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 I1: $\Rightarrow \mathrm{I}_{1}=\int_{0}^{\frac{\pi}{2}} \sin \mathrm{x} \cdot \cos ^{2} \mathrm{x} \mathrm{d} \mathrm{x}$ 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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