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

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

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

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

- \frac{7}{2} tan^{2} (x) + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

: \int u d u

= - 7 \cdot \int u d u

Apply the Power Rule

: \frac{u ^{2}}{2}

= - 7 \cdot \frac{u ^{2}}{2}

Substitute back

u = tan (x)

= - 7 \cdot \frac{tan ^{2} ( x )}{2}

Simplify

- 7 \cdot \frac{tan ^{2} ( x )}{2} : - \frac{7}{2} tan^{2} (x)

= - \frac{7}{2} tan^{2} (x)

Add a constant to the solution

= - \frac{7}{2} tan^{2} (x) + C

\frac{d}{d x} (\frac{5}{x + 1} - x^{4})

n = 1 \sum \infty n (x - 3)^{n}

\int x^{3} 9 x^{4} + 1 d x

\int 6 x sec^{2} (x) d x

x y + 2 x \frac{d y}{d x} = y

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