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Integrals — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIntegrals1 Mark
Question
Prove $\int_{0}^{\frac{\pi}{2}} \sin ^{3} x d x=\frac{2}{3}$
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Answer

Given integral is:$ \int_{0}^{\frac{\pi}{2}} \sin ^{3} x d x$
To prove:$\int_{0}^{\frac{\pi}{2}} \sin ^{3} x d x=\frac{2}{3}$
Let$I=\int_{0}^{\frac{\pi}{2}} \sin ^{3} x d x ...(i)$
$= \int_{0}^{\frac{\pi}{2}} \sin x \cdot \sin ^{2} x d x$
$= \int_{0}^{\frac{\pi}{2}} \sin x \cdot\left(1-\cos ^{2} x\right) d x$
$ \Rightarrow \mathrm{I}=\int_{0}^{\frac{\pi}{2}} \sin \mathrm{x} \mathrm{dx}-\int_{0}^{\frac{\pi}{2}} \sin \mathrm{x} \cdot \cos ^{2} \mathrm{x} d \mathrm{x}$
$ \Rightarrow \mathrm{I}=[-\cos \mathrm{x}]_{0}^{\pi / 2}-\mathrm{I}_{1} ...(ii)$
Now, we solve$I_1:$
$ \Rightarrow \mathrm{I}_{1}=\int_{0}^{\frac{\pi}{2}} \sin \mathrm{x} \cdot \cos ^{2} \mathrm{x} \mathrm{d} \mathrm{x}$
Let \cos$x = t \Rightarrow -\sin x dx = dt$
$\Rightarrow sinx dx = -dt$
When$x = 0$ then$t = 1$ and when$x = \frac{\pi}{2}$ then$t = 0$
$ \Rightarrow \mathrm{I}_{1}=\int_{1}^{0} \mathrm{t}^{2}(-\mathrm{dt})$
$= -\int_{1}^{0} t^{2}(d t)$
$= -\left[\frac{t^{3}}{3}\right]_{1}^{0}$
$= -\left\{-\frac{1}{3}\right\}$
$ \Rightarrow \mathrm{I}_{1}=\frac{1}{3}$
Using this value in equation$(ii)$
$ \Rightarrow \mathrm{I}=[-\cos \mathrm{x}]_{0}^{\pi / 2}-\frac{1}{3}$
$ \Rightarrow \mathrm{I}=-\left\{\cos \frac{\pi}{2}-\cos 0\right\}-\frac{1}{3}$
$ \Rightarrow \mathrm{I}=1-\frac{1}{3}$
$ \Rightarrow \mathrm{I}=\frac{2}{3}$
Hence Proved.

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