What is the integral of 16sec^2(x)tan^3(x) ?

The integral of 16sec^2(x)tan^3(x) is 4tan^4(x)+C

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integral of 16sec^2(x)tan^3(x)

\int 16 sec^{2} (x) tan^{3} (x) d x

4 tan^{4} (x) + C

Solution steps

\int 16 sec^{2} (x) tan^{3} (x) d x

Take the constant out:

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

= 16 \cdot \int sec^{2} (x) tan^{3} (x) d x

Rewrite using trig identities

= 16 \cdot \int (1 + tan^{2} (x)) tan^{3} (x) d x

Apply u-substitution

: \int u^{3} d u

= 16 \cdot \int u^{3} d u

Apply the Power Rule

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

= 16 \cdot \frac{u ^{4}}{4}

Substitute back

u = tan (x)

= 16 \cdot \frac{tan ^{4} ( x )}{4}

Simplify

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

= 4 tan^{4} (x)

Add a constant to the solution

= 4 tan^{4} (x) + C

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What is the integral of 16sec^2(x)tan^3(x) ?
The integral of 16sec^2(x)tan^3(x) is 4tan^4(x)+C