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

Question Number 66325 by mathmax by abdo last updated on 12/Aug/19

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 Prithwish sen last updated on 13/Aug/19

\frac{1}{8} \int_{0}^{\frac{π}{4}} (\cos 2 x + 1) \sin^{2} 2 xdx = \frac{1}{8} \int_{0}^{\frac{π}{4}} {\cos 2 xsin}^{2} 2 xdx + \frac{1}{16} \int_{0}^{\frac{π}{4}} (1 - \cos 4 x) dx

= {[\frac{\sin^{3} 2 x}{48} + \frac{x}{16} - \frac{\sin 4 x}{64}]}_{0}^{\frac{π}{4}} = \frac{1}{48} + \frac{π}{64} please check .

Commented by mathmax by abdo last updated on 13/Aug/19

l e t I = \int_{0}^{\frac{π}{4}} {c o s}^{4} x {s i n}^{2} x \Rightarrow I = \int_{0}^{\frac{π}{4}} {(\frac{1 + c o s (2 x)}{2})}^{2} (\frac{1 - c o s (2 x)}{2}) d x

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

= \frac{1}{16} \int_{0}^{\frac{π}{4}} (2 + 4 c o s (2 x) + 1 + c o s (4 x)) (1 - c o s (2 x)) d x

= \frac{1}{16} \int_{0}^{\frac{π}{4}} (3 + 4 c o s (2 x) + c o s (4 x)) (1 - c o s (2 x)) d x

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

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

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

= \frac{3 π}{4} + \frac{1}{2} {[s i n (2 x)]}_{0}^{\frac{π}{4}} + \frac{1}{4} {[s i n (4 x)]}_{0}^{\frac{π}{4}} - \frac{π}{2} - \frac{1}{2} {[s i n (4 x)]}_{0}^{\frac{π}{4}}

= \frac{3 π}{4} + \frac{1}{2} - \frac{π}{2} = \frac{π}{4} + \frac{1}{2}

Commented by mathmax by abdo last updated on 13/Aug/19

e r r o r a t f i n a l l i n e

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

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

= \frac{1}{6} s i n (\frac{3 π}{2}) + \frac{1}{2} = \frac{1}{2} - \frac{1}{6} \Rightarrow

16 I = \frac{π}{4} + \frac{1}{2} - \frac{1}{4} + \frac{1}{12} = \frac{π}{4} + \frac{1}{4} + \frac{1}{12} = \frac{π}{4} + \frac{1}{3} \Rightarrow

I = \frac{π}{64} + \frac{1}{48} .