Question

cos⁴(2x).dx
Integrate the function

Answer 1

cos⁴2x=(cos²2x)²=[(1+cos2x)/2]²dx1
=1/4[1+cos²2x+2cos2x]dx
=1/4[1+(1+cos2x)/2+2cos2x]dx
1/8[2+1+cos2x+4cos2x]dx
1/8[3dx+cos2xdx+4cos2xdx]
1/8[3x+sin2x/2+2sin2x]+c

Answer 2

HELLO DEAR,

Given function is integral of(cos⁴2x).dx

$\sf{\int{cos^22x*cos^22x}\,dx}$

[tex]\sf{\Rightarrow1/4[\int{(1+cos4x)(1+
cos4x)}\,dx]}[/tex]

[tex]\sf{\Rightarrow1/4[\int{1+cos^24x+
2cos4x}\,dx]}[/tex]

[tex]\sf{\Rightarrow1/4[\int{1}\,dx+1/2\int{(1+
cos8x)}\,dx+2int{cos4x}\,dx]}[/tex]

[tex]\sf{\Rightarrow\frac{1}{4}[x+\frac{1}{2}x+
\frac{1}{16}sin8x+\frac{1}{2}sin4x]+c}[/tex]

[tex]\sf{\Rightarrow\frac{3}{8}x+\frac{1}{64}sin8x+
\frac{1}{8}sin4x+c}[/tex]

I HOPE ITS HELP YOU DEAR,
THANKS