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

The integral of x/(cos^2(2x^2)) is 1/4 tan(2x^2)+C

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

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

\frac{1}{4} tan (2 x^{2}) + C

Solution steps

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

Use the basic trigonometric identity:

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply u-substitution

: \int sec^{2} (v) \frac{1}{2} d v

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

Take the constant out:

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

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

Use the common integral:

\int sec^{2} (v) d v = tan (v)

= \frac{1}{2} \cdot \frac{1}{2} tan (v)

Substitute back

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

Simplify

\frac{1}{2} \cdot \frac{1}{2} tan (2 x^{2}) : \frac{1}{4} tan (2 x^{2})

= \frac{1}{4} tan (2 x^{2})

Add a constant to the solution

= \frac{1}{4} tan (2 x^{2}) + C

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What is the integral of x/(cos^2(2x^2)) ?
The integral of x/(cos^2(2x^2)) is 1/4 tan(2x^2)+C