What is the integral of (50x)/(sec(5x^2+5)) ?

The integral of (50x)/(sec(5x^2+5)) is 5sin(5x^2+5)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (50x)/(sec(5x^2+5))

\int \frac{50 x}{sec ( 5 x ^{2} + 5 )} d x

5 sin (5 x^{2} + 5) + C

Solution steps

\int \frac{50 x}{sec ( 5 x ^{2} + 5 )} d x

Take the constant out:

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

= 50 \cdot \int \frac{x}{sec ( 5 x ^{2} + 5 )} d x

Apply u-substitution

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

= 50 \cdot \int \frac{1}{2 sec ( 5 u + 5 )} d u

Take the constant out:

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

= 50 \cdot \frac{1}{2} \cdot \int \frac{1}{sec ( 5 u + 5 )} d u

Use the basic trigonometric identity:

\frac{1}{sec ( x )} = cos (x)

= 50 \cdot \frac{1}{2} \cdot \int cos (5 u + 5) d u

Apply u-substitution

: \int cos (v) \frac{1}{5} d v

= 50 \cdot \frac{1}{2} \cdot \int cos (v) \frac{1}{5} d v

Take the constant out:

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

= 50 \cdot \frac{1}{2} \cdot \frac{1}{5} \cdot \int cos (v) d v

Use the common integral:

\int cos (v) d v = sin (v)

= 50 \cdot \frac{1}{2} \cdot \frac{1}{5} sin (v)

Substitute back

= 50 \cdot \frac{1}{2} \cdot \frac{1}{5} sin (5 x^{2} + 5)

Simplify

50 \cdot \frac{1}{2} \cdot \frac{1}{5} sin (5 x^{2} + 5) : 5 sin (5 x^{2} + 5)

= 5 sin (5 x^{2} + 5)

Add a constant to the solution

= 5 sin (5 x^{2} + 5) + C

x \to 0 lim (\frac{( 8 ^{x} - 5 ^{x} )}{x})

\frac{1}{e ^{x^{2}}}

y^{^{'}} = x y^{3} - x y

x^{5} - x^{3} + x, 1

\frac{d y}{d x} = cos (3 x)

What is the integral of (50x)/(sec(5x^2+5)) ?
The integral of (50x)/(sec(5x^2+5)) is 5sin(5x^2+5)+C