Find out the answer to "Trigonometric integrals. Evaluate the following integrals. ∫0π(1−cos⁡2x)32dx"

High School
Calculus and Analysis
Calculus 2
Applications of integrals

Paganellash

Answered question

2021-12-03

Trigonometric integrals. Evaluate the following integrals.

\int_{0}^{π} {(1 - \cos 2 x)}^{\frac{3}{2}} d x

Answer & Explanation

Elizabeth Witte

Beginner2021-12-04Added 24 answers

Step 1
To evaluate the given integrals , $\int_{0}^{π} {(1 - \cos 2 x)}^{\frac{3}{2}} d x$ .
Solution:
Let us denote the given integral by I as,
$I = \int_{0}^{π} {(1 - \cos 2 x)}^{\frac{3}{2}} d x$
Since we know the identity as, $1 - \cos 2 x = \sin^{2} x$ ,
So the integral becomes,
$I = \int_{0}^{π} {(\sin^{2} x)}^{\frac{3}{2}} d x$
So further solving the integral gives,
Step 2
$I = \int_{0}^{π} (\sin^{3} x) d x$
$= \int_{0}^{π} (\sin^{2} x) \sin x d x$
$= \int_{0}^{π} (1 - \cos^{2} x) (\sin x) d x$
Now using Substitution method, Put $c o s x = t$ , we get $- \sin x d x = d t$
Therefore, the given integral changes with upper limit, $\cos π = - 1$ and lower limit, $\cos 0 = 1$ as,
$I = - \int_{1}^{- 1} (1 - t^{2}) d t$
$I = - {(t - \frac{t^{3}}{3})}_{1}^{- 1}$
Step 3
Further, solving the integral gives,
$I = {(\frac{t^{3}}{3} - t)}_{1}^{- 1}$
$= (\frac{- 1}{3} + 1) - (\frac{1}{3} - 1)$
$= \frac{2}{3} + \frac{2}{3}$
$= \frac{4}{3}$
Hence, the value of the integral is $\frac{4}{3}$ .

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