What is the integral of (cos(x)ln(cos(x)))/(csc(x)) ?

The integral of (cos(x)ln(cos(x)))/(csc(x)) is -1/2 cos^2(x)ln(cos(x))+1/4 cos^2(x)+C

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integral of (cos(x)ln(cos(x)))/(csc(x))

\int \frac{cos ( x ) ln ( cos ( x ) )}{csc ( x )} d x

- \frac{1}{2} cos^{2} (x) ln (cos (x)) + \frac{1}{4} cos^{2} (x) + C

Solution steps

\int \frac{cos ( x ) ln ( cos ( x ) )}{csc ( x )} d x

Apply u-substitution

: \int - u ln (u) d u

= \int - u ln (u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int u ln (u) d u

Apply Integration By Parts

: \frac{1}{2} u^{2} ln (u) - \int \frac{u}{2} d u

= - (\frac{1}{2} u^{2} ln (u) - \int \frac{u}{2} d u)

\int \frac{u}{2} d u = \frac{u ^{2}}{4}

= - (\frac{1}{2} u^{2} ln (u) - \frac{u ^{2}}{4})

Substitute back

u = cos (x)

= - (\frac{1}{2} cos^{2} (x) ln (cos (x)) - \frac{cos ^{2} ( x )}{4})

Simplify

- (\frac{1}{2} cos^{2} (x) ln (cos (x)) - \frac{cos ^{2} ( x )}{4}) : - \frac{1}{2} cos^{2} (x) ln (cos (x)) + \frac{1}{4} cos^{2} (x)

= - \frac{1}{2} cos^{2} (x) ln (cos (x)) + \frac{1}{4} cos^{2} (x)

Add a constant to the solution

= - \frac{1}{2} cos^{2} (x) ln (cos (x)) + \frac{1}{4} cos^{2} (x) + C

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What is the integral of (cos(x)ln(cos(x)))/(csc(x)) ?
The integral of (cos(x)ln(cos(x)))/(csc(x)) is -1/2 cos^2(x)ln(cos(x))+1/4 cos^2(x)+C