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

The integral of (sin(x))/(cos(2x)) is 1/(2sqrt(2))(ln|sqrt(2)cos(x)+1|-ln|sqrt(2)cos(x)-1|)+C

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

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

\frac{1}{2 2} (ln 2 cos (x) + 1 - ln 2 cos (x) - 1) + C

Solution steps

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

Rewrite using trig identities

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integral Substitution

= - \int \frac{1}{2 ( - 1 + v ^{2} )} d v

Take the constant out:

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

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

Factor

- 1 + v^{2} : - (1 - v^{2})

= - \frac{1}{2} \cdot \int \frac{1}{- ( 1 - v ^{2} )} d v

Take the constant out:

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

= - \frac{1}{2} (- \int \frac{1}{1 - v ^{2}} d v)

Use the common integral:

\int \frac{1}{1 - v ^{2}} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= - \frac{1}{2} (- (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}))

Substitute back

= - \frac{1}{2} (- (\frac{ln 2 cos ( x ) + 1}{2} - \frac{ln 2 cos ( x ) - 1}{2}))

Simplify

- \frac{1}{2} (- (\frac{ln 2 cos ( x ) + 1}{2} - \frac{ln 2 cos ( x ) - 1}{2})) : \frac{1}{2 2} (ln 2 cos (x) + 1 - ln 2 cos (x) - 1)

= \frac{1}{2 2} (ln 2 cos (x) + 1 - ln 2 cos (x) - 1)

Add a constant to the solution

= \frac{1}{2 2} (ln 2 cos (x) + 1 - ln 2 cos (x) - 1) + C

e^{t^{2}} y^{^{'}} + (t + t y^{2}) = 0

(7 x^{2} + 2) (6 x + x)

y^{^{'}} = - 0.000631594 \cdot y

x \to 1 + lim (\frac{4 x - 1}{x - 1})

f (x) = 2 cos (x)

What is the integral of (sin(x))/(cos(2x)) ?
The integral of (sin(x))/(cos(2x)) is 1/(2sqrt(2))(ln|sqrt(2)cos(x)+1|-ln|sqrt(2)cos(x)-1|)+C