What is the integral of sin(7x)sec^5(7x) ?

The integral of sin(7x)sec^5(7x) is 1/28 sec^4(7x)+C

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integral of sin(7x)sec^5(7x)

\int sin (7 x) sec^{5} (7 x) d x

\frac{1}{28} sec^{4} (7 x) + C

Solution steps

\int sin (7 x) sec^{5} (7 x) d x

Rewrite using trig identities

= \int \frac{tan ( 7 x )}{cos ^{4} ( 7 x )} d x

Use the basic trigonometric identity:

\frac{1}{cos ( x )} = sec (x)

= \int sec^{4} (7 x) tan (7 x) d x

Apply u-substitution

: \int sec^{4} (u) tan (u) \frac{1}{7} d u

= \int sec^{4} (u) tan (u) \frac{1}{7} d u

Take the constant out:

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

= \frac{1}{7} \cdot \int sec^{4} (u) tan (u) d u

Apply u-substitution

: \int v^{3} d v

= \frac{1}{7} \cdot \int v^{3} d v

Apply the Power Rule

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

= \frac{1}{7} \cdot \frac{v ^{4}}{4}

Substitute back

= \frac{1}{7} \cdot \frac{sec ^{4} ( 7 x )}{4}

Simplify

\frac{1}{7} \cdot \frac{sec ^{4} ( 7 x )}{4} : \frac{1}{28} sec^{4} (7 x)

= \frac{1}{28} sec^{4} (7 x)

Add a constant to the solution

= \frac{1}{28} sec^{4} (7 x) + C

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What is the integral of sin(7x)sec^5(7x) ?
The integral of sin(7x)sec^5(7x) is 1/28 sec^4(7x)+C