What is the integral of (cot(bx)+tan(bx))^2 ?

The integral of (cot(bx)+tan(bx))^2 is 1/b (-cot(bx)+tan(bx))+C

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integral of (cot(bx)+tan(bx))^2

\int (cot (b x) + tan (b x))^{2} d x

\frac{1}{b} (- cot (b x) + tan (b x)) + C

Solution steps

\int (cot (b x) + tan (b x))^{2} d x

Apply u-substitution

= \int \frac{( cot ( u ) + tan ( u ) ) ^{2}}{b} d u

Take the constant out:

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

= \frac{1}{b} \cdot \int (cot (u) + tan (u))^{2} d u

Apply u-substitution

= \frac{1}{b} \cdot \int \frac{v ^{2} + 1}{v ^{2}} d v

\frac{v ^{2} + 1}{v ^{2}} = \frac{1}{v ^{2}} + 1

= \frac{1}{b} \cdot \int \frac{1}{v ^{2}} + 1 d v

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{1}{v ^{2}} = v^{- 2}

= \frac{1}{b} \cdot \int v^{- 2} + 1 d v

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}{b} (\int v^{- 2} d v + \int 1 d v)

\int v^{- 2} d v = - \frac{1}{v}

\int 1 d v = v

= \frac{1}{b} (- \frac{1}{v} + v)

Substitute back

= \frac{1}{b} (- \frac{1}{tan ( b x )} + tan (b x))

Use the basic trigonometric identity:

\frac{1}{tan ( x )} = cot (x)

= \frac{1}{b} (- cot (b x) + tan (b x))

Add a constant to the solution

= \frac{1}{b} (- cot (b x) + tan (b x)) + C

What is the integral of (cot(bx)+tan(bx))^2 ?
The integral of (cot(bx)+tan(bx))^2 is 1/b (-cot(bx)+tan(bx))+C