ज्ञात करें `int cos^4 x dx`

Explanation

`because" "cos^2x=(1+cos2x)/2`
`:." "cos^4x=((1+cos2x)/2)=1/4(1+2cos2x+cos^22x)`
`=1/4(1+2cos2x+(1+cos4x)/(2))=1/4(1+2cos2x+1/2+1/2cos4x)`
`=1/4(3/2+2cos2x+1/2cos4x)=3/8+1/2cos2x+1/8cos4x`
अब,`intcos^4 xdx=int(3/8+1/2cos2x1/8cos4x)dx`
`3/4intdx+1/2 int "cos 2x dx" + 1/8int"cos 4x dx"`
`=3/4x+1/2*(sin2x)/2+1/8*("sin 4x")/4+c=3/8x+("sin 2x")2+("sin 4x")/32+c`