What is the integral of-7cos^2(x)tan^3(x) ?

The integral of-7cos^2(x)tan^3(x) is -7(-ln|cos(x)|+(cos^2(x))/2)+C

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integral of-7cos^2(x)tan^3(x)

\int - 7 cos^{2} (x) tan^{3} (x) d x

- 7 (- ln ∣ cos (x) ∣ + \frac{cos ^{2} ( x )}{2}) + C

Solution steps

\int - 7 cos^{2} (x) tan^{3} (x) d x

Take the constant out:

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

= - 7 \cdot \int cos^{2} (x) tan^{3} (x) d x

Rewrite using trig identities

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

Apply u-substitution

: \int - \frac{1 - u ^{2}}{u} d u

= - 7 \cdot \int - \frac{1 - u ^{2}}{u} d u

Expand

- \frac{1 - u ^{2}}{u} : - \frac{1}{u} + u

= - 7 \cdot \int - \frac{1}{u} + u d u

Apply the Sum Rule:

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

= - 7 (- \int \frac{1}{u} d u + \int u d u)

\int \frac{1}{u} d u = ln ∣ u ∣

\int u d u = \frac{u ^{2}}{2}

= - 7 (- ln ∣ u ∣ + \frac{u ^{2}}{2})

Substitute back

u = cos (x)

= - 7 (- ln ∣ cos (x) ∣ + \frac{cos ^{2} ( x )}{2})

Add a constant to the solution

= - 7 (- ln ∣ cos (x) ∣ + \frac{cos ^{2} ( x )}{2}) + C

\frac{d ^{2}}{d x ^{2}} (x^{x})

\int x ln (2 - x) d x

\int \frac{1}{3 + sin ^{2} ( x )} d x

f (x) = 7 sin (x) + 3 cos (x)

\frac{1}{2 x}

What is the integral of-7cos^2(x)tan^3(x) ?
The integral of-7cos^2(x)tan^3(x) is -7(-ln|cos(x)|+(cos^2(x))/2)+C