What is the integral of 4tan^3(x)sec^2(x) ?

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

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integral of 4tan^3(x)sec^2(x)

\int 4 tan^{3} (x) sec^{2} (x) d x

tan^{4} (x) + C

Solution steps

\int 4 tan^{3} (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^{3} (x) sec^{2} (x) d x

Rewrite using trig identities

= 4 \cdot \int (1 + tan^{2} (x)) tan^{3} (x) d x

Apply u-substitution

: \int u^{3} d u

= 4 \cdot \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

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

Substitute back

u = tan (x)

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

Simplify

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

= tan^{4} (x)

Add a constant to the solution

= tan^{4} (x) + C

\frac{d}{d x} ((x^{2} + 1) arctan (x))

\int \frac{1}{x ^{2}} - \frac{4}{x ^{3}} d x

x \to 0 - lim (\frac{1}{x ^{3}})

x \to 0 - lim (x^{x})

\int \frac{3 x ^{5} - 2 x ^{3} + 3}{x ^{2}} d x

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