What is the integral of (3cos(x))/(1+cos(x)) ?

The integral of (3cos(x))/(1+cos(x)) is 3(-csc(x)+cot(x)+x)+C

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integral of (3cos(x))/(1+cos(x))

\int \frac{3 cos ( x )}{1 + cos ( x )} d x

3 (- csc (x) + cot (x) + x) + C

Solution steps

\int \frac{3 cos ( x )}{1 + cos ( x )} d x

Take the constant out:

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

= 3 \cdot \int \frac{cos ( x )}{1 + cos ( x )} d x

Multiply by the conjugate of

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

= 3 \cdot \int \frac{cos ( x )}{\frac{s i n ^{2} ( x )}{1 - c o s ( x )}} d x

Expand

\frac{cos ( x )}{\frac{s i n ^{2} ( x )}{1 - c o s ( x )}} : \frac{cos ( x )}{sin ^{2} ( x )} - \frac{cos ^{2} ( x )}{sin ^{2} ( x )}

= 3 \cdot \int \frac{cos ( x )}{sin ^{2} ( x )} - \frac{cos ^{2} ( x )}{sin ^{2} ( x )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 (\int \frac{cos ( x )}{sin ^{2} ( x )} d x - \int \frac{cos ^{2} ( x )}{sin ^{2} ( x )} d x)

\int \frac{cos ( x )}{sin ^{2} ( x )} d x = - csc (x)

\int \frac{cos ^{2} ( x )}{sin ^{2} ( x )} d x = - cot (x) - x

= 3 (- csc (x) - (- cot (x) - x))

Simplify

= 3 (- csc (x) + cot (x) + x)

Add a constant to the solution

= 3 (- csc (x) + cot (x) + x) + C

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What is the integral of (3cos(x))/(1+cos(x)) ?
The integral of (3cos(x))/(1+cos(x)) is 3(-csc(x)+cot(x)+x)+C