What is the integral of 1/(sqrt(cos(x)+cos^2(x))) ?

The integral of 1/(sqrt(cos(x)+cos^2(x))) is sqrt(2)arcsin(tan(x/2))+C

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integral of 1/(sqrt(cos(x)+cos^2(x)))

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

2 arcsin (tan (\frac{x}{2})) + C

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int \frac{1}{2} d v

= 2 \cdot \int \frac{1}{2} d v

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 2 \cdot \frac{1}{2} \cdot 1 \cdot v

Substitute back

= 2 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (\frac{2}{2} tan (\frac{x}{2}))

Simplify

2 \cdot \frac{1}{2} \cdot 1 \cdot arcsin (\frac{2}{2} tan (\frac{x}{2})) : 2 arcsin (tan (\frac{x}{2}))

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

Add a constant to the solution

= 2 arcsin (tan (\frac{x}{2})) + C

\frac{\partial}{\partial z} (ln (1 + x y) - z)

f (x) = x + 8

\int t ln (1 + t) d t

\int_{- 8}^{0} (\frac{y}{8} + y + 9) d y

x \to \infty lim (x + 9)

What is the integral of 1/(sqrt(cos(x)+cos^2(x))) ?
The integral of 1/(sqrt(cos(x)+cos^2(x))) is sqrt(2)arcsin(tan(x/2))+C