Question

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

$\frac{d}{dx}\left({\left({\int }_{\sin x}^{\pi } \frac{t}{\cos t}dt\right)}^3\right)$

Short answer

Ans: Given integral is unable to find the derivative.

Step-by-step solution

Step 1. Given information.

given,

$\frac{d}{dx}\left({\left({\int }_{\sin x}^{\pi } \frac{t}{\cos t}dt\right)}^3\right)$

Step 2. The objective is to find the above derivative using the Second Fundamental Theorem of Calculus.

Note that, if $f$is continuous on $[a,b]$and $u\left(x\right)$is differentiable on $[a,b]$, then for all $x\in [a,b]$

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

Use the power rule and chain rule,

$\frac{d}{dx}\left({\left({\int }_{\sin x}^{\pi } \frac{t}{\cos t}dt\right)}^3\right)=3{\left({\int }_{\sin x}^{\pi } \frac{t}{\cos t}dt\right)}^2\cdot \frac{d}{dx}\left({\int }_{\sin x}^{\pi } \frac{t}{\cos t}dt\right).....\left(1\right)$

consider, $\frac{d}{dx}\left({\int }_{\sin x}^{\pi } \frac{t}{\cos t}dt\right)$. Rewrite this as,

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

Let $u\left(t\right)=\sin x$. then $u\text{'}\left(x\right)=\cos x$

And, let $f\left(t\right)=\frac{t}{\cos t},\text{so}f\left(u\right(x\left)\right)=\frac{\sin x}{\cos (\sin x)}$.

Use the Second Fundamental Theorem of calculus to obtained that,

$\begin{array}{r}\frac{d}{dx}{\int }_a^{u\left(x\right)} f\left(t\right)dt=f\left(u\right(x\left)\right)u^{\mathrm{\prime }}\left(x\right)\\ \frac{d}{dx}\left({\int }_{\sin x}^{\pi } \frac{t}{\cos t}dt\right)=-\frac{\sin x}{\cos (\sin x)}\cdot \cos x\\ =-\frac{\sin x\cos x}{\cos (\sin x)}\end{array}$

Step 3. Substitute this value in the equation (1) to obtain that,

$\begin{array}{r}\frac{d}{dx}\left({\left({\int }_{\sin x}^{\pi } \frac{t}{\cos t}dt\right)}^3\right)=3{\left({\int }_{\sin x}^{\pi } \frac{t}{\cos t}dt\right)}^2\cdot -\frac{\sin x\cos x}{\cos (\sin x)}\\ =-\frac{3\sin x\cos x}{\cos (\sin x)}{\left({\int }_{\sin x}^{\pi } \frac{t}{\cos t}dt\right)}^2\end{array}$

Here the derivate is unable to find in the given integral.