What is the integral of sin(8x)sec^5(8x) ?

The integral of sin(8x)sec^5(8x) is 1/32 sec^4(8x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of sin(8x)sec^5(8x)

\int sin (8 x) sec^{5} (8 x) d x

\frac{1}{32} sec^{4} (8 x) + C

Solution steps

\int sin (8 x) sec^{5} (8 x) d x

Apply u-substitution

: \int sin (u) sec^{5} (u) \frac{1}{8} d u

= \int sin (u) sec^{5} (u) \frac{1}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int sin (u) sec^{5} (u) d u

Rewrite using trig identities

= \frac{1}{8} \cdot \int tan (u) sec^{4} (u) d u

Apply u-substitution

: \int v^{3} d v

= \frac{1}{8} \cdot \int v^{3} d v

Apply the Power Rule

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

= \frac{1}{8} \cdot \frac{v ^{4}}{4}

Substitute back

= \frac{1}{8} \cdot \frac{sec ^{4} ( 8 x )}{4}

Simplify

\frac{1}{8} \cdot \frac{sec ^{4} ( 8 x )}{4} : \frac{1}{32} sec^{4} (8 x)

= \frac{1}{32} sec^{4} (8 x)

Add a constant to the solution

= \frac{1}{32} sec^{4} (8 x) + C

\int tan^{2} (arctan (d) x) d x

\int (1 + 2 x)^{\frac{1}{2}} d x

\int \frac{1}{4 + 2 x} d x

\int \frac{z + 5}{z ( z ^{2} + 49 )} d z

\int \frac{w}{1 - w ^{2}} d w

What is the integral of sin(8x)sec^5(8x) ?
The integral of sin(8x)sec^5(8x) is 1/32 sec^4(8x)+C