Evaluate the integrals ∫−ππ(1−cos2t)32dt

Question

Evaluate the integrals   ∫−ππ(1−cos2t)32dt

Caren · Accepted Answer

We have to evaluate the definite integral:
$\int_{- π}^{π} {(1 - \cos^{2} t)}^{\frac{3}{2}} d t$
We know the trigonometric identity,
$\cos^{2} x + \sin^{2} x = 1$
$\Rightarrow \sin^{2} x = 1 - \cos^{2} x$
So applying above identity for the given integral, we get
$\int_{- π}^{π} {(1 - \cos^{2} t)}^{\frac{3}{2}} d t = \int_{- π}^{π} {(\sin^{2} t)}^{\frac{3}{2}} d t$
$= \int_{- π}^{π} {(\sin t)}^{2 \times \frac{3}{2}} d t$
$= \int_{- π}^{π} {(1 - \cos^{2} t)}^{\frac{3}{2}} d t = \int_{- π}^{π} {(\sin^{2} t)}^{\frac{3}{2}} d t$
$= \int_{- π}^{π} {(\sin t)}^{3} d t$
$= \int_{- π}^{π} \sin^{3} t d t$
We know the property of definite integral,
$\int_{- a}^{a} f (x) = 2 \int_{0}^{a} f (x) if f (x) i s e v e n ∖ 0 if f (x) i s o d d$
For odd function we need to satisfy,
$f (- x) = - f (x)$
and for even function,
$f (- x) = f (x)$
According to question,
$f (t) = \sin^{3} t$
checking for odd or even function,
$f (t) = \sin^{3} t$
$f (- t) = \sin^{3} (- t)$
$= \sin^{3} t$
$= - f (t)$
This function is odd.
Applying the properties od definite integral for odd function, we get
$\int_{- π}^{π} \sin^{3} t = 0$
Hence, value of the integral is 0.

Evaluate the integrals \int_{-\pi}^{\pi}(1-\cos^2t)^{3/2}dt

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