What is the integral of xsec^2(4x) ?

The integral of xsec^2(4x) is 1/2 x^2sec^2(4x)-1/16 (8x^2tan^2(4x)-4xtan(4x)-ln|cos(4x)|+8x^2)+C

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integral of xsec^2(4x)

\int x sec^{2} (4 x) d x

\frac{1}{2} x^{2} sec^{2} (4 x) - \frac{1}{16} (8 x^{2} tan^{2} (4 x) - 4 x tan (4 x) - ln ∣ cos (4 x) ∣ + 8 x^{2}) + C

Solution steps

\int x sec^{2} (4 x) d x

Rewrite using trig identities

= \int x (1 + tan^{2} (4 x)) d x

Apply Integration By Parts

: \frac{1}{2} x^{2} sec^{2} (4 x) - \int 4 x^{2} sec^{2} (4 x) tan (4 x) d x

= \frac{1}{2} x^{2} sec^{2} (4 x) - \int 4 x^{2} sec^{2} (4 x) tan (4 x) d x

\int 4 x^{2} sec^{2} (4 x) tan (4 x) d x = \frac{1}{16} (8 x^{2} tan^{2} (4 x) - 4 x tan (4 x) - ln ∣ cos (4 x) ∣ + 8 x^{2})

= \frac{1}{2} x^{2} sec^{2} (4 x) - \frac{1}{16} (8 x^{2} tan^{2} (4 x) - 4 x tan (4 x) - ln ∣ cos (4 x) ∣ + 8 x^{2})

Add a constant to the solution

= \frac{1}{2} x^{2} sec^{2} (4 x) - \frac{1}{16} (8 x^{2} tan^{2} (4 x) - 4 x tan (4 x) - ln ∣ cos (4 x) ∣ + 8 x^{2}) + C

f (x) = (x^{2} - 15)^{7}, at x = 4

x \frac{d y}{d x} + 2 y = 6 x^{2} y

\frac{d}{d x} (a cos^{3} (x))

y = x^{2} ln (3 x)

y^{^{'}} = \frac{( 1 - y ) ^{2}}{x + 1}, y (0) = 5

What is the integral of xsec^2(4x) ?
The integral of xsec^2(4x) is 1/2 x^2sec^2(4x)-1/16 (8x^2tan^2(4x)-4xtan(4x)-ln|cos(4x)|+8x^2)+C