find-the-values-of-I-0-pi-cos-4-dx-and-J-0-pi-sin-4-dx-

Question Number 36406 by abdo mathsup 649 cc last updated on 01/Jun/18

f i n d t h e v a l u e s o f I = \int_{0}^{π} {c o s}^{4} d x a n d

J = \int_{0}^{π} {s i n}^{4} d x .

Commented by abdo.msup.com last updated on 05/Jun/18

w e h a v e I + J = \int_{0}^{π} ({c o s}^{4} x + {s i n}^{4} x) d x

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

= π - 2 \int_{0}^{π} \frac{1}{4} {(s i n (2 x))}^{2} d x

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

= π - \frac{π}{4} + \frac{1}{16} {[s i n (4 x)]}_{0}^{π} = \frac{3 π}{4} a l s o

I - J = \int_{0}^{π} ({c o s}^{4} x - {s i n}^{4} x) d x

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

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

I = J \Rightarrow 2 I = \frac{3 π}{4} \Rightarrow I = \frac{3 π}{8} a n d J = \frac{3 π}{8}

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Jun/18

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

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

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

= \frac{1}{4} \int_{0}^{Π} d x + \frac{1}{2} \int_{0}^{Π} c o s 2 x d x + \frac{1}{8} \int_{0}^{Π} d x + \frac{1}{8} \int_{0}^{Π} c o s 4 x

= \frac{Π}{4} + \frac{Π}{8}

= \frac{3 Π}{8}

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Jun/18

j = \int_{0}^{Π (} {(\frac{1 - c o s 2 x}{2})}^{2}

= \frac{1}{4} \int_{0}^{Π} 1 - 2 c o s 2 x + \frac{1 + c o s 4 x}{2} d x

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

= \frac{Π}{4} + \frac{Π}{8} = \frac{3 Π}{8}