What is the integral from 0 to pi/4 of 1+sin^2(x) ?

The integral from 0 to pi/4 of 1+sin^2(x) is (3pi-2)/8

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integral from 0 to pi/4 of 1+sin^2(x)

\int_{0}^{\frac{π}{4}} 1 + sin^{2} (x) d x

\frac{3 π - 2}{8}

Decimal

0.92809 \dots

Solution steps

\int_{0}^{\frac{π}{4}} 1 + sin^{2} (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}} 1 d x + \int_{0}^{\frac{π}{4}} sin^{2} (x) d x

\int_{0}^{\frac{π}{4}} 1 d x = \frac{π}{4}

\int_{0}^{\frac{π}{4}} sin^{2} (x) d x = \frac{π - 2}{8}

= \frac{π}{4} + \frac{π - 2}{8}

Simplify

\frac{π}{4} + \frac{π - 2}{8} : \frac{3 π - 2}{8}

= \frac{3 π - 2}{8}

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What is the integral from 0 to pi/4 of 1+sin^2(x) ?
The integral from 0 to pi/4 of 1+sin^2(x) is (3pi-2)/8