What is the integral of 14sec^2(x)tan^6(x) ?

The integral of 14sec^2(x)tan^6(x) is 2tan^7(x)+C

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integral of 14sec^2(x)tan^6(x)

\int 14 sec^{2} (x) tan^{6} (x) d x

2 tan^{7} (x) + C

Solution steps

\int 14 sec^{2} (x) tan^{6} (x) d x

Take the constant out:

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

= 14 \cdot \int sec^{2} (x) tan^{6} (x) d x

Rewrite using trig identities

= 14 \cdot \int (1 + tan^{2} (x)) tan^{6} (x) d x

Apply u-substitution

: \int u^{6} d u

= 14 \cdot \int u^{6} d u

Apply the Power Rule

: \frac{u ^{7}}{7}

= 14 \cdot \frac{u ^{7}}{7}

Substitute back

u = tan (x)

= 14 \cdot \frac{tan ^{7} ( x )}{7}

Simplify

14 \cdot \frac{tan ^{7} ( x )}{7} : 2 tan^{7} (x)

= 2 tan^{7} (x)

Add a constant to the solution

= 2 tan^{7} (x) + C

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What is the integral of 14sec^2(x)tan^6(x) ?
The integral of 14sec^2(x)tan^6(x) is 2tan^7(x)+C