How do you evaluate this trig integral $\int 340 {cos}^{4} (20 x) d x$ $int 340cos^4(20x) dx$ ?

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Answer 1 · 2018-05-15T16:24:32

$I = \frac{17}{32} [240 x + 8 sin (40 x) + 4 sin (80 x)] + c$ $I=17/32[240x+8sin(40x)+4sin(80x)]+c$

We know that,

$(1) {cos}^{2} θ = \frac{1 + cos 2 θ}{2}$ $color(red)((1)cos^2theta=(1+cos2theta)/2$

$(2) \int cos A x d x = \frac{1}{A} sin A x + c$ $color(blue)((2)intcosAxdx=1/AsinAx+c$

Here,

$I = \int 340 {cos}^{4} (20 x) d x$ $I=int340cos^4(20x)dx$

Subst. , $20 x = u \Rightarrow x = \frac{u}{20} \Rightarrow d x = \frac{1}{20} d u$ $20x=u=>x=u/20=>dx=1/20du$

So,

$I = \int 340 {cos}^{4} u \times \frac{1}{20} d u$ $I=int340cos^4uxx1/20du$

$=17int (cos^2u)^2du...toApply(1)$

$=17int((1+cos2u)/2)^2$

$=17/4int(1+2cos2u+cos^2 2u)du$

$=17/4int[1+2cos2u+(1+cos4u)/2]$

$=17/8int[2+4cos2u+1+cos4u]du$

$=17/8int[3+4cos2u+cos4u]du...toApply(2)$

$=17/8[3u+(4sin2u)/2+(sin4u)/4]+c$

$=17/32[12u+8sin2u+sin4u]+c$

Subst. back , $u=20x$

$I=17/32[12(20x)+8sin2(20x)+4sin4(20x)]+c$

$I=17/32[240x+8sin(40x)+4sin(80x)]+c$

How do you evaluate this trig integral int 340cos^4(20x) dx∫340cos4(20x)dxint 340cos^4(20x) dx?