What is the integral of 3xy^2sec^2(2xy) ?

The integral of 3xy^2sec^2(2xy) is 3/4 (2yxtan(2yx)+ln|cos(2yx)|)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 3xy^2sec^2(2xy)

\int 3 x y^{2} sec^{2} (2 x y) d x

\frac{3}{4} (2 y x tan (2 y x) + ln ∣ cos (2 y x) ∣) + C

Solution steps

\int 3 x y^{2} sec^{2} (2 x y) d x

Take the constant out:

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

= 3 y^{2} \cdot \int sec^{2} (2 y x) x d x

Apply u-substitution

: \int \frac{u sec ^{2} ( u )}{4 y ^{2}} d u

= 3 y^{2} \cdot \int \frac{u sec ^{2} ( u )}{4 y ^{2}} d u

Take the constant out:

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

= 3 y^{2} \frac{1}{4 y ^{2}} \cdot \int u sec^{2} (u) d u

Apply Integration By Parts

: u tan (u) - \int tan (u) d u

= 3 y^{2} \frac{1}{4 y ^{2}} (u tan (u) - \int tan (u) d u)

\int tan (u) d u = - ln ∣ cos (u) ∣

= 3 y^{2} \frac{1}{4 y ^{2}} (u tan (u) - (- ln ∣ cos (u) ∣))

Substitute back

u = 2 y x

= 3 y^{2} \frac{1}{4 y ^{2}} (2 y x tan (2 y x) - (- ln ∣ cos (2 y x) ∣))

Simplify

3 y^{2} \frac{1}{4 y ^{2}} (2 y x tan (2 y x) - (- ln ∣ cos (2 y x) ∣)) : \frac{3}{4} (2 y x tan (2 y x) + ln ∣ cos (2 y x) ∣)

= \frac{3}{4} (2 y x tan (2 y x) + ln ∣ cos (2 y x) ∣)

Add a constant to the solution

= \frac{3}{4} (2 y x tan (2 y x) + ln ∣ cos (2 y x) ∣) + C

\int 5 x^{\frac{4}{3}} + 2 x^{\frac{5}{4}} d x

\frac{d}{d x} (4 + 3 x - 2 x^{3})

\frac{d}{d x} (\frac{12 x}{( x ^{2} + 2 ) ^{2}})

f (x) = ln (x^{2} + 4)

y^{^{'}} - 2 y = 0

What is the integral of 3xy^2sec^2(2xy) ?
The integral of 3xy^2sec^2(2xy) is 3/4 (2yxtan(2yx)+ln|cos(2yx)|)+C