What is the integral of (7(1-tan^2(x)))/(sec^2(x)) ?

The integral of (7(1-tan^2(x)))/(sec^2(x)) is 7/2 sin(2x)+C

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integral of (7(1-tan^2(x)))/(sec^2(x))

\int \frac{7 ( 1 - tan ^{2} ( x ) )}{sec ^{2} ( x )} d x

\frac{7}{2} sin (2 x) + C

Solution steps

\int \frac{7 ( 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

= 7 \cdot \int \frac{1 - tan ^{2} ( x )}{sec ^{2} ( x )} d x

Rewrite using trig identities

= 7 \cdot \int cos (2 x) d x

Apply u-substitution

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

= 7 \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

= 7 \cdot \frac{1}{2} \cdot \int cos (u) d u

Use the common integral:

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

= 7 \cdot \frac{1}{2} sin (u)

Substitute back

u = 2 x

= 7 \cdot \frac{1}{2} sin (2 x)

Simplify

7 \cdot \frac{1}{2} sin (2 x) : \frac{7}{2} sin (2 x)

= \frac{7}{2} sin (2 x)

Add a constant to the solution

= \frac{7}{2} sin (2 x) + C

y = 3 x^{2} + 4

\int (x + 2) d x

\frac{\partial}{\partial x} (5 x^{7 y})

\int 56 x e^{7 x^{2}} d x

y^{^{'}} + 2 y^{2} cos (x + 1) = 0

What is the integral of (7(1-tan^2(x)))/(sec^2(x)) ?
The integral of (7(1-tan^2(x)))/(sec^2(x)) is 7/2 sin(2x)+C