What is the integral of 20sec(5x)tan(5x)sec^3(5x) ?

The integral of 20sec(5x)tan(5x)sec^3(5x) is sec^4(5x)+C

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integral of 20sec(5x)tan(5x)sec^3(5x)

\int 20 sec (5 x) tan (5 x) sec^{3} (5 x) d x

sec^{4} (5 x) + C

Solution steps

\int 20 sec (5 x) tan (5 x) sec^{3} (5 x) d x

Take the constant out:

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

= 20 \cdot \int sec (5 x) tan (5 x) sec^{3} (5 x) d x

Simplify

sec (5 x) tan (5 x) sec^{3} (5 x) : sec^{4} (5 x) tan (5 x)

= 20 \cdot \int sec^{4} (5 x) tan (5 x) d x

Apply u-substitution

: \int sec^{4} (u) tan (u) \frac{1}{5} d u

= 20 \cdot \int sec^{4} (u) tan (u) \frac{1}{5} d u

Take the constant out:

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

= 20 \cdot \frac{1}{5} \cdot \int sec^{4} (u) tan (u) d u

Apply u-substitution

: \int v^{3} d v

= 20 \cdot \frac{1}{5} \cdot \int v^{3} d v

Apply the Power Rule

: \frac{v ^{4}}{4}

= 20 \cdot \frac{1}{5} \cdot \frac{v ^{4}}{4}

Substitute back

= 20 \cdot \frac{1}{5} \cdot \frac{sec ^{4} ( 5 x )}{4}

Simplify

20 \cdot \frac{1}{5} \cdot \frac{sec ^{4} ( 5 x )}{4} : sec^{4} (5 x)

= sec^{4} (5 x)

Add a constant to the solution

= sec^{4} (5 x) + C

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What is the integral of 20sec(5x)tan(5x)sec^3(5x) ?
The integral of 20sec(5x)tan(5x)sec^3(5x) is sec^4(5x)+C