What is the integral of (sin(5x))/(cos^2(5x)) ?

The integral of (sin(5x))/(cos^2(5x)) is 1/5 sec(5x)+C

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integral of (sin(5x))/(cos^2(5x))

\int \frac{sin ( 5 x )}{cos ^{2} ( 5 x )} d x

\frac{1}{5} sec (5 x) + C

Solution steps

\int \frac{sin ( 5 x )}{cos ^{2} ( 5 x )} d x

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

= \int \frac{tan ( 5 x )}{cos ( 5 x )} d x

Apply u-substitution

: \int \frac{tan ( u )}{5 cos ( u )} d u

= \int \frac{tan ( u )}{5 cos ( u )} 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 \frac{tan ( u )}{cos ( u )} d u

Rewrite using trig identities

= \frac{1}{5} \cdot \int tan (u) sec (u) d u

Apply u-substitution

: \int 1 d v

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

Integral of a constant:

\int a d x = a x

= \frac{1}{5} \cdot 1 \cdot v

Substitute back

= \frac{1}{5} \cdot 1 \cdot sec (5 x)

Simplify

= \frac{1}{5} sec (5 x)

Add a constant to the solution

= \frac{1}{5} sec (5 x) + C

x \to + 10 lim (\frac{8}{( x - 10 ) ^{2}})

f (x) = 4 x^{3} - \frac{2}{x ^{3}}

\frac{d y}{( y ^{2} + 0.6 y )} = \frac{1}{x} d x

\frac{d}{d x} (e^{- x} (x - 1))

f (x) = 3^{3} - 3 (3)^{2} + 4 (3) + 5 + x

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