# Evaluate the Integral

Discussion in 'Calculus' started by iwntchckknnggts, Nov 28, 2021.

1. ### iwntchckknnggts

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i need help

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iwntchckknnggts, Nov 28, 2021
2. ### MathLover1

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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)

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

MathLover1, Nov 28, 2021
nycmathguy and iwntchckknnggts like this.
3. ### iwntchckknnggts

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THANK YOU SO MUCH

iwntchckknnggts, Nov 28, 2021
4. ### nycmathguy

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

nycmathguy, Nov 28, 2021
5. ### MathLover1

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

MathLover1, Nov 28, 2021
nycmathguy likes this.

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