What is the integral from 0 to pi/3 of tan^3(x)sec^2(x) ?

The integral from 0 to pi/3 of tan^3(x)sec^2(x) is 9/4

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integral from 0 to pi/3 of tan^3(x)sec^2(x)

\int_{0}^{\frac{π}{3}} tan^{3} (x) sec^{2} (x) d x

\frac{9}{4}

Decimal

2.25

Solution steps

\int_{0}^{\frac{π}{3}} tan^{3} (x) sec^{2} (x) d x

Rewrite using trig identities

= \int_{0}^{\frac{π}{3}} (1 + tan^{2} (x)) tan^{3} (x) d x

Apply u-substitution

: \int_{0}^{3} u^{3} d u

= \int_{0}^{3} u^{3} d u

Apply the Power Rule

: [\frac{u ^{4}}{4}]_{0}^{3}

= [\frac{u ^{4}}{4}]_{0}^{3}

Compute the boundaries:

\frac{9}{4}

= \frac{9}{4}

\int_{3}^{9} x d x

\int_{3}^{9} \frac{9}{x} d x

\int_{0}^{2} (\frac{3 x - 6}{2}) d x

\int_{5}^{7} x^{π - 1} d x

\int_{- 3}^{3} x + 2 ∣ x ∣ d x

What is the integral from 0 to pi/3 of tan^3(x)sec^2(x) ?
The integral from 0 to pi/3 of tan^3(x)sec^2(x) is 9/4