What is the integral of (sin^2(3x))/(cos(3x)) ?

The integral of (sin^2(3x))/(cos(3x)) is 1/3 ln|tan(3x)+sec(3x)|-1/3 sin(3x)+C

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integral of (sin^2(3x))/(cos(3x))

\int \frac{sin ^{2} ( 3 x )}{cos ( 3 x )} d x

\frac{1}{3} ln ∣ tan (3 x) + sec (3 x) ∣ - \frac{1}{3} sin (3 x) + C

Solution steps

\int \frac{sin ^{2} ( 3 x )}{cos ( 3 x )} d x

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

= \int sin (3 x) tan (3 x) d x

Apply Integration By Parts

: - \int - sec^{2} (3 x) cos (3 x) d x - \frac{1}{3} sin (3 x)

= - \int - sec^{2} (3 x) cos (3 x) d x - \frac{1}{3} sin (3 x)

\int - sec^{2} (3 x) cos (3 x) d x = - \frac{1}{3} ln ∣ tan (3 x) + sec (3 x) ∣

= - (- \frac{1}{3} ln ∣ tan (3 x) + sec (3 x) ∣) - \frac{1}{3} sin (3 x)

Simplify

= \frac{1}{3} ln ∣ tan (3 x) + sec (3 x) ∣ - \frac{1}{3} sin (3 x)

Add a constant to the solution

= \frac{1}{3} ln ∣ tan (3 x) + sec (3 x) ∣ - \frac{1}{3} sin (3 x) + C

\int \frac{1}{( 169 - x ^{2} ) ^{\frac{3}{2}}} d x

1 - x^{3}

6 x^{2}, x^{2} + 5

\int - \frac{1}{u} d u

\int \frac{( x + 4 ) ( x - 7 )}{x ^{2}} d x

What is the integral of (sin^2(3x))/(cos(3x)) ?
The integral of (sin^2(3x))/(cos(3x)) is 1/3 ln|tan(3x)+sec(3x)|-1/3 sin(3x)+C