What is the integral of 4tan^2(x)sec^2(x) ?

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

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integral of 4tan^2(x)sec^2(x)

\int 4 tan^{2} (x) sec^{2} (x) d x

\frac{4}{3} tan^{3} (x) + C

Solution steps

\int 4 tan^{2} (x) sec^{2} (x) d x

Take the constant out:

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

= 4 \cdot \int tan^{2} (x) sec^{2} (x) d x

Rewrite using trig identities

= 4 \cdot \int (1 + tan^{2} (x)) tan^{2} (x) d x

Apply u-substitution

: \int u^{2} d u

= 4 \cdot \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= 4 \cdot \frac{u ^{3}}{3}

Substitute back

u = tan (x)

= 4 \cdot \frac{tan ^{3} ( x )}{3}

Simplify

4 \cdot \frac{tan ^{3} ( x )}{3} : \frac{4}{3} tan^{3} (x)

= \frac{4}{3} tan^{3} (x)

Add a constant to the solution

= \frac{4}{3} tan^{3} (x) + C

y = 3 x, y = \frac{5}{7} x, y = 54 - x^{2}

\int arcsin (\frac{1}{x}) d x

f (x) = 8.8 x

\frac{d}{d x} (e^{2 x} x (a cos (x) + b sin (x)))

y^{^{''}} + 3 y^{^{'}} - 4 y = - 8 e^{x} cos (2 x)

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