Intergration?

Question

Do i use intergration by parts?

Evaltulate the intergral of sin^n (x) cos^3 (x) dx between x=0 & x=pie/2, where n is a non negaative interger.

Answer 1

No need to use integration by parts ... I will tell you a simpler method ....

INTEGRATION SIGN IS DENOTED BY ; int( )

Steps :

1 . Write the integral
2. Write cos(cube) x as cos^2 multiplied with cos x
3. Now split the cos^2 (x) as 1- sin^2 (x)
4. Take sin x = t
5. Differentiate both sides
6. find dx in terms of dt
7. You get a simple integral and solve it.
8. All steps are done below with answer.

int ( sin^n(x) cos^3x dx
= int ( sin^n(x) cos^2(x) cos (x) dx
= int ( sin^n(x) [ 1- sin^2(x) ] cos (x) dx

take sin x = t
cos x dx = dt

= int ( t^n [ 1 - t^2 ] dt
= int ( t^n ) + int ( t^ (n+2))

= [{t^(n+1)} / n+1 ] + [{ t^(n+3) } / n+3 ]

substituting t as sin x

= [{(sin x)^(n+1)} / n+1 ] + [{ (sinx )^(n+3) } / n+3 ]


Putting limits 0 to pie/2

= [ { (1)^ (n+1) } / n+1 ] + [ { (1) ^ (n+3) } / (n+3)] - 0

= [1/(n+1)] + [1/(n+3)] Answer

Answer 2

Use integration by parts with sin^n (x) as the function to be differentiated and set up a reduction formula, then all you need is to substitute a value for n and you can evaluate for any integer.

Alternatively use the substitution method by stripping out one cosine and converting the rest to sines and use u= sin x

Answer 3

No by substtution it can be done
because cos is having odd power
cos^3x = cos^2x.cos x
=(1-sin^2x) cos x
so number = sin^n(x)cosx - sin^(n+2) cos x
if sinx -=t cosxdx = dt
so we have
intgrate t^ndt -t^(n+2) dt


this is t^(n+1)/(n+1)-t^(n+3)/(n+3)

t = sin x
at lowe limit it is 0
at pi/2 is 1
so valulue = 1/(n+1) -1/(n+3) or 2/((n+1)(n+3))

Answer 4

yes we can use integration by parts