What is the integral of tan^3(5x)sec^2(5x) ?

The integral of tan^3(5x)sec^2(5x) is 1/20 tan^4(5x)+C

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integral of tan^3(5x)sec^2(5x)

\int tan^{3} (5 x) sec^{2} (5 x) d x

\frac{1}{20} tan^{4} (5 x) + C

Solution steps

\int tan^{3} (5 x) sec^{2} (5 x) d x

Apply u-substitution

: \int tan^{3} (u) sec^{2} (u) \frac{1}{5} d u

= \int tan^{3} (u) sec^{2} (u) \frac{1}{5} d u

Take the constant out:

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

= \frac{1}{5} \cdot \int tan^{3} (u) sec^{2} (u) d u

Rewrite using trig identities

= \frac{1}{5} \cdot \int (1 + tan^{2} (u)) tan^{3} (u) d u

Apply u-substitution

: \int v^{3} d v

= \frac{1}{5} \cdot \int v^{3} d v

Apply the Power Rule

: \frac{v ^{4}}{4}

= \frac{1}{5} \cdot \frac{v ^{4}}{4}

Substitute back

= \frac{1}{5} \cdot \frac{tan ^{4} ( 5 x )}{4}

Simplify

\frac{1}{5} \cdot \frac{tan ^{4} ( 5 x )}{4} : \frac{1}{20} tan^{4} (5 x)

= \frac{1}{20} tan^{4} (5 x)

Add a constant to the solution

= \frac{1}{20} tan^{4} (5 x) + C

\frac{d}{d x} (\frac{1 + cos ( x )}{1 - cos ( x )})

\frac{d}{d x} (\frac{sin ( x )}{1 - cos ( x )})

\frac{d}{d θ} (sin (θ))

x \to - 5 lim (\frac{( x - 5 )}{x + 5})

8 t^{4}

What is the integral of tan^3(5x)sec^2(5x) ?
The integral of tan^3(5x)sec^2(5x) is 1/20 tan^4(5x)+C