pi-2-pi-4-sin-3-x-cos-2-x-dx-

Question Number 154877 by SANOGO last updated on 22/Sep/21

\int_{\frac{π}{2}}^{\frac{π}{4}} {s i n}^{3} (x) {c o s}^{2} (x) d x

Commented by saly last updated on 24/Sep/21

t h a n k y o u

Answered by Mr.D.N. last updated on 22/Sep/21

\int_{\frac{π}{2}}^{\frac{π}{4}} \sin^{3} (x) \cos^{2} (x) dx

\int_{\frac{π}{2}}^{\frac{π}{4}} \sin^{2} x . \cos^{2} x . \sin x dx

\int_{\frac{π}{2}}^{\frac{π}{4}} (1 - \cos^{2} x) \cos^{2} x sinx dx

put \cos x = t

- \sin x dx = dt

\sin x dx = - dt

change limit,

if x = \frac{π}{4} then t = \frac{1}{\sqrt{2}}

if x = \frac{π}{2} then t = 0

- \int_{0}^{\frac{1}{\sqrt{2}}} (1 - t^{2}) t^{2} dt

\int_{0}^{\frac{1}{\sqrt{2}}} (t^{4} - t^{2}) dt

= {[\frac{t^{5}}{5} - \frac{t^{3}}{3}]}_{0}^{\frac{1}{\sqrt{2}}}

= {[t^{3} (\frac{1}{5} t^{2} - \frac{1}{3})]}_{0}^{1 / \sqrt{2}}

= [{(\frac{1}{\sqrt{2}})}^{3} {\frac{1}{5} {(\sqrt{2})}^{2} - \frac{1}{3}}] - 0

= \frac{1}{\sqrt{2}} \times \frac{1}{2} (\frac{2}{5} - \frac{1}{3})

= \frac{1}{2 \sqrt{2}} (\frac{6 - 5}{15}) = \frac{1}{30 \sqrt{2}} / / .

Commented by SANOGO last updated on 23/Sep/21

m e r c i b i e n

Answered by Ar Brandon last updated on 23/Sep/21

I = \int_{\frac{π}{2}}^{\frac{π}{4}} \sin^{3} x \cos^{2} x d x = \int_{\frac{π}{2}}^{\frac{π}{4}} \sin x (\cos^{2} x - \cos^{4} x) d x

= \int_{\frac{π}{2}}^{\frac{π}{4}} (\cos^{4} x - \cos^{2} x) d (\cos x) = {[\frac{\cos^{5} x}{5} - \frac{\cos^{3} x}{3}]}_{\frac{π}{2}}^{\frac{π}{4}}

= \frac{1}{5 \sqrt{2^{5}}} - \frac{1}{3 \sqrt{2^{3}}} = \frac{1}{20 \sqrt{2}} - \frac{1}{6 \sqrt{2}} = - \frac{7 \sqrt{2}}{120}