0-pi-2-sin-3-x-sin-x-cos-x-dx-

Question Number 175351 by cortano1 last updated on 28/Aug/22

\underset{0}{\int^{π / 2}} \frac{\sin^{3} x}{\sin x + \cos x} d x = ?

Answered by som(math1967) last updated on 28/Aug/22

I = \int_{0}^{\frac{π}{2}} \frac{{s i n}^{3} (\frac{π}{2} + 0 - x) d x}{s i n (\frac{π}{2} + 0 - x) + c o s (\frac{π}{2} + 0 - x)}

= \int_{0}^{\frac{π}{2}} \frac{{c o s}^{3} x d x}{c o s x + s i n x}

2 I = \int_{0}^{\frac{π}{2}} \frac{{s i n}^{3} x + {c o s}^{3} x}{s i n x + c o s x} d x

2 I = \int_{0}^{\frac{π}{2}} ({s i n}^{2} x - s i n x c o x + {c o s}^{2} x) d x

2 I = \int_{0}^{\frac{π}{2}} d x - \frac{1}{2} \int_{0}^{\frac{π}{2}} s i n 2 x d x

2 I = {[x + \frac{c o s 2 x}{4}]}_{0}^{\frac{π}{2}}

2 I = (\frac{π}{2} - \frac{1}{4}) - (0 + \frac{1}{4})

I = \frac{π}{4} - \frac{1}{4}

Commented by cortano1 last updated on 28/Aug/22

b y K i n g F o r m u l a

Commented by som(math1967) last updated on 28/Aug/22

y e s