Gujarat BoardEnglish MediumSTD 12 ScienceMathsIntegrals1 Mark
Question
Prove $\int_{0}^{\frac{\pi}{2}} \sin ^{3} x d x=\frac{2}{3}$
✓
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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