What is the integral of 3/(sec^2(5x)) ?

The integral of 3/(sec^2(5x)) is 3/2 (x+1/10 sin(10x))+C

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integral of 3/(sec^2(5x))

\int \frac{3}{sec ^{2} ( 5 x )} d x

\frac{3}{2} (x + \frac{1}{10} sin (10 x)) + C

Solution steps

\int \frac{3}{sec ^{2} ( 5 x )} d x

Take the constant out:

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

= 3 \cdot \int \frac{1}{sec ^{2} ( 5 x )} d x

Use the basic trigonometric identity:

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

= 3 \cdot \int cos^{2} (5 x) d x

Rewrite using trig identities

= 3 \cdot \int \frac{1}{2} (1 + cos (10 x)) d x

Take the constant out:

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

= 3 \cdot \frac{1}{2} \cdot \int 1 + cos (10 x) d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 \cdot \frac{1}{2} (\int 1 d x + \int cos (10 x) d x)

\int 1 d x = x

\int cos (10 x) d x = \frac{1}{10} sin (10 x)

= 3 \cdot \frac{1}{2} (x + \frac{1}{10} sin (10 x))

Simplify

3 \cdot \frac{1}{2} (x + \frac{1}{10} sin (10 x)) : \frac{3}{2} (x + \frac{1}{10} sin (10 x))

= \frac{3}{2} (x + \frac{1}{10} sin (10 x))

Add a constant to the solution

= \frac{3}{2} (x + \frac{1}{10} sin (10 x)) + C

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What is the integral of 3/(sec^2(5x)) ?
The integral of 3/(sec^2(5x)) is 3/2 (x+1/10 sin(10x))+C