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

The integral of x/(cos^2(4x^2+1/a)) is 1/8 tan(4x^2+1/a)+C

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integral of x/(cos^2(4x^2+1/a))

\int \frac{x}{cos ^{2} ( 4 x ^{2} + \frac{1}{a} )} d x

\frac{1}{8} tan (4 x^{2} + \frac{1}{a}) + C

Solution steps

\int \frac{x}{cos ^{2} ( 4 x ^{2} + \frac{1}{a} )} d x

Apply u-substitution

: \int \frac{1}{8 cos ^{2} ( u )} d u

= \int \frac{1}{8 cos ^{2} ( u )} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{1}{cos ^{2} ( u )} d u

Use the common integral:

\int \frac{1}{cos ^{2} ( u )} d u = tan (u)

= \frac{1}{8} tan (u)

Substitute back

u = 4 x^{2} + \frac{1}{a}

= \frac{1}{8} tan (4 x^{2} + \frac{1}{a})

Add a constant to the solution

= \frac{1}{8} tan (4 x^{2} + \frac{1}{a}) + C

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What is the integral of x/(cos^2(4x^2+1/a)) ?
The integral of x/(cos^2(4x^2+1/a)) is 1/8 tan(4x^2+1/a)+C