What is the integral of 7tan^3(x)sec^3(x) ?

The integral of 7tan^3(x)sec^3(x) is 7((sec^5(x))/5-(sec^3(x))/3)+C

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integral of 7tan^3(x)sec^3(x)

\int 7 tan^{3} (x) sec^{3} (x) d x

7 (\frac{sec ^{5} ( x )}{5} - \frac{sec ^{3} ( x )}{3}) + C

Solution steps

\int 7 tan^{3} (x) sec^{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 tan^{3} (x) sec^{3} (x) d x

Rewrite using trig identities

= 7 \cdot \int (- 1 + sec^{2} (x)) tan (x) sec^{3} (x) d x

Apply u-substitution

: \int u^{2} (u^{2} - 1) d u

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

Expand

u^{2} (u^{2} - 1) : u^{4} - u^{2}

= 7 \cdot \int u^{4} - u^{2} 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 u^{4} d u - \int u^{2} d u)

\int u^{4} d u = \frac{u ^{5}}{5}

\int u^{2} d u = \frac{u ^{3}}{3}

= 7 (\frac{u ^{5}}{5} - \frac{u ^{3}}{3})

Substitute back

u = sec (x)

= 7 (\frac{sec ^{5} ( x )}{5} - \frac{sec ^{3} ( x )}{3})

Add a constant to the solution

= 7 (\frac{sec ^{5} ( x )}{5} - \frac{sec ^{3} ( x )}{3}) + C

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

f (x) = x^{2} + 4, (4, 20)

x \to 2 - lim (e^{\frac{1}{x - 2}})

y^{^{'}} = y (\frac{1}{x} - y)

y = x^{2} - 4 x - 5, y = 2 x - 5

What is the integral of 7tan^3(x)sec^3(x) ?
The integral of 7tan^3(x)sec^3(x) is 7((sec^5(x))/5-(sec^3(x))/3)+C