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

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

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

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

- \frac{1}{2} ln cos (2 x) + 2 cos (x) + 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 - 2 + 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}{- 2 + v ^{2}} d v

Apply Trigonometric Substitution

: \int sec (w) d w

= - \frac{1}{2} \cdot \int sec (w) d w

Use the common integral:

\int sec (w) d w = ln ∣ tan (w) + sec (w) ∣

= - \frac{1}{2} ln ∣ tan (w) + sec (w) ∣

Substitute back

= - \frac{1}{2} ln tan (arcsec (\frac{1}{2} \cdot 2 cos (x))) + sec (arcsec (\frac{1}{2} \cdot 2 cos (x)))

Simplify

- \frac{1}{2} ln tan (arcsec (\frac{1}{2} \cdot 2 cos (x))) + sec (arcsec (\frac{1}{2} \cdot 2 cos (x))) : - \frac{1}{2} ln cos (2 x) + 2 cos (x)

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

Add a constant to the solution

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

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