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

The integral of (1+sec(pix))^2sec(pix)tan(pix) is 1/(3pi)(1+sec(pix))^3+C

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integral of (1+sec(pix))^2sec(pix)tan(pix)

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

\frac{1}{3 π} (1 + sec (π x))^{3} + C

Solution steps

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

Apply u-substitution

: \int \frac{u ^{2}}{π} d u

= \int \frac{u ^{2}}{π} d u

Take the constant out:

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

= \frac{1}{π} \cdot \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= \frac{1}{π} \cdot \frac{u ^{3}}{3}

Substitute back

u = 1 + sec (π x)

= \frac{1}{π} \cdot \frac{( 1 + sec ( π x ) ) ^{3}}{3}

Simplify

\frac{1}{π} \cdot \frac{( 1 + sec ( π x ) ) ^{3}}{3} : \frac{1}{3 π} (1 + sec (π x))^{3}

= \frac{1}{3 π} (1 + sec (π x))^{3}

Add a constant to the solution

= \frac{1}{3 π} (1 + sec (π x))^{3} + C

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