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

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

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

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

2 tan^{2} (x) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int u d u

= 4 \cdot \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= 4 \cdot \frac{u ^{2}}{2}

Substitute back

u = tan (x)

= 4 \cdot \frac{tan ^{2} ( x )}{2}

Simplify

4 \cdot \frac{tan ^{2} ( x )}{2} : 2 tan^{2} (x)

= 2 tan^{2} (x)

Add a constant to the solution

= 2 tan^{2} (x) + C

f (x) = x^{3} - 2

\frac{d}{d x} (lo g_{4} (x^{2} - 7 x))

\int x^{2} + 9 d x

y^{^{'}} - 2 t y = 9 t^{2} e^{t^{2}}

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