What is the integral of pi((sin(x))/x)^2 ?

The integral of pi((sin(x))/x)^2 is pi(-1/x+(cos^2(x))/x+\Si(2x))+C

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integral of pi((sin(x))/x)^2

\int π (\frac{sin ( x )}{x})^{2} d x

π (- \frac{1}{x} + \frac{cos ^{2} ( x )}{x} + Si (2 x)) + C

Solution steps

\int π (\frac{sin ( x )}{x})^{2} d x

Take the constant out:

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

= π \cdot \int (\frac{sin ( x )}{x})^{2} d x

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

= π \cdot \int \frac{sin ^{2} ( x )}{x ^{2}} d x

Rewrite using trig identities

= π \cdot \int \frac{1 - cos ^{2} ( x )}{x ^{2}} d x

Expand

\frac{1 - cos ^{2} ( x )}{x ^{2}} : \frac{1}{x ^{2}} - \frac{cos ^{2} ( x )}{x ^{2}}

= π \cdot \int \frac{1}{x ^{2}} - \frac{cos ^{2} ( x )}{x ^{2}} 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

= π (\int \frac{1}{x ^{2}} d x - \int \frac{cos ^{2} ( x )}{x ^{2}} d x)

\int \frac{1}{x ^{2}} d x = - \frac{1}{x}

\int \frac{cos ^{2} ( x )}{x ^{2}} d x = - \frac{cos ^{2} ( x )}{x} - Si (2 x)

= π (- \frac{1}{x} - (- \frac{cos ^{2} ( x )}{x} - Si (2 x)))

Simplify

= π (- \frac{1}{x} + \frac{cos ^{2} ( x )}{x} + Si (2 x))

Add a constant to the solution

= π (- \frac{1}{x} + \frac{cos ^{2} ( x )}{x} + Si (2 x)) + C

x \to \frac{π}{2} lim (3 sin (x))

\int_{- 1}^{1} \frac{5 x}{( x + 4 ) ^{2}} d x

y = ∣ x - 6 ∣, y = \frac{x}{2}

\int \frac{x + 9}{36 - ( x - 6 ) ^{2}} d x

n = 1 \sum \infty \frac{2 n}{5 n - 3}

What is the integral of pi((sin(x))/x)^2 ?
The integral of pi((sin(x))/x)^2 is pi(-1/x+(cos^2(x))/x+\Si(2x))+C