What is the integral from 0 to pi/4 of 6x^2+tan(x) ?

The integral from 0 to pi/4 of 6x^2+tan(x) is (pi^3)/(32)+1/2 ln(2)

Popular >

integral from 0 to pi/4 of 6x^2+tan(x)

\int_{0}^{\frac{π}{4}} 6 x^{2} + tan (x) d x

\frac{π ^{3}}{32} + \frac{1}{2} ln (2)

Decimal

1.31551 \dots

Solution steps

\int_{0}^{\frac{π}{4}} 6 x^{2} + tan (x) d x

Apply the Sum Rule:

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

= \int_{0}^{\frac{π}{4}} 6 x^{2} d x + \int_{0}^{\frac{π}{4}} tan (x) d x

\int_{0}^{\frac{π}{4}} 6 x^{2} d x = \frac{π ^{3}}{32}

\int_{0}^{\frac{π}{4}} tan (x) d x = \frac{1}{2} ln (2)

= \frac{π ^{3}}{32} + \frac{1}{2} ln (2)

\int_{- 1}^{1} x^{3} e^{x} d x

\int_{0}^{2} (x^{3} + x) d x

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

\int_{\frac{π}{2}}^{x} cos (t) d t

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

What is the integral from 0 to pi/4 of 6x^2+tan(x) ?
The integral from 0 to pi/4 of 6x^2+tan(x) is (pi^3)/(32)+1/2 ln(2)