Evaluate the Integral

[iwntchckknnggts]

i need help

 

[MathLover1]

int(sin(x)*ln(1+sin(x))dx) (I am using int() instead symbol ∫)

apply integration by parts:

Using the product rule, we have that

int(udv)=uv-int(vdu)

let u=ln(1+sin(x)) and v=sin(x)dx

Then du=(ln(sin(x)+1))′dx=(cos(x)/(sin(x)+1))dx
and

v=int(sin(x)dx)=-cos(x)

The integral can be rewritten as

int(ln(sin(x)+1)sin(x)dx)
=(ln(sin(x)+1)*(-cos(x))-int(-cos(x))*(cos(x)/(sin(x)+1))dx)
=(-ln(sin(x)+1)cos(x)-int((-cos^2(x)/(sin(x)+1))dx)

Rewrite the cosine in terms of the sine, rewrite the numerator further, use the formula for difference of squares, and simplify:

-ln(sin(x)+1)cos(x)- int((-cos^2(x)/(sin(x)+1))dx)
=-ln(sin(x)+1)cos(x)- int((sin(x)-1)dx)

Integrate term by term:

-ln(sin(x)+1)cos(x)- int((sin(x)-1)dx)
=-ln(sin(x)+1)cos(x)-(-int(1dx)+int(sin(x)dx)

Apply the constant rule int(c)dx=cx with c=1:

-ln(sin(x)+1)cos(x)-int(sin(x)dx)+int(1dx)
=-ln(sin(x)+1)cos(x)-int(sin(x)dx)+x

The integral of the sine is int(sin(x)dx)=-cos(x):

x-ln(sin(x)+1)cos(x)-int(sin(x)dx)=x-ln(sin(x)+1)cos(x)-(-cos(x))

Therefore,

int(ln(sin(x)+1)sin(x)dx)=x-ln(sin(x)+1)cos(x)+cos(x)

Add the constant of integration:

int(ln(sin(x)+1)sin(x)dx)=x-ln(sin(x)+1)cos(x)+cos(x)+C
Answer:
int(ln(sin(x)+1)sin(x)dx)=x-ln(sin(x)+1)cos(x)+cos(x)+C

 

[iwntchckknnggts]

int(sin(x)*ln(1+sin(x))dx) (I am using int() instead symbol ∫)

apply integration by parts:

Using the product rule, we have that

int(udv)=uv-int(vdu)

let u=ln(1+sin(x)) and v=sin(x)dx

Then du=(ln(sin(x)+1))′dx=(cos(x)/(sin(x)+1))dx
and

v=int(sin(x)dx)=-cos(x)

The integral can be rewritten as

int(ln(sin(x)+1)sin(x)dx)
=(ln(sin(x)+1)*(-cos(x))-int(-cos(x))*(cos(x)/(sin(x)+1))dx)
=(-ln(sin(x)+1)cos(x)-int((-cos^2(x)/(sin(x)+1))dx)

Rewrite the cosine in terms of the sine, rewrite the numerator further, use the formula for difference of squares, and simplify:

-ln(sin(x)+1)cos(x)- int((-cos^2(x)/(sin(x)+1))dx)
=-ln(sin(x)+1)cos(x)- int((sin(x)-1)dx)

Integrate term by term:

-ln(sin(x)+1)cos(x)- int((sin(x)-1)dx)
=-ln(sin(x)+1)cos(x)-(-int(1dx)+int(sin(x)dx)

Apply the constant rule int(c)dx=cx with c=1:

-ln(sin(x)+1)cos(x)-int(sin(x)dx)+int(1dx)
=-ln(sin(x)+1)cos(x)-int(sin(x)dx)+x

The integral of the sine is int(sin(x)dx)=-cos(x):

x-ln(sin(x)+1)cos(x)-int(sin(x)dx)=x-ln(sin(x)+1)cos(x)-(-cos(x))

Therefore,

int(ln(sin(x)+1)sin(x)dx)=x-ln(sin(x)+1)cos(x)+cos(x)

Add the constant of integration:

int(ln(sin(x)+1)sin(x)dx)=x-ln(sin(x)+1)cos(x)+cos(x)+C
Answer:
int(ln(sin(x)+1)sin(x)dx)=x-ln(sin(x)+1)cos(x)+cos(x)+C

THANK YOU SO MUCH

 

[nycmathguy]

int(sin(x)*ln(1+sin(x))dx) (I am using int() instead symbol ∫)

apply integration by parts:

Using the product rule, we have that

int(udv)=uv-int(vdu)

let u=ln(1+sin(x)) and v=sin(x)dx

Then du=(ln(sin(x)+1))′dx=(cos(x)/(sin(x)+1))dx
and

v=int(sin(x)dx)=-cos(x)

The integral can be rewritten as

int(ln(sin(x)+1)sin(x)dx)
=(ln(sin(x)+1)*(-cos(x))-int(-cos(x))*(cos(x)/(sin(x)+1))dx)
=(-ln(sin(x)+1)cos(x)-int((-cos^2(x)/(sin(x)+1))dx)

Rewrite the cosine in terms of the sine, rewrite the numerator further, use the formula for difference of squares, and simplify:

-ln(sin(x)+1)cos(x)- int((-cos^2(x)/(sin(x)+1))dx)
=-ln(sin(x)+1)cos(x)- int((sin(x)-1)dx)

Integrate term by term:

-ln(sin(x)+1)cos(x)- int((sin(x)-1)dx)
=-ln(sin(x)+1)cos(x)-(-int(1dx)+int(sin(x)dx)

Apply the constant rule int(c)dx=cx with c=1:

-ln(sin(x)+1)cos(x)-int(sin(x)dx)+int(1dx)
=-ln(sin(x)+1)cos(x)-int(sin(x)dx)+x

The integral of the sine is int(sin(x)dx)=-cos(x):

x-ln(sin(x)+1)cos(x)-int(sin(x)dx)=x-ln(sin(x)+1)cos(x)-(-cos(x))

Therefore,

int(ln(sin(x)+1)sin(x)dx)=x-ln(sin(x)+1)cos(x)+cos(x)

Add the constant of integration:

int(ln(sin(x)+1)sin(x)dx)=x-ln(sin(x)+1)cos(x)+cos(x)+C
Answer:
int(ln(sin(x)+1)sin(x)dx)=x-ln(sin(x)+1)cos(x)+cos(x)+C

Can you please use the math symbols method for easy reading? You know, the wolfram stuff.

 

[MathLover1]

Can you please use the math symbols method for easy reading? You know, the wolfram stuff.

I was trying, for some reason it didn't work

 

[nycmathguy]

I was trying, for some reason it didn't work

Ok. Thanks anyway.