What is the integral of 5tan^3(sin(x))cos(x) ?

The integral of 5tan^3(sin(x))cos(x) is 5(-ln|sec(sin(x))|+(sec^2(sin(x)))/2)+C

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integral of 5tan^3(sin(x))cos(x)

\int 5 tan^{3} (sin (x)) cos (x) d x

5 (- ln ∣ sec (sin (x)) ∣ + \frac{sec ^{2} ( sin ( x ) )}{2}) + C

Solution steps

\int 5 tan^{3} (sin (x)) cos (x) d x

Take the constant out:

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

= 5 \cdot \int tan^{3} (sin (x)) cos (x) d x

Apply u-substitution

: \int tan^{3} (u) d u

= 5 \cdot \int tan^{3} (u) d u

Rewrite using trig identities

= 5 \cdot \int (- 1 + sec^{2} (u)) tan (u) d u

Apply u-substitution

: \int \frac{- 1 + v ^{2}}{v} d v

= 5 \cdot \int \frac{- 1 + v ^{2}}{v} d v

Expand

\frac{- 1 + v ^{2}}{v} : - \frac{1}{v} + v

= 5 \cdot \int - \frac{1}{v} + v 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

= 5 (- \int \frac{1}{v} d v + \int v d v)

\int \frac{1}{v} d v = ln ∣ v ∣

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

= 5 (- ln ∣ v ∣ + \frac{v ^{2}}{2})

Substitute back

= 5 (- ln ∣ sec (sin (x)) ∣ + \frac{sec ^{2} ( sin ( x ) )}{2})

Add a constant to the solution

= 5 (- ln ∣ sec (sin (x)) ∣ + \frac{sec ^{2} ( sin ( x ) )}{2}) + C

- x^{2}, [0, 2]

\frac{\partial}{\partial x} (tan (x^{2}))

\frac{\partial}{\partial x} (x + t)

\frac{d y}{d x} + 3 y = 40, y (1) = 70

x \to (7) + lim (\frac{1}{( x - 7 ) ^{2}})

What is the integral of 5tan^3(sin(x))cos(x) ?
The integral of 5tan^3(sin(x))cos(x) is 5(-ln|sec(sin(x))|+(sec^2(sin(x)))/2)+C