What is the integral of (9(1-tan^2(x)))/(sec^2(x)) ?

The integral of (9(1-tan^2(x)))/(sec^2(x)) is 9/2 sin(2x)+C

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integral of (9(1-tan^2(x)))/(sec^2(x))

\int \frac{9 ( 1 - tan ^{2} ( x ) )}{sec ^{2} ( x )} d x

\frac{9}{2} sin (2 x) + C

Solution steps

\int \frac{9 ( 1 - tan ^{2} ( x ) )}{sec ^{2} ( x )} d x

Take the constant out:

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

= 9 \cdot \int \frac{1 - tan ^{2} ( x )}{sec ^{2} ( x )} d x

Rewrite using trig identities

= 9 \cdot \int cos (2 x) d x

Apply u-substitution

: \int cos (u) \frac{1}{2} d u

= 9 \cdot \int cos (u) \frac{1}{2} d u

Take the constant out:

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

= 9 \cdot \frac{1}{2} \cdot \int cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= 9 \cdot \frac{1}{2} sin (u)

Substitute back

u = 2 x

= 9 \cdot \frac{1}{2} sin (2 x)

Simplify

9 \cdot \frac{1}{2} sin (2 x) : \frac{9}{2} sin (2 x)

= \frac{9}{2} sin (2 x)

Add a constant to the solution

= \frac{9}{2} sin (2 x) + C

\frac{d}{d x} (csc (x) + cot (x))

\frac{ln ( x )}{5 x ^{3}}

(4 x^{3} y^{2} - 6 x^{2} y - 2 x - 3) d x + (2 x^{4} y - 2 x^{3}) d y = 0

e^{- t} sin (3 t)

(\frac{2 y}{x} + 2 x) + \frac{( 2 ln ( x ) - 3 ) d y}{d x} = 0

What is the integral of (9(1-tan^2(x)))/(sec^2(x)) ?
The integral of (9(1-tan^2(x)))/(sec^2(x)) is 9/2 sin(2x)+C