Q. 46 - Calculus

Question

Use the Second Fundamental Theorem of Calculus, if needed, to calculate each of the derivatives given below.  

       $\frac{d}{dx}\left(x^2{\int }_0^{\sin x} \sqrt{t}dt\right)$

Step-by-Step Solution

Answer

Ans:   $\frac{d}{dx}\left(x^2{\int }_0^{\sin x} \sqrt{t}dt\right) = 2x\sqrt{\sin x}\cos x+x^2\frac{{\cos }^{2}x}{2\sqrt{\sin x}}-x^2\sin x\sqrt{\sin x}$

Step 1. Given information.

given expression,

               $\frac{d}{dx}\left(x^2{\int }_0^{\sin x} \sqrt{t}dt\right)$

Step 2. The objective is to calculate the derivative.

Now, if $f$ is continuous on $[a,b]$ then for all  $x\in [a,b]$, 

     $\frac{d}{dx}{\int }_a^{u\left(x\right)} f\left(t\right)dt=f\left(u\right(x\left)\right)u^{\mathrm{\prime }}\left(x\right)$

So,

    $\begin{array}{r}f\left(u\right(t\left)\right)=\sqrt{\sin t}\\ f\left(u\right(x\left)\right)=\sqrt{\sin x}\\ u^{\mathrm{\prime }}\left(x\right)=\cos x\\ f\left(u\right(x\left)\right)u^{\mathrm{\prime }}\left(x\right)=\sqrt{\sin x}\cos x\end{array}$

Step 3. The derivate expression can be written as,

   $\begin{array}{r}\frac{d}{dx}\left(x^2{\int }_0^{\sin x} \sqrt{t}dt\right)\\ =\frac{d}{dx}\left(x^2\sqrt{\sin x}\cos x\right)\\ =2x\sqrt{\sin x}\cos x+x^2\left(\frac{d}{dx}\sqrt{\sin x}\cos x\right)\\ =2x\sqrt{\sin x}\cos x+x^2\left(\frac{\cos x}{2\sqrt{\sin x}}\cos x+\sqrt{\sin x}(-\sin x)\right)\\ =2x\sqrt{\sin x}\cos x+x^2\frac{{\cos }^{2}x}{2\sqrt{\sin x}}-x^2\sin x\sqrt{\sin x}\end{array}$

Therefore, the answer is $2x\sqrt{\sin x}\cos x+x^2\frac{{\cos }^{2}x}{2\sqrt{\sin x}}-x^2\sin x\sqrt{\sin x}$