91影视

given,

$\frac{d}{dx}\left({\left({鈭�}_{\sin 鈦�x}^{蟺}鈥�\frac{t}{\cos 鈦�t}dt\right)}^3\right)$

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

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

Use the power rule and chain rule,

$\frac{d}{dx}\left({\left({鈭�}_{\sin 鈦�x}^{蟺}鈥�\frac{t}{\cos 鈦�t}dt\right)}^3\right)=3{\left({鈭�}_{\sin 鈦�x}^{蟺}鈥�\frac{t}{\cos 鈦�t}dt\right)}^2鈰�\frac{d}{dx}\left({鈭�}_{\sin 鈦�x}^{蟺}鈥�\frac{t}{\cos 鈦�t}dt\right).....\left(1\right)$

consider, $\frac{d}{dx}\left({鈭�}_{\sin 鈦�x}^{蟺}鈥�\frac{t}{\cos 鈦�t}dt\right)$. Rewrite this as,

$\begin{array}{r}\frac{d}{dx}\left({鈭�}_{\sin 鈦�x}^{蟺}鈥�\frac{t}{\cos 鈦�t}dt\right)=\frac{d}{dx}\left(鈭�{鈭�}_{蟺}^{\sin 鈦�x}鈥�\frac{t}{\cos 鈦�t}dt\right)\\ =鈭�\frac{d}{dx}\left({鈭�}_{蟺}^{\sin 鈦�x}鈥�\frac{t}{\cos 鈦�t}dt\right)\end{array}$

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

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

Use the Second Fundamental Theorem of calculus to obtained that,

$\begin{array}{r}\frac{d}{dx}{鈭�}_a^{u\left(x\right)}鈥�f\left(t\right)dt=f\left(u\right(x\left)\right)u^{\mathrm{鈥�}}\left(x\right)\\ \frac{d}{dx}\left({鈭�}_{\sin 鈦�x}^{蟺}鈥�\frac{t}{\cos 鈦�t}dt\right)=鈭�\frac{\sin 鈦�x}{\cos 鈦�(\sin 鈦�x)}鈰�\cos 鈦�x\\ =鈭�\frac{\sin 鈦�x\cos 鈦�x}{\cos 鈦�(\sin 鈦�x)}\end{array}$

$\begin{array}{r}\frac{d}{dx}\left({\left({鈭�}_{\sin 鈦�x}^{蟺}鈥�\frac{t}{\cos 鈦�t}dt\right)}^3\right)=3{\left({鈭�}_{\sin 鈦�x}^{蟺}鈥�\frac{t}{\cos 鈦�t}dt\right)}^2鈰�鈭�\frac{\sin 鈦�x\cos 鈦�x}{\cos 鈦�(\sin 鈦�x)}\\ =鈭�\frac{3\sin 鈦�x\cos 鈦�x}{\cos 鈦�(\sin 鈦�x)}{\left({鈭�}_{\sin 鈦�x}^{蟺}鈥�\frac{t}{\cos 鈦�t}dt\right)}^2\end{array}$

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