Anonymous

# Evaluate the indefinite integral: ∫ cos(θ) sin^6(θ) d(θ)?

Relevance

I'll use the | symbol to represent the integral symbol.

I'll use x for theta

Let's do a u-substitution:

u = sin(x)

du = cos(x) dx

dx = du/cosx

|cosx(u^6)du/cosx the cosx cancels out

|u^6 du

(1/7)u^7 + C replace u with sinx

(1/7)(sin^7x) + C

voila

• Anonymous

ok, the method that u are going to use for this problem is called a u substitution.

∫ cos(θ) sin^6(θ) d(θ)

first let u = sin (θ)

implicitly differentiate u to with respect to θ

d/dθ [ u = sin (θ) ]

du / dθ = cos (θ)

now cross multiply.

du = cos (θ) dθ

so u = sin (θ) and du = cos (θ) dθ. plug in u for sin (θ) and du for cos (θ) dθ. notice that cos (θ) dθ is conviently a part of the integrand

∫ cos (θ) sin^6(θ) dθ

= ∫ [ sin (θ) ]^6 [ cos (θ) dθ ]

the sin (θ) inside the bracket is u. the cos (θ) dθ inside the other bracket is du

= ∫ u^6 du

= (u^7 / 7) + C

now remember when we let u = sin θ? now we have to plug sin θ back in for u

= (sin^7(θ) / 7) + C

and theres the answer. the hardest part of a u substitution is figuring out which substitution to make. wen u get really good at u substitutions u can do the problem in your head. like before i even did this problem, i already knew the answer was going to be sin^7(θ) / 7 + C.

• Anonymous

You substitute a variable for the trig function that has the higher power usually. That is the common method in dealing with those integrals

z = sin(t)

dz = cos(t) dt

INT z^6 dz = z^7/ 7 = sin^7(t) /7 + C

Hello ,

If u look well , u can realize that ∫sin(θ) is already existed but without

(-)

so :

∫ cos(θ) sin^6(θ) d(θ) = - ∫ sin^6(θ) *(-cos(θ) d(θ)) = - (1/7)sin^7(θ)+C

Good Luck

• 3 years ago

this indefinite fundamental could be solved utilising u substitution. i'm utilising x as theta reason i don't have that image set u=sinx then Du= cosx u get the fundamental of a million/u^2 Du then carry u to the coolest making it the fundamental of u^-2 Du utilising integration rules the fundamental is u^-a million/-a million then in basic terms substitute sinx for u and you get -(a million/sinx)

• Como
Lv 7

Let u = sin Ө

du = cos Ө dӨ

I = ∫ u^6 du

I = u^7 / 7 + C

I = ( sin Ө )^7 / 7 + C

∫ cos(θ) sin^6(θ) d(θ)?

let u = sin (θ)

=> du = cos(θ) d(θ)

∫ cos(θ) sin^6(θ) d(θ)

= ∫ sin^6(θ) cos(θ)d(θ)

= ∫ u^6 du

= u^7 /7

= sin^7(θ) / 7 + C

QED