What is the integral of 5/(6+4sec(x)) ?

The integral of 5/(6+4sec(x)) is

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 5/(6+4sec(x))

\int \frac{5}{6 + 4 sec ( x )} d x

5 (- \frac{1}{3 5} ln tan (\frac{x}{2}) + 5 + \frac{1}{3 5} ln tan (\frac{x}{2}) - 5 + \frac{1}{3} arctan (tan (\frac{x}{2}))) + C

Solution steps

\int \frac{5}{6 + 4 sec ( 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{1}{6 + 4 sec ( x )} d x

Rewrite using trig identities

= 5 \cdot \int \frac{cos ( x )}{6 cos ( x ) + 4} d x

Apply u-substitution

: \int \frac{1 - u ^{2}}{( - u ^{2} + 5 ) ( u ^{2} + 1 )} d u

= 5 \cdot \int \frac{1 - u ^{2}}{( - u ^{2} + 5 ) ( u ^{2} + 1 )} d u

Take the partial fraction of

\frac{1 - u ^{2}}{( - u ^{2} + 5 ) ( u ^{2} + 1 )} : - \frac{1}{3 5 ( u + 5 )} + \frac{1}{3 5 ( u - 5 )} + \frac{1}{3 ( u ^{2} + 1 )}

= 5 \cdot \int - \frac{1}{3 5 ( u + 5 )} + \frac{1}{3 5 ( u - 5 )} + \frac{1}{3 ( 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

= 5 (- \int \frac{1}{3 5 ( u + 5 )} d u + \int \frac{1}{3 5 ( u - 5 )} d u + \int \frac{1}{3 ( u ^{2} + 1 )} d u)

\int \frac{1}{3 5 ( u + 5 )} d u = \frac{1}{3 5} ln u + 5

\int \frac{1}{3 5 ( u - 5 )} d u = \frac{1}{3 5} ln u - 5

\int \frac{1}{3 ( u ^{2} + 1 )} d u = \frac{1}{3} arctan (u)

= 5 (- \frac{1}{3 5} ln u + 5 + \frac{1}{3 5} ln u - 5 + \frac{1}{3} arctan (u))

Substitute back

u = tan (\frac{x}{2})

= 5 (- \frac{1}{3 5} ln tan (\frac{x}{2}) + 5 + \frac{1}{3 5} ln tan (\frac{x}{2}) - 5 + \frac{1}{3} arctan (tan (\frac{x}{2})))

Add a constant to the solution

= 5 (- \frac{1}{3 5} ln tan (\frac{x}{2}) + 5 + \frac{1}{3 5} ln tan (\frac{x}{2}) - 5 + \frac{1}{3} arctan (tan (\frac{x}{2}))) + C

\int e^{- x} - x e^{- x} d x

\int csc^{2} (5 x) cot^{5} (5 x) d x

\int \frac{1}{x ( 1 + 2 x )} d x

\int \frac{4 + t}{t ^{3}} d t

\int 9 sin^{3} (x) cos (x) d x