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$$\int_{\sin x}^{\cos x} {{t+\sin t^5}\over1+t^2}dt$$ I would use formula : $${d\over dx}\int_{a}^{x}f(t)dt = f(x)$$ but unfortunately I don't know how to find $sinx$ and $cosx$ in order to get a constant 'a'.

Paul Frost
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retne
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1 Answers1

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HINT

Imagine the anti-derivative is some $F(t)$, then $$ g(x) = F(\cos x) - F(\sin x), $$ and you need to find $g'(x)$ given $F'(t)$ is the integrand in question. Just apply the chain rule and simplify.

UPDATE

Here is a more detailed explanation $$ \begin{split} g'(x) &= F'(\cos x) \frac{d[\cos x]}{dx}-F'(\sin x) \frac{d[\sin x]}{dx} \qquad &\text{by Chain Rule}\\ &= f(\cos x) (-\sin x)-f(\sin x)\cos x \qquad &\text{since } F'(t) = f(t) \end{split} $$ Can you now plug in what $f(\cdot)$ is and simplify?

gt6989b
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  • So basically I have to calculate integral of the given function. When I do it, I shall use cosx and sinx in order to calculate g(x). In the end I get g'(x). Am I right? – retne Apr 04 '19 at 19:46
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    You don't have to calculate the integral of the given function. $g'(x)=f(\cos{(x)})(-\sin{(x)})-f(\sin{(x)})(\cos{(x)})$ where $f(x)$ is the given function. – Peter Foreman Apr 04 '19 at 19:49
  • @Michael please see the update – gt6989b Apr 04 '19 at 20:07
  • I substituted t with $cosx$ and $sinx$ respectively but I ended up with nothing. – retne Apr 04 '19 at 20:29