how to solve integrals of trig functions?

∫ csc^2 4x cot 4x dx

Let u = cot 4x

then du = - csc^2 4x • 4 dx

so (-1/4) du = csc^2 x dx

(-1/4) ∫ u du = (-1/8) u^2 + C = (-1/8) cot^2 4x + C

∫ csc^2( 4x) cot (4x) dx

let cot(4x) = u

= - 4 csc^2(4x) dx = du

csc^2(4x) dx = - du/4

now the integral becomes

- 1/4 ∫ u du

= -(1/8 ) u^2 + C

back substitute u = cot(4x)

= - (1/8) cot^2(4x) + C

I = ∫ csc ² (4x) cot (4x) dx

Let u = cot (4x)

du/dx = - 4 csc ² (4x)

- du/4 = csc ² (4x) dx

I = (-1/4) ∫ u du

I = (-1/8) u ² + C

I = (-1/8) cot ² (4x) + C

∫cosec² 4x cot² 4x dx

let u = cot 4x

so, du = - 4 cosec² 4x dx

∫cosec² 4x cot² 4x dx

=∫ (-u/4) du

= (-1/4) (u²/2) + c

= - [(cot² 4x)/8] +c

∫ csc²(4x) cot(4x) dx

Let u = cot(4x) then du = -4csc²(4x) dx

(-1/4) ∫ cot(4x) [-4csc²(4x) dx]

(-1/4) ∫ u du

(-1/4) u²/2

(-1/4) cot²(4x) / 2

(-1/8) cot²(4x) + C

a million. ?(t² - sint) dt = ?t²dt - ?sintdt = t³/3 + value + C 2. ?sinxdx/(a million - sin²x) = ?d(cosx)/cos²x = -a million/cosx + C 3. ? dx/?(x) = ?x^(-a million/4)dx = (4/3)x^(3/4) + C = (4/3)?(x³) + C

= ( - 1 / 4) ln abs ( sin 4x ) + C