Question

1/sin x√cos³x ,Integrate the given function defined on proper domain w.r.t. x.

Answer 1

Answer:

$\bf\int{sinx\,\sqrt{cos^3x}}\,dx=-\frac{2}{5}\,cos^{\frac{5}{2}}x+c$

Step-by-step explanation:

I have applied substituition method to solve the given integral.

$\bf\int{sinx\,\sqrt{cos^3x}}\,dx$

$\text{Let I}=\int{sinx\,\sqrt{cos^3x}}\,dx$

$\text{Take }t=cosx$

$\frac{dt}{dx}=-sinx$

$-dt=sinx\,dx$

$\text{Now}$

$I=\int{\sqrt{t^3}}\,(-dt)$

$I=-\int{t^{\frac{3}{2}}}\,dt$

$I=-\frac{t^{\frac{5}{2}}}{\frac{5}{2}}+c$

$I=-\frac{2}{5}t^{\frac{5}{2}}+c$

$I=-\frac{2}{5}\,cos^{\frac{5}{2}}x+c$

Answer 2

Given that,

The function is

$\sin x\sqrt{\cos^{3}x}$

We need to find the integration

Using substitute method  

$I=\int{\sin x\sqrt{\cos^{3}x}}$...(I)

Let $t=\cos x$

On differentiating w.r.to x

$\dfrac{dt}{dx}=-\sin x$

$dt=-\sin x dx$

$-dt=\sin x dx$

Put the value in the equation (I)

$I=\int{(t)^3(-dt)}$

$I=-\int{t^{\frac{3}{2}dt}$

$I=-\dfrac{t^{\frac{5}{2}}}{\dfrac{5}{2}}+C$

$I=-\dfrac{2}{5}t^{\frac{5}{2}}+C$

Put the value of t

$I=-\dfrac{2}{5}\cos^{\frac{5}{2}}x+C$

Hence,  The integration is $-\dfrac{2}{5}\cos^{\frac{5}{2}}x+C$