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

The integral of x/(cos^2(x)) is xtan(x)+ln|cos(x)|+C

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

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

x tan (x) + ln ∣ cos (x) ∣ + C

Solution steps

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

Rewrite using trig identities

= \int \frac{2 x}{1 + cos ( 2 x )} d x

Take the constant out:

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

= 2 \cdot \int \frac{x}{1 + cos ( 2 x )} d x

Apply Integration By Parts

: \frac{1}{2} x tan (x) - \int \frac{1}{2} tan (x) d x

= 2 (\frac{1}{2} x tan (x) - \int \frac{1}{2} tan (x) d x)

\int \frac{1}{2} tan (x) d x = - \frac{1}{2} ln ∣ cos (x) ∣

= 2 (\frac{1}{2} x tan (x) - (- \frac{1}{2} ln ∣ cos (x) ∣))

Simplify

2 (\frac{1}{2} x tan (x) - (- \frac{1}{2} ln ∣ cos (x) ∣)) : x tan (x) + ln ∣ cos (x) ∣

= x tan (x) + ln ∣ cos (x) ∣

Add a constant to the solution

= x tan (x) + ln ∣ cos (x) ∣ + C

n = 0 \sum \infty \frac{2 n - 1}{n !}

\frac{d}{d x} (- \frac{x}{4})

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

y = ∣ x ∣, y = x^{2} - 6

\frac{d}{d x} 9 x^{2} y^{2} = 1

What is the integral of x/(cos^2(x)) ?
The integral of x/(cos^2(x)) is xtan(x)+ln|cos(x)|+C