Prove the following question- -0-2sin 3 x dx -2-3

Let I = ∫_0^π/2sin^3 x dx=∫_0^π/2sin^2 x. sin x dx =∫_0^π/2(1 cos^2x)sin x dx (∵ sin^2x=1 cos^2x) Put, cos x=t ⇒ sin x dx=dt When x=0 ⇒ t=cos 0=1,w ...

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Standard XII

Mathematics

Property 4

Prove the fol...

Question

Prove the following question.

$\int_{0}^{\frac{π}{2}} s i n^{3} x d x = \frac{2}{3} .$

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Solution

$L e t I = \int_{0}^{\frac{π}{2}} s i n^{3} x d x = \int_{0}^{\frac{π}{2}} s i n^{2} x . s i n x d x = \int_{0}^{\frac{π}{2}} (1 - c o s^{2} x) s i n x d x (∵ s i n^{2} x = 1 - c o s^{2} x) P u t, c o s x = t \Rightarrow - s i n x d x = d t W h e n x = 0 \Rightarrow t = c o s 0 = 1, w h e n x = \frac{π}{2} = 0 ∴ I = \int_{0}^{\frac{π}{2}} (1 - c o s^{2} x) s i n x d x = \int_{1}^{0} (1 - t^{2}) (- d t) = - {[t - \frac{t^{3}}{3}]}_{1}^{0} = - {(0 - 0) - (1 - \frac{1}{3})} = \frac{2}{3} . H e n c e p r o v e d .$

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Prove the following question. ∫π20sin3 x dx=23.

Let I = ∫π20sin3 x dx=∫π20sin2 x. sin x dx =∫π20(1−cos2x)sin x dx (∵ sin2x=1−cos2x)Put,cos x=t ⇒−sin x dx=dtWhen x=0⇒ t=cos 0=1,when x=π2=0∴ I=∫π20(1−cos2x)sin x dx=∫01(1−t2)(−dt)=−[t−t33]01=−{(0−0)−(1−13)}=23. Hence proved.

Prove the following question.

$\int_{0}^{\frac{π}{2}} s i n^{3} x d x = \frac{2}{3} .$