What is the integral of 15sec^2(x)tan^4(x) ?

The integral of 15sec^2(x)tan^4(x) is 3tan^5(x)+C

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integral of 15sec^2(x)tan^4(x)

\int 15 sec^{2} (x) tan^{4} (x) d x

3 tan^{5} (x) + C

Solution steps

\int 15 sec^{2} (x) tan^{4} (x) d x

Take the constant out:

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

= 15 \cdot \int sec^{2} (x) tan^{4} (x) d x

Rewrite using trig identities

= 15 \cdot \int (1 + tan^{2} (x)) tan^{4} (x) d x

Apply u-substitution

= 15 \cdot \int u^{4} d u

Apply the Power Rule

= 15 \cdot \frac{u ^{5}}{5}

Substitute back

u = tan (x)

= 15 \cdot \frac{tan ^{5} ( x )}{5}

Simplify

15 \cdot \frac{tan ^{5} ( x )}{5} : 3 tan^{5} (x)

= 3 tan^{5} (x)

Add a constant to the solution

= 3 tan^{5} (x) + C

What is the integral of 15sec^2(x)tan^4(x) ?
The integral of 15sec^2(x)tan^4(x) is 3tan^5(x)+C