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

The integral from 0 to 1 of (sin^2(x))/(x^2) is diverges

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

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

diverges

Solution steps

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

Rewrite using trig identities

= \int_{0}^{1} \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}}

= \int_{0}^{1} \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

= \int_{0}^{1} \frac{1}{x ^{2}} d x - \int_{0}^{1} \frac{cos ^{2} ( x )}{x ^{2}} d x

\int_{0}^{1} \frac{1}{x ^{2}} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\int_{- 2 π}^{2 π} x^{2} cos (2 x) d x

\int_{1}^{2} 4 d x

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

\int_{0}^{1} k x^{2} d x

\int_{- 2}^{0} - 1 d x

What is the integral from 0 to 1 of (sin^2(x))/(x^2) ?
The integral from 0 to 1 of (sin^2(x))/(x^2) is diverges