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

The integral of (sin(x))/(cos^3(x)+cos(x)) is -ln|cos(x)|+1/2 ln|cos^2(x)+1|+C

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integral of (sin(x))/(cos^3(x)+cos(x))

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

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

Solution steps

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

Apply u-substitution

: \int - \frac{1}{u ^{3} + u} d u

= \int - \frac{1}{u ^{3} + u} d u

Take the constant out:

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

= - \int \frac{1}{u ^{3} + u} d u

Take the partial fraction of

\frac{1}{u ^{3} + u} : \frac{1}{u} - \frac{u}{u ^{2} + 1}

= - \int \frac{1}{u} - \frac{u}{u ^{2} + 1} d u

Apply the Sum Rule:

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

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

\int \frac{1}{u} d u = ln ∣ u ∣

\int \frac{u}{u ^{2} + 1} d u = \frac{1}{2} ln u^{2} + 1

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

Substitute back

u = cos (x)

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

Simplify

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

Add a constant to the solution

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

\frac{d ^{3}}{d x ^{3}} (x)

e^{4 x}

\int \frac{cos ( ln ( 4 x ^{2} ) )}{x} d x

6 y^{^{'}} = y^{2}, y (0) = 5

(2 x - 1)^{^{'}}

What is the integral of (sin(x))/(cos^3(x)+cos(x)) ?
The integral of (sin(x))/(cos^3(x)+cos(x)) is -ln|cos(x)|+1/2 ln|cos^2(x)+1|+C