Mathematics

# Solve $\displaystyle\int\limits_0^{\pi /2} {{{\sin }^4}x{{\cos }^3}xdx}$

$\dfrac {2}{35}$

##### SOLUTION
$\int_{0}^{\pi/2}\sin^{4} x\cos^{3} x dx$
$\int_{0}^{\pi/2}\sin^{4} x(1-\sin^{2} x)\cos x dx$       $\because \cos^{2} x=1-\sin^{2} x$
$\int_{0}^{\pi/2}(\sin^{4} x-\sin^{6} x)\cos x dx$
substitute $\sin x=t$  hence, $dt=\cos x\ dx$
and at $x=0; \ t=\sin (0)=0 ;$ at $x=\pi/2; \ t=\sin(\pi/2)=1$
$\int_{0}^{1}(t^{4} -t^{6} ) dt$
$\bigg[\dfrac{t^{5}}{5}-\dfrac{t^{7}}{7}\bigg]_{0}^{1}$
$\bigg[\dfrac{1}{5}-\dfrac{1}{7} \bigg]=\dfrac{2}{35}$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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