What is the integral of (sin(3t))/(cos(3t-1)) ?

The integral of (sin(3t))/(cos(3t-1)) is 1/3 (sin(1)(3t-1)-cos(1)ln|cos(3t-1)|)+C

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integral of (sin(3t))/(cos(3t-1))

\int \frac{sin ( 3 t )}{cos ( 3 t - 1 )} d t

\frac{1}{3} (sin (1) (3 t - 1) - cos (1) ln ∣ cos (3 t - 1) ∣) + C

Solution steps

\int \frac{sin ( 3 t )}{cos ( 3 t - 1 )} d t

Apply u-substitution

: \int \frac{sin ( u + 1 )}{3 cos ( u )} d u

= \int \frac{sin ( u + 1 )}{3 cos ( u )} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{sin ( u + 1 )}{cos ( u )} d u

Rewrite using trig identities

= \frac{1}{3} \cdot \int \frac{cos ( u ) sin ( 1 ) + cos ( 1 ) sin ( u )}{cos ( u )} d u

Expand

\frac{cos ( u ) sin ( 1 ) + cos ( 1 ) sin ( u )}{cos ( u )} : sin (1) + \frac{cos ( 1 ) sin ( u )}{cos ( u )}

= \frac{1}{3} \cdot \int sin (1) + \frac{cos ( 1 ) sin ( u )}{cos ( u )} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3} (\int sin (1) d u + \int \frac{cos ( 1 ) sin ( u )}{cos ( u )} d u)

\int sin (1) d u = sin (1) u

\int \frac{cos ( 1 ) sin ( u )}{cos ( u )} d u = - cos (1) ln ∣ cos (u) ∣

= \frac{1}{3} (sin (1) u - cos (1) ln ∣ cos (u) ∣)

Substitute back

u = 3 t - 1

= \frac{1}{3} (sin (1) (3 t - 1) - cos (1) ln ∣ cos (3 t - 1) ∣)

Add a constant to the solution

= \frac{1}{3} (sin (1) (3 t - 1) - cos (1) ln ∣ cos (3 t - 1) ∣) + C

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What is the integral of (sin(3t))/(cos(3t-1)) ?
The integral of (sin(3t))/(cos(3t-1)) is 1/3 (sin(1)(3t-1)-cos(1)ln|cos(3t-1)|)+C