What is the solution for y^'= 1/(sec^2(x)) ?

The solution for y^'= 1/(sec^2(x)) is y= 1/2 (x+1/2 sin(2x))+c_{1}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

y^'= 1/(sec^2(x))

y^{^{'}} = \frac{1}{sec ^{2} ( x )}

y = \frac{1}{2} (x + \frac{1}{2} sin (2 x)) + c_{1}

Solution steps

y^{'} (x) = \frac{1}{sec ^{2} ( x )}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int \frac{1}{sec ^{2} ( x )} d x

\int \frac{1}{sec ^{2} ( x )} d x = \frac{1}{2} (x + \frac{1}{2} sin (2 x)) + c_{1}

y = \frac{1}{2} (x + \frac{1}{2} sin (2 x)) + c_{1}

\frac{d}{d x} (x^{2} + y^{2} + z^{2})

y = cos (x), y = 2 - cos (x), 0 \leq x \leq 2 π

\frac{d}{d y} (5 x + 4 y)

\int_{- 1}^{4} (2 x + 4)^{2} d x

\frac{d}{d x} (ln (\frac{2 x}{3}))

What is the solution for y^'= 1/(sec^2(x)) ?
The solution for y^'= 1/(sec^2(x)) is y= 1/2 (x+1/2 sin(2x))+c_{1}