What is the integral of (1+tan(x))^3sec^2(x) ?

The integral of (1+tan(x))^3sec^2(x) is ((1+tan(x))^4)/4+C

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integral of (1+tan(x))^3sec^2(x)

\int (1 + tan (x))^{3} sec^{2} (x) d x

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

Solution steps

\int (1 + tan (x))^{3} sec^{2} (x) d x

Apply u-substitution

: \int u^{3} d u

= \int u^{3} d u

Apply the Power Rule

: \frac{u ^{4}}{4}

= \frac{u ^{4}}{4}

Substitute back

u = 1 + tan (x)

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

Add a constant to the solution

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

\int - \frac{1}{2} cos (2 t) d t

\int 3 sec (2 x) tan (2 x) d x

\int x cos (7 x) d x

\frac{d}{d x} (3 - 2 x)

x \to \infty lim (\frac{1}{4})

What is the integral of (1+tan(x))^3sec^2(x) ?
The integral of (1+tan(x))^3sec^2(x) is ((1+tan(x))^4)/4+C