What is the integral of (cos^4(x))/(sin(x)) ?

The integral of (cos^4(x))/(sin(x)) is 1/3 cos^3(x)+cos(x)-1/2 ln|cos(x)+1|+1/2 ln|cos(x)-1|+C

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integral of (cos^4(x))/(sin(x))

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

\frac{1}{3} cos^{3} (x) + cos (x) - \frac{1}{2} ln ∣ cos (x) + 1 ∣ + \frac{1}{2} ln ∣ cos (x) - 1 ∣ + C

Solution steps

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

Apply Trigonometric Substitution

: \int - \frac{u ^{4}}{1 - u ^{2}} d u

= \int - \frac{u ^{4}}{1 - u ^{2}} d u

Take the constant out:

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

= - \int \frac{u ^{4}}{1 - u ^{2}} d u

Long division

\frac{u ^{4}}{1 - u ^{2}} : - u^{2} - 1 + \frac{1}{- u ^{2} + 1}

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

\int u^{2} d u = \frac{u ^{3}}{3}

\int 1 d u = u

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

= - (- \frac{u ^{3}}{3} - u + \frac{ln ∣ u + 1 ∣}{2} - \frac{ln ∣ u - 1 ∣}{2})

Substitute back

u = cos (x)

= - (- \frac{cos ^{3} ( x )}{3} - cos (x) + \frac{ln ∣ cos ( x ) + 1 ∣}{2} - \frac{ln ∣ cos ( x ) - 1 ∣}{2})

Simplify

- (- \frac{cos ^{3} ( x )}{3} - cos (x) + \frac{ln ∣ cos ( x ) + 1 ∣}{2} - \frac{ln ∣ cos ( x ) - 1 ∣}{2}) : \frac{1}{3} cos^{3} (x) + cos (x) - \frac{1}{2} ln ∣ cos (x) + 1 ∣ + \frac{1}{2} ln ∣ cos (x) - 1 ∣

= \frac{1}{3} cos^{3} (x) + cos (x) - \frac{1}{2} ln ∣ cos (x) + 1 ∣ + \frac{1}{2} ln ∣ cos (x) - 1 ∣

Add a constant to the solution

= \frac{1}{3} cos^{3} (x) + cos (x) - \frac{1}{2} ln ∣ cos (x) + 1 ∣ + \frac{1}{2} ln ∣ cos (x) - 1 ∣ + C

y^{^{'}} = x e^{x^{2} - 4}

y = ln (x^{3} + 4)

\frac{\partial}{\partial x} (6 x - 2 y)

y^{^{'}} = 4 x + y, y (0) = 7

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = 1

What is the integral of (cos^4(x))/(sin(x)) ?
The integral of (cos^4(x))/(sin(x)) is 1/3 cos^3(x)+cos(x)-1/2 ln|cos(x)+1|+1/2 ln|cos(x)-1|+C