Question

Integrate 2 cos8x sin4x with respect to x

Answer 1

Hey there!

∫ 2 cos8x sin4x dx

= 2 ∫ cos8x sin4x dx

Solve :  ∫ cos8x sin4x first.

Now, let u = 4x , dx = 1/4du

= 1/4 ∫ cos2u sinu du

Now, Solve : ∫ cos2u sinu du.

= ∫ cos2u sinu du

= ∫ - ( 1 - 2cos²u) sinu du

v = cos u.

du = -1/ sinu dv.

= - ∫ ( 2v² - 1 ) dv

Now solving, ∫ ( 2v² - 1 ) dv

= ∫ 2v² dv - ∫ 1 dv

= 2 ∫ v² dv - ∫ 1 dv

Now solving,

= ∫ v² dv

[ ∫ v^n dv = v ^ n + 1 / n+1 ]

Here n = 2 .

= v^3/3

Also, ∫ 1 dv = v.

So, 2 ∫ v² dv - ∫ 1 dv

= 2v³/3 - v

Now,

- ∫ ( 2v² - 1 ) dv

= v - 2v³/3

{ We have v = cosu }

= cosu - 2cos³u/3

Now,

1/4 ∫ cos2u sinu du

= 1/4 ( cosu - 2 cos³u / 3 )

= cosu /4 - cos³u/6

We know that, u = 4x .

= cos4x/4 - cos³4x/6

Now,

∫ 2 cos8x sin4x

= cos4x/2 - cos³4x/3

Finally,

$\therefore$ ∫ 2 cos8x sin4x dx = cos4x/2 - cos³4x/3 + C.

Hope helped !

Answer 2

$\bold{\int{2cos8x.sin4x}\,dx}$

we know, formula,
cos2Θ = 2cos²Θ - 1 , use it here
∴ cos8x = 2cos²4x -1
Now, ∫2(2cos²4x -1)sin4x.dx
= ∫(4cos²4x.sin4x - 2sin4x)dx
= ∫(4cos²4x.sin4x)dx - ∫(2sin4x)dx
=- ∫(cos4x)².{-4sin4x}dx - 2∫sin4x.dx

We know, one important thing,
∫f(x)ⁿf'(x)dx = f(x)ⁿ⁺¹/(n + 1) , use this concept in place of ∫(cos4x)²{-4sin4x}dx
Here if cos4x = f(x) then, -4sin4x = f'(x)
Then, ∫(cos4x)² {-4sin4x}dx = (cos4x)³/3

Hence, I = -(cos4x)³/3 + 2cos4x/4 + K [ ∵∫sinxdx = -cosx ]

So, answer is [ cos4x/2 - cos³4x/3 + K ] , here K is constant