What is the integral of (sin(4x))/(sin(2x)cos^2(2x)) ?

The integral of (sin(4x))/(sin(2x)cos^2(2x)) is ln|tan(2x)+sec(2x)|+C

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integral of (sin(4x))/(sin(2x)cos^2(2x))

\int \frac{sin ( 4 x )}{sin ( 2 x ) cos ^{2} ( 2 x )} d x

ln ∣ tan (2 x) + sec (2 x) ∣ + C

Solution steps

\int \frac{sin ( 4 x )}{sin ( 2 x ) cos ^{2} ( 2 x )} d x

Apply u-substitution

: \int \frac{sin ( 2 u )}{2 cos ^{2} ( u ) sin ( u )} d u

= \int \frac{sin ( 2 u )}{2 cos ^{2} ( u ) sin ( u )} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{sin ( 2 u )}{cos ^{2} ( u ) sin ( u )} d u

Rewrite using trig identities

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

Take the constant out:

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

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

Rewrite using trig identities

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

Use the common integral:

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

= \frac{1}{2} \cdot 2 ln ∣ tan (u) + sec (u) ∣

Substitute back

u = 2 x

= \frac{1}{2} \cdot 2 ln ∣ tan (2 x) + sec (2 x) ∣

Simplify

\frac{1}{2} \cdot 2 ln ∣ tan (2 x) + sec (2 x) ∣ : ln ∣ tan (2 x) + sec (2 x) ∣

= ln ∣ tan (2 x) + sec (2 x) ∣

Add a constant to the solution

= ln ∣ tan (2 x) + sec (2 x) ∣ + C

\int (\frac{1}{x - 5}) d x

\int 2 sec^{6} (10 x) d x

\int b 1 - \frac{x ^{2}}{a ^{2}} d x

\int 4 sin (5 x + 7) d x

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

What is the integral of (sin(4x))/(sin(2x)cos^2(2x)) ?
The integral of (sin(4x))/(sin(2x)cos^2(2x)) is ln|tan(2x)+sec(2x)|+C