What is the integral from 0 to (3pi)/4 of sec^4(x/3) ?

The integral from 0 to (3pi)/4 of sec^4(x/3) is 4

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integral from 0 to (3pi)/4 of sec^4(x/3)

\int_{0}^{\frac{3 π}{4}} sec^{4} (\frac{x}{3}) d x

4

Solution steps

\int_{0}^{\frac{3 π}{4}} sec^{4} (\frac{x}{3}) d x

Apply u-substitution

: \int_{0}^{\frac{π}{4}} sec^{4} (u) \cdot 3 d u

= \int_{0}^{\frac{π}{4}} sec^{4} (u) \cdot 3 d u

Take the constant out:

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

= 3 \cdot \int_{0}^{\frac{π}{4}} sec^{4} (u) d u

Rewrite using trig identities

= 3 \cdot \int_{0}^{\frac{π}{4}} (1 + tan^{2} (u)) sec^{2} (u) d u

Apply u-substitution

: \int_{0}^{1} 1 + v^{2} d v

= 3 \cdot \int_{0}^{1} 1 + v^{2} d v

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 (\int_{0}^{1} 1 d v + \int_{0}^{1} v^{2} d v)

\int_{0}^{1} 1 d v = 1

\int_{0}^{1} v^{2} d v = \frac{1}{3}

= 3 (1 + \frac{1}{3})

Simplify

3 (1 + \frac{1}{3}) : 4

= 4

\int 2 r 1 - r^{2} d r

\frac{d}{d x} (\frac{x ^{3} + 3 x}{x ^{2} + 1})

x y^{^{'}} + 4 y = x^{3} - x

\int t \cdot e^{- t} d t

What is the integral from 0 to (3pi)/4 of sec^4(x/3) ?
The integral from 0 to (3pi)/4 of sec^4(x/3) is 4