What is the integral of 16sec^2(x)tan^7(x) ?

The integral of 16sec^2(x)tan^7(x) is 2tan^8(x)+C

integral of 16sec^2(x)tan^7(x)

\int 16 sec^{2} (x) tan^{7} (x) d x

2 tan^{8} (x) + C

Solution steps

\int 16 sec^{2} (x) tan^{7} (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^{7} (x) d x

Rewrite using trig identities

= 16 \cdot \int (1 + tan^{2} (x)) tan^{7} (x) d x

Apply u-substitution

= 16 \cdot \int u^{7} d u

Apply the Power Rule

= 16 \cdot \frac{u ^{8}}{8}

Substitute back

u = tan (x)

= 16 \cdot \frac{tan ^{8} ( x )}{8}

Simplify

16 \cdot \frac{tan ^{8} ( x )}{8} : 2 tan^{8} (x)

= 2 tan^{8} (x)

Add a constant to the solution

= 2 tan^{8} (x) + C

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What is the integral of 16sec^2(x)tan^7(x) ?
The integral of 16sec^2(x)tan^7(x) is 2tan^8(x)+C