Question

if (h(x)=int_{-\frac{pi}{4}}^{sin^{2}x}csc(t^{4}+1)dt) for (-\frac{pi}{4}leq xleq\frac{pi}{4}), then (h(x)=)
(a) (csc(x^{4}+1))
(b) (csc(sin^{8}x + 1))
(c) ((sin^{2}x)(csc(x^{4}+1)))
(d) ((2sin xcos x)(csc(sin^{8}x + 1)))

Explanation:

Step1: Apply the fundamental theorem of calculus and chain - rule

If $h(x)=\int_{a}^{u(x)}f(t)dt$, then $h'(x)=f(u(x))\cdot u'(x)$ by the fundamental theorem of calculus and chain - rule. Here, $a =-\frac{\pi}{4}$, $u(x)=\sin^{2}x$, and $f(t)=\csc(t^{4}+1)$.

Step2: Find the derivative of $u(x)$

The derivative of $u(x)=\sin^{2}x$ using the chain - rule. Let $y = u^{2}$ and $u=\sin x$. Then $\frac{dy}{du}=2u$ and $\frac{du}{dx}=\cos x$. So, $u'(x)=2\sin x\cos x$.

Step3: Substitute $u(x)$ and $u'(x)$ into the formula

Substitute $u(x)=\sin^{2}x$ into $f(t)$ to get $f(u(x))=\csc((\sin^{2}x)^{4}+1)=\csc(\sin^{8}x + 1)$, and $u'(x)=2\sin x\cos x$. Then $h'(x)=(2\sin x\cos x)\csc(\sin^{8}x + 1)$.

Answer:

D. $(2\sin x\cos x)(\csc(\sin^{8}x + 1))$