What is the integral of x^2sec^2(x)tan(x) ?

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

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integral of x^2sec^2(x)tan(x)

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

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

Solution steps

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

Apply Integration By Parts

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

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

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

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

Simplify

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

Add a constant to the solution

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

y = x^{2} - 4, y = 2 - 5 x^{2}

x \to 0 lim (\frac{6 sin ( 2 x )}{x})

(2, - 14), (1, - 11)

\int \frac{5}{x ^{3} + 2 x ^{2} - 35 x} d x

u^{9}

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