

Then,

f (t) = f (\frac{π}{2} - t) = \frac{\sin^{3} (\frac{π}{2} - t)}{\sin^{3} (\frac{π}{2} - t) + \cos^{3} (\frac{π}{2} - t)} = \frac{\cos^{3} t}{\cos^{3} t + \sin^{3} t}

= \frac{\cos^{3} t}{\cos^{3} t + \sin^{3} t}

Thus,

I = \int_{0}^{\frac{π}{2}} \frac{\cos^{3} t}{\cos^{3} t + \sin^{3} t} d t

So we have

2 I = \int_{0}^{\frac{π}{2}} \frac{\sin^{3} t}{\sin^{3} t + \cos^{3} t} d t + \int_{0}^{\frac{π}{2}} \frac{\cos^{3} t}{\cos^{3} t + \sin^{3} t} d t

= \int_{0}^{\frac{π}{2}} d t

= \frac{π}{2}

I = \frac{π}{4}

Note that this is true for any natural number n

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