What is the integral of 2xsec(x^2) ?

The integral of 2xsec(x^2) is ln|tan(x^2)+sec(x^2)|+C

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integral of 2xsec(x^2)

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

ln tan (x^{2}) + sec (x^{2}) + C

Solution steps

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

Take the constant out:

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

= 2 \cdot \int x sec (x^{2}) d x

Apply u-substitution

: \int \frac{sec ( u )}{2} d u

= 2 \cdot \int \frac{sec ( u )}{2} d u

Take the constant out:

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

= 2 \cdot \frac{1}{2} \cdot \int sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= 2 \cdot \frac{1}{2} ln ∣ tan (u) + sec (u) ∣

Substitute back

u = x^{2}

= 2 \cdot \frac{1}{2} ln tan (x^{2}) + sec (x^{2})

Simplify

2 \cdot \frac{1}{2} ln tan (x^{2}) + sec (x^{2}) : ln tan (x^{2}) + sec (x^{2})

= ln tan (x^{2}) + sec (x^{2})

Add a constant to the solution

= ln tan (x^{2}) + sec (x^{2}) + C

\int (9 x^{2} + 12) (2 x^{3} + 8 x)^{11} d x

\int (x - 2 ∣ x ∣) d x

\int \frac{8}{( y - 9 ) ^{3}} d y

\int 2 x \cdot ln (x) d x

What is the integral of 2xsec(x^2) ?
The integral of 2xsec(x^2) is ln|tan(x^2)+sec(x^2)|+C