What is the integral of (5cos^2(x))/(sin(x)) ?

The integral of (5cos^2(x))/(sin(x)) is 5(ln|tan(x/2)|+cos(x))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (5cos^2(x))/(sin(x))

\int \frac{5 cos ^{2} ( x )}{sin ( x )} d x

5 (ln tan (\frac{x}{2}) + cos (x)) + C

Solution steps

\int \frac{5 cos ^{2} ( x )}{sin ( x )} d x

Take the constant out:

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

= 5 \cdot \int \frac{cos ^{2} ( x )}{sin ( x )} d x

Rewrite using trig identities

= 5 \cdot \int \frac{1 - sin ^{2} ( x )}{sin ( x )} d x

Expand

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

= 5 \cdot \int \frac{1}{sin ( x )} - sin (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

= 5 (\int \frac{1}{sin ( x )} d x - \int sin (x) d x)

\int \frac{1}{sin ( x )} d x = ln tan (\frac{x}{2})

\int sin (x) d x = - cos (x)

= 5 (ln tan (\frac{x}{2}) - (- cos (x)))

Simplify

= 5 (ln tan (\frac{x}{2}) + cos (x))

Add a constant to the solution

= 5 (ln tan (\frac{x}{2}) + cos (x)) + C

t an g e n t y = \frac{8 x}{( x ^{2} + 1 )}, at

\int sec (6 x) tan (6 x) d x

\frac{\partial}{\partial x} (\frac{1}{1 + x ^{2}})

\frac{d}{d x} (\frac{1}{16} e^{2 x^{2}})

a re a y = 2 x + 2, y = x^{2} + 2, x = - 1, x = 3

What is the integral of (5cos^2(x))/(sin(x)) ?
The integral of (5cos^2(x))/(sin(x)) is 5(ln|tan(x/2)|+cos(x))+C