What is the integral of (sin(a)x)/(cos(a)x+b) ?

The integral of (sin(a)x)/(cos(a)x+b) is sec^2(a)sin(a)(xcos(a)+b-bln|xcos(a)+b|)+C

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integral of (sin(a)x)/(cos(a)x+b)

\int \frac{sin ( a ) x}{cos ( a ) x + b} d x

sec^{2} (a) sin (a) (x cos (a) + b - b ln ∣ x cos (a) + b ∣) + C

Solution steps

\int \frac{sin ( a ) x}{cos ( a ) x + b} d x

Take the constant out:

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

= sin (a) \cdot \int \frac{x}{cos ( a ) x + b} d x

Apply u-substitution

= sin (a) \cdot \int \frac{u - b}{u cos ^{2} ( a )} d u

Take the constant out:

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

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

Expand

\frac{u - b}{u} : 1 - \frac{b}{u}

= sin (a) \frac{1}{cos ^{2} ( a )} \cdot \int 1 - \frac{b}{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

= sin (a) \frac{1}{cos ^{2} ( a )} (\int 1 d u - \int \frac{b}{u} d u)

\int 1 d u = u

\int \frac{b}{u} d u = b ln ∣ u ∣

= sin (a) \frac{1}{cos ^{2} ( a )} (u - b ln ∣ u ∣)

Substitute back

u = cos (a) x + b

= sin (a) \frac{1}{cos ^{2} ( a )} (cos (a) x + b - b ln ∣ cos (a) x + b ∣)

Use the basic trigonometric identity:

\frac{1}{cos ( x )} = sec (x)

= sec^{2} (a) sin (a) (x cos (a) + b - b ln ∣ x cos (a) + b ∣)

Add a constant to the solution

= sec^{2} (a) sin (a) (x cos (a) + b - b ln ∣ x cos (a) + b ∣) + C

What is the integral of (sin(a)x)/(cos(a)x+b) ?
The integral of (sin(a)x)/(cos(a)x+b) is sec^2(a)sin(a)(xcos(a)+b-bln|xcos(a)+b|)+C