What is the integral of tan(5x)sec^2(5x) ?

The integral of tan(5x)sec^2(5x) is 1/10 tan^2(5x)+C

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integral of tan(5x)sec^2(5x)

\int tan (5 x) sec^{2} (5 x) d x

\frac{1}{10} tan^{2} (5 x) + C

Solution steps

\int tan (5 x) sec^{2} (5 x) d x

Apply u-substitution

: \int \frac{u}{5} d u

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

Take the constant out:

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

= \frac{1}{5} \cdot \int u d u

Apply the Power Rule

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

= \frac{1}{5} \cdot \frac{u ^{2}}{2}

Substitute back

u = tan (5 x)

= \frac{1}{5} \cdot \frac{tan ^{2} ( 5 x )}{2}

Simplify

\frac{1}{5} \cdot \frac{tan ^{2} ( 5 x )}{2} : \frac{1}{10} tan^{2} (5 x)

= \frac{1}{10} tan^{2} (5 x)

Add a constant to the solution

= \frac{1}{10} tan^{2} (5 x) + C

x - 2, x, [0, 4]

[x + (x + sin^{2} (x))^{7}]^{5}

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

\int \frac{3}{x ^{7}} d x

x \to \infty lim (e^{- 5})

What is the integral of tan(5x)sec^2(5x) ?
The integral of tan(5x)sec^2(5x) is 1/10 tan^2(5x)+C