calculate-0-pi-4-cos-4-x-sin-2-xdx-

Question Number 40130 by maxmathsup by imad last updated on 16/Jul/18

c a l c u l a t e \int_{0}^{\frac{π}{4}} {c o s}^{4} x {s i n}^{2} x d x

Commented by math khazana by abdo last updated on 21/Jul/18

l e t I = \int_{0}^{\frac{π}{4}} {c o s}^{4} x {s i n}^{2} x d x b y p a r t s

I = \int_{0}^{\frac{π}{4}} s i n x {c o s}^{4} x s i n x d x

= {[- \frac{1}{5} {c o s}^{5} x s i n x]}_{0}^{\frac{π}{4}} - \int_{0}^{\frac{π}{4}} (- \frac{1}{5}) {c o s}^{5} x c o s x d x

= - \frac{1}{5} {(\frac{\sqrt{2}}{2})}^{6} + \frac{1}{5} \int_{0}^{\frac{π}{4}} {c o s}^{6} x d x b u t

\int_{0}^{\frac{π}{4}} {c o s}^{6} x d x = \int_{0}^{\frac{π}{4}} {(\frac{1 + c o s (2 x)}{2})}^{3} d x

= \frac{1}{8} \int_{0}^{\frac{π}{4}} {{c o s}^{3} (2 x) + 3 {c o s}^{2} (2 x) + 3 c o s (2 x) + 1)} d x

= \frac{1}{8} \int_{0}^{\frac{π}{4}} {c o s}^{3} (2 x) d x + \frac{3}{8} \int_{0}^{\frac{π}{4}} \frac{1 + c o s (4 x)}{2} d x

+ \frac{3}{8} \int_{0}^{\frac{π}{4}} c o s (2 x) d x + \frac{π}{32} b u t

\int_{0}^{\frac{π}{4}} c o s (2 x) d x = {[\frac{1}{2} s i n (2 x)]}_{0}^{\frac{π}{4}} = \frac{1}{2}

\int_{0}^{\frac{π}{4}} \frac{1 + c o s (4 x)}{2} d x = \frac{π}{8} + \frac{1}{8} {[s i n (4 x)]}_{0}^{\frac{π}{4}}

= \frac{π}{8}

\int_{0}^{\frac{π}{4}} {c o s}^{3} (2 x) d x = \int_{0}^{\frac{π}{4}} c o s (2 x) \frac{1 + c o s (4 x)}{2} d x

= \frac{1}{2} \int_{0}^{\frac{π}{4}} c o s (2 x) + \frac{1}{2} \int_{0}^{\frac{π}{4}} c o s (2 x) c o s (4 x) d x

= \frac{1}{4} {[s i n (2 x)]}_{0}^{\frac{π}{4}} + \frac{1}{4} \int_{0}^{\frac{π}{4}} {c o s (6 x) + c o s (2 x)) d x

= \frac{1}{4} + \frac{1}{24} {[s i n (6 x)]}_{0}^{\frac{π}{4}} + \frac{1}{8} {[s i n (2 x)]}_{0}^{\frac{π}{4}}

= \frac{1}{4} - \frac{1}{24} + \frac{1}{8} = \frac{6 - 1 + 3}{24} = \frac{8}{24} = \frac{1}{3}

s o t h e v a l u e o f I i s d e t e r m i n e d .