Integrate sin(x)^2 from 0a to 1b

\int_{0\cdota}^{1b}\sin\left(x\right)^2dx

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Answer

$\frac{1}{4}\sin\left(2x\right)-\frac{1}{2}x-\frac{1}{2}\cos\left(x\right)\sin\left(x\right)+\frac{1}{2}x$

Step by step solution

Problem

$\int_{0\cdota}^{1b}\sin\left(x\right)^2dx$
1

Any expression multiplied by $0$ is equal to $0$

$\int_{0}^{1b}\sin\left(x\right)^2dx$
2

Any expression multiplied by $1$ is equal to itself

$\int_{0}^{b}\sin\left(x\right)^2dx$
3

Applying a sine identity in order to reduce the exponent: $\displaystyle\sin(\theta)=\sqrt{\frac{1-\cos(2\theta)}{2}}$

$\int_{0}^{b}\frac{1-\cos\left(2x\right)}{2}dx$
4

Taking the constant out of the integral

$\frac{1}{2}\int_{0}^{b}\left(1-\cos\left(2x\right)\right)dx$
5

The integral of a sum of two or more functions is equal to the sum of their integrals

$\frac{1}{2}\left(\int_{0}^{b}-\cos\left(2x\right)dx+\int_{0}^{b}1dx\right)$
6

The integral of a constant is equal to the constant times the integral's variable

$\frac{1}{2}\left(\int_{0}^{b}-\cos\left(2x\right)dx+x\right)$
7

Taking the constant out of the integral

$\frac{1}{2}\left(x-\int_{0}^{b}\cos\left(2x\right)dx\right)$
8

Apply the formula: $\int\cos\left(x\cdot a\right)dx$$=\frac{1}{a}\sin\left(x\cdot a\right)$, where $a=2$

$\left[\frac{1}{2}\left(x-1\cdot \frac{1}{2}\sin\left(2x\right)\right)\right]_{0}^{b}$
9

Multiply $\frac{1}{2}$ times $-1$

$\left[\frac{1}{2}\left(x-\frac{1}{2}\sin\left(2x\right)\right)\right]_{0}^{b}$
10

Evaluate the definite integral

$\frac{1}{2}\left(x-\frac{1}{2}\sin\left(2x\right)\right)-1\cdot \frac{1}{2}\left(x-\frac{1}{2}\sin\left(2x\right)\right)$
11

Multiply $\frac{1}{2}$ times $-1$

$\frac{1}{2}\left(x-\frac{1}{2}\sin\left(2x\right)\right)-\frac{1}{2}\left(x-\frac{1}{2}\sin\left(2x\right)\right)$
12

Multiply $\left(x+-\frac{1}{2}\sin\left(2x\right)\right)$ by $\frac{1}{2}$

$\frac{1}{4}\sin\left(2x\right)-\frac{1}{2}x-\frac{1}{4}\sin\left(2x\right)+\frac{1}{2}x$
13

Using the sine double-angle identity

$\frac{1}{4}\sin\left(2x\right)-\frac{1}{2}x-\frac{1}{4}\cdot 2\cos\left(x\right)\sin\left(x\right)+\frac{1}{2}x$
14

Multiply $2$ times $-\frac{1}{4}$

$\frac{1}{4}\sin\left(2x\right)-\frac{1}{2}x-\frac{1}{2}\cos\left(x\right)\sin\left(x\right)+\frac{1}{2}x$

Answer

$\frac{1}{4}\sin\left(2x\right)-\frac{1}{2}x-\frac{1}{2}\cos\left(x\right)\sin\left(x\right)+\frac{1}{2}x$

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Problem Analysis

Main topic:

Integration by substitution

Time to solve it:

0.27 seconds

Views:

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