What is the integral of 4cos^2(x)tan^3(x) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 4cos^2(x)tan^3(x)

\int 4 cos^{2} (x) tan^{3} (x) d x

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

Solution steps

\int 4 cos^{2} (x) tan^{3} (x) d x

Take the constant out:

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

= 4 \cdot \int cos^{2} (x) tan^{3} (x) d x

Rewrite using trig identities

= 4 \cdot \int \frac{( 1 - cos ^{2} ( x ) ) sin ( x )}{cos ( x )} d x

Apply u-substitution

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

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

Expand

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

= 4 \cdot \int - \frac{1}{u} + u 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

= 4 (- \int \frac{1}{u} d u + \int u d u)

\int \frac{1}{u} d u = ln ∣ u ∣

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

= 4 (- ln ∣ u ∣ + \frac{u ^{2}}{2})

Substitute back

u = cos (x)

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

Add a constant to the solution

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

\frac{\partial}{\partial x} ((x + y) \cdot e^{y})

y^{^{''}} - 6 y^{^{'}} + 9 y = e^{3 x} ln (x)

1 - 6 x

y = 3 x, y = \frac{3 x}{4}

\frac{d}{d x} (arcsec (3 x^{2} + 1))

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