Intergration by parts?

Question

How to integral In(sinx)/tanx dx by using intergration by parts?

Answer 1

i'd rather use substitution

∫ln(sinx) * (cosx/sinx) dx

let t = sinx

dt = cosx dx

substitute:

∫ln(t)/t dt

let u = ln(t)

du = 1/t dt

∫u du

(1/2) u² + C

(1/2) Ln²(t) + C

(1/2) Ln²(sinx) + C

Answer 2

int(In(sinx)/tanx dx)

int(In(sinx)cotx dx)

int(In(sinx)cosx/sinx dx)

why do you want to do this by parts, substitution is easier

u=ln(sinx)

du=cosx/sinx dx

int(In(sinx)cosx/sinx dx)

int(udu)

=u^2/2

=0.5(ln(sinx))^2

i will still do it by parts too

int(In(sinx)cotx dx)

u=ln(sinx)                  dv=cotx dx

du=cosx dx/sinx          v=ln(sinx)

du=cotx dx

int(In(sinx)cotx dx)=(ln(sinx))^2-int(ln(sinx)cotx dx)

int(In(sinx)cotx dx) +int(ln(sinx)cotx dx)=(ln(sinx))^2

2int(In(sinx)cotx dx)=(ln(sinx))^2

int(In(sinx)cotx dx)=0.5(ln(sinx))^2

Answer 3

∫ In(sinx) / tanx dx =

rewrite it as:

∫ In(sinx) / tanx dx =

∫  In(sinx) (1/ tanx) dx =

∫  In(sinx) (cotx) dx =

let:

ln(sinx) = u → (cosx/sinx) dx = du → cotx dx = du

cotx dx = dv → (cosx/sinx) dx = dv → d(sinx)/sinx = dv → ln(sinx) = v

thus, integrating by parts, you get:

∫  In(sinx) (cotx) dx = ln(sinx) ln(sinx) - ∫ ln(sinx) cotx dx →

having got the same unknown integral at both sides, collect it at left side as:

∫ In(sinx) (cotx) dx + ∫ ln(sinx) cotx dx = ln²(sinx) →

2 ∫ ln(sinx) cotx dx = ln²(sinx) →

∫ ln(sinx) cotx dx = (1/2) ln²(sinx) + C

in conclusion,

∫ [ln(sinx) / tanx] dx = ∫ ln(sinx) cotx dx = (1/2) ln²(sinx) + C

I hope it helps...

Bye!

Answer 4

The two prior answers are right - substitution is easier.  However, I'm guessing that your instructor requires you to solve this problem by IBP.  

Basic formula:

int(u dv) = uv - int(v du)

define:

u = ln(|sin(x)|) ==> du = cos(x)/sin(x) = 1/tan(x)

dv = 1/tan(x) dx ==> v = int(cot(x) dx) = ln|sin(x)|

Applying the formula yields:

int(ln|sin(x)|/tan(x) dx)

= (ln|sin(x)|)*(ln|sin(x)|) - int(ln|sin(x)|/tan(x))

= (ln|sin(x)|)^2 - int(ln|sin(x)|/tan(x))

=[(ln|sin(x)|)^2] / 2 + C

Don't forget than the natural logarithm must be positive, hence the |sin(x)| rather than just sin(x).

Answer 5

int In(sinx)/tanx dx=

int In(sinx)*cosxdx/sinx=

= int In(sinx)*d(sinx)/sinx=

=int In(sinx)*d[In(sinx)]=

=1/2*ln²(sinx)+c